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Convolution solutions (Sect. 6.6).
I Convolution of two functions.
I Properties of convolutions.
I Laplace Transform of a convolution.
I Impulse response solution.
I Solution decomposition theorem.
Convolution solutions (Sect. 6.6).
I Convolution of two functions.
I Properties of convolutions.
I Laplace Transform of a convolution.
I Impulse response solution.
I Solution decomposition theorem.
Convolution of two functions.
DefinitionThe convolution of piecewise continuous functions f , g : R → R isthe function f ∗ g : R → R given by
(f ∗ g)(t) =
∫ t
0
f (τ)g(t − τ) dτ.
Remarks:
I f ∗ g is also called the generalized product of f and g .
I The definition of convolution of two functions also holds inthe case that one of the functions is a generalized function,like Dirac’s delta.
Convolution of two functions.
DefinitionThe convolution of piecewise continuous functions f , g : R → R isthe function f ∗ g : R → R given by
(f ∗ g)(t) =
∫ t
0
f (τ)g(t − τ) dτ.
Remarks:
I f ∗ g is also called the generalized product of f and g .
I The definition of convolution of two functions also holds inthe case that one of the functions is a generalized function,like Dirac’s delta.
Convolution of two functions.
DefinitionThe convolution of piecewise continuous functions f , g : R → R isthe function f ∗ g : R → R given by
(f ∗ g)(t) =
∫ t
0
f (τ)g(t − τ) dτ.
Remarks:
I f ∗ g is also called the generalized product of f and g .
I The definition of convolution of two functions also holds inthe case that one of the functions is a generalized function,like Dirac’s delta.
Convolution of two functions.
Example
Find the convolution of f (t) = e−t and g(t) = sin(t).
Solution: By definition: (f ∗ g)(t) =
∫ t
0
e−τ sin(t − τ) dτ .
Integrate by parts twice:
∫ t
0
e−τ sin(t − τ) dτ =[e−τ cos(t − τ)
]∣∣∣t0
−[e−τ sin(t − τ)
]∣∣∣t0
−∫ t
0
e−τ sin(t − τ) dτ,
2
∫ t
0
e−τ sin(t − τ) dτ =[e−τ cos(t − τ)
]∣∣∣t0
−[e−τ sin(t − τ)
]∣∣∣t0
,
2(f ∗ g)(t) = e−t − cos(t)− 0 + sin(t).
We conclude: (f ∗ g)(t) =1
2
[e−t + sin(t)− cos(t)
]. C
Convolution of two functions.
Example
Find the convolution of f (t) = e−t and g(t) = sin(t).
Solution: By definition: (f ∗ g)(t) =
∫ t
0
e−τ sin(t − τ) dτ .
Integrate by parts twice:
∫ t
0
e−τ sin(t − τ) dτ =[e−τ cos(t − τ)
]∣∣∣t0
−[e−τ sin(t − τ)
]∣∣∣t0
−∫ t
0
e−τ sin(t − τ) dτ,
2
∫ t
0
e−τ sin(t − τ) dτ =[e−τ cos(t − τ)
]∣∣∣t0
−[e−τ sin(t − τ)
]∣∣∣t0
,
2(f ∗ g)(t) = e−t − cos(t)− 0 + sin(t).
We conclude: (f ∗ g)(t) =1
2
[e−t + sin(t)− cos(t)
]. C
Convolution of two functions.
Example
Find the convolution of f (t) = e−t and g(t) = sin(t).
Solution: By definition: (f ∗ g)(t) =
∫ t
0
e−τ sin(t − τ) dτ .
Integrate by parts twice:
∫ t
0
e−τ sin(t − τ) dτ =[e−τ cos(t − τ)
]∣∣∣t0
−[e−τ sin(t − τ)
]∣∣∣t0
−∫ t
0
e−τ sin(t − τ) dτ,
2
∫ t
0
e−τ sin(t − τ) dτ =[e−τ cos(t − τ)
]∣∣∣t0
−[e−τ sin(t − τ)
]∣∣∣t0
,
2(f ∗ g)(t) = e−t − cos(t)− 0 + sin(t).
We conclude: (f ∗ g)(t) =1
2
[e−t + sin(t)− cos(t)
]. C
Convolution of two functions.
Example
Find the convolution of f (t) = e−t and g(t) = sin(t).
Solution: By definition: (f ∗ g)(t) =
∫ t
0
e−τ sin(t − τ) dτ .
Integrate by parts twice:
∫ t
0
e−τ sin(t − τ) dτ =[e−τ cos(t − τ)
]∣∣∣t0
−[e−τ sin(t − τ)
]∣∣∣t0
−∫ t
0
e−τ sin(t − τ) dτ,
2
∫ t
0
e−τ sin(t − τ) dτ =[e−τ cos(t − τ)
]∣∣∣t0
−[e−τ sin(t − τ)
]∣∣∣t0
,
2(f ∗ g)(t) = e−t − cos(t)− 0 + sin(t).
We conclude: (f ∗ g)(t) =1
2
[e−t + sin(t)− cos(t)
]. C
Convolution of two functions.
Example
Find the convolution of f (t) = e−t and g(t) = sin(t).
Solution: By definition: (f ∗ g)(t) =
∫ t
0
e−τ sin(t − τ) dτ .
Integrate by parts twice:
∫ t
0
e−τ sin(t − τ) dτ =[e−τ cos(t − τ)
]∣∣∣t0
−[e−τ sin(t − τ)
]∣∣∣t0
−∫ t
0
e−τ sin(t − τ) dτ,
2
∫ t
0
e−τ sin(t − τ) dτ =[e−τ cos(t − τ)
]∣∣∣t0
−[e−τ sin(t − τ)
]∣∣∣t0
,
2(f ∗ g)(t) = e−t − cos(t)− 0 + sin(t).
We conclude: (f ∗ g)(t) =1
2
[e−t + sin(t)− cos(t)
]. C
Convolution of two functions.
Example
Find the convolution of f (t) = e−t and g(t) = sin(t).
Solution: By definition: (f ∗ g)(t) =
∫ t
0
e−τ sin(t − τ) dτ .
Integrate by parts twice:
∫ t
0
e−τ sin(t − τ) dτ =[e−τ cos(t − τ)
]∣∣∣t0
−[e−τ sin(t − τ)
]∣∣∣t0
−∫ t
0
e−τ sin(t − τ) dτ,
2
∫ t
0
e−τ sin(t − τ) dτ =[e−τ cos(t − τ)
]∣∣∣t0
−[e−τ sin(t − τ)
]∣∣∣t0
,
2(f ∗ g)(t) = e−t − cos(t)− 0 + sin(t).
We conclude: (f ∗ g)(t) =1
2
[e−t + sin(t)− cos(t)
]. C
Convolution of two functions.
Example
Find the convolution of f (t) = e−t and g(t) = sin(t).
Solution: By definition: (f ∗ g)(t) =
∫ t
0
e−τ sin(t − τ) dτ .
Integrate by parts twice:
∫ t
0
e−τ sin(t − τ) dτ =[e−τ cos(t − τ)
]∣∣∣t0
−[e−τ sin(t − τ)
]∣∣∣t0
−∫ t
0
e−τ sin(t − τ) dτ,
2
∫ t
0
e−τ sin(t − τ) dτ =[e−τ cos(t − τ)
]∣∣∣t0
−[e−τ sin(t − τ)
]∣∣∣t0
,
2(f ∗ g)(t) = e−t − cos(t)− 0 + sin(t).
We conclude: (f ∗ g)(t) =1
2
[e−t + sin(t)− cos(t)
]. C
Convolution solutions (Sect. 6.6).
I Convolution of two functions.
I Properties of convolutions.
I Laplace Transform of a convolution.
I Impulse response solution.
I Solution decomposition theorem.
Properties of convolutions.
Theorem (Properties)
For every piecewise continuous functions f , g , and h, hold:
(i) Commutativity: f ∗ g = g ∗ f ;
(ii) Associativity: f ∗ (g ∗ h) = (f ∗ g) ∗ h;
(iii) Distributivity: f ∗ (g + h) = f ∗ g + f ∗ h;
(iv) Neutral element: f ∗ 0 = 0;
(v) Identity element: f ∗ δ = f .
Proof:
(v): (f ∗ δ)(t) =
∫ t
0
f (τ) δ(t − τ) dτ = f (t).
Properties of convolutions.
Theorem (Properties)
For every piecewise continuous functions f , g , and h, hold:
(i) Commutativity: f ∗ g = g ∗ f ;
(ii) Associativity: f ∗ (g ∗ h) = (f ∗ g) ∗ h;
(iii) Distributivity: f ∗ (g + h) = f ∗ g + f ∗ h;
(iv) Neutral element: f ∗ 0 = 0;
(v) Identity element: f ∗ δ = f .
Proof:
(v): (f ∗ δ)(t) =
∫ t
0
f (τ) δ(t − τ) dτ
= f (t).
Properties of convolutions.
Theorem (Properties)
For every piecewise continuous functions f , g , and h, hold:
(i) Commutativity: f ∗ g = g ∗ f ;
(ii) Associativity: f ∗ (g ∗ h) = (f ∗ g) ∗ h;
(iii) Distributivity: f ∗ (g + h) = f ∗ g + f ∗ h;
(iv) Neutral element: f ∗ 0 = 0;
(v) Identity element: f ∗ δ = f .
Proof:
(v): (f ∗ δ)(t) =
∫ t
0
f (τ) δ(t − τ) dτ = f (t).
Properties of convolutions.
Proof:(1): Commutativity: f ∗ g = g ∗ f .
The definition of convolution is,
(f ∗ g)(t) =
∫ t
0
f (τ) g(t − τ) dτ.
Change the integration variable: τ̂ = t − τ , hence d τ̂ = −dτ ,
(f ∗ g)(t) =
∫ 0
tf (t − τ̂) g(τ̂)(−1) d τ̂
(f ∗ g)(t) =
∫ t
0
g(τ̂) f (t − τ̂) d τ̂
We conclude: (f ∗ g)(t) = (g ∗ f )(t).
Properties of convolutions.
Proof:(1): Commutativity: f ∗ g = g ∗ f .
The definition of convolution is,
(f ∗ g)(t) =
∫ t
0
f (τ) g(t − τ) dτ.
Change the integration variable: τ̂ = t − τ , hence d τ̂ = −dτ ,
(f ∗ g)(t) =
∫ 0
tf (t − τ̂) g(τ̂)(−1) d τ̂
(f ∗ g)(t) =
∫ t
0
g(τ̂) f (t − τ̂) d τ̂
We conclude: (f ∗ g)(t) = (g ∗ f )(t).
Properties of convolutions.
Proof:(1): Commutativity: f ∗ g = g ∗ f .
The definition of convolution is,
(f ∗ g)(t) =
∫ t
0
f (τ) g(t − τ) dτ.
Change the integration variable: τ̂ = t − τ ,
hence d τ̂ = −dτ ,
(f ∗ g)(t) =
∫ 0
tf (t − τ̂) g(τ̂)(−1) d τ̂
(f ∗ g)(t) =
∫ t
0
g(τ̂) f (t − τ̂) d τ̂
We conclude: (f ∗ g)(t) = (g ∗ f )(t).
Properties of convolutions.
Proof:(1): Commutativity: f ∗ g = g ∗ f .
The definition of convolution is,
(f ∗ g)(t) =
∫ t
0
f (τ) g(t − τ) dτ.
Change the integration variable: τ̂ = t − τ , hence d τ̂ = −dτ ,
(f ∗ g)(t) =
∫ 0
tf (t − τ̂) g(τ̂)(−1) d τ̂
(f ∗ g)(t) =
∫ t
0
g(τ̂) f (t − τ̂) d τ̂
We conclude: (f ∗ g)(t) = (g ∗ f )(t).
Properties of convolutions.
Proof:(1): Commutativity: f ∗ g = g ∗ f .
The definition of convolution is,
(f ∗ g)(t) =
∫ t
0
f (τ) g(t − τ) dτ.
Change the integration variable: τ̂ = t − τ , hence d τ̂ = −dτ ,
(f ∗ g)(t) =
∫ 0
tf (t − τ̂) g(τ̂)(−1) d τ̂
(f ∗ g)(t) =
∫ t
0
g(τ̂) f (t − τ̂) d τ̂
We conclude: (f ∗ g)(t) = (g ∗ f )(t).
Properties of convolutions.
Proof:(1): Commutativity: f ∗ g = g ∗ f .
The definition of convolution is,
(f ∗ g)(t) =
∫ t
0
f (τ) g(t − τ) dτ.
Change the integration variable: τ̂ = t − τ , hence d τ̂ = −dτ ,
(f ∗ g)(t) =
∫ 0
tf (t − τ̂) g(τ̂)(−1) d τ̂
(f ∗ g)(t) =
∫ t
0
g(τ̂) f (t − τ̂) d τ̂
We conclude: (f ∗ g)(t) = (g ∗ f )(t).
Properties of convolutions.
Proof:(1): Commutativity: f ∗ g = g ∗ f .
The definition of convolution is,
(f ∗ g)(t) =
∫ t
0
f (τ) g(t − τ) dτ.
Change the integration variable: τ̂ = t − τ , hence d τ̂ = −dτ ,
(f ∗ g)(t) =
∫ 0
tf (t − τ̂) g(τ̂)(−1) d τ̂
(f ∗ g)(t) =
∫ t
0
g(τ̂) f (t − τ̂) d τ̂
We conclude: (f ∗ g)(t) = (g ∗ f )(t).
Convolution solutions (Sect. 6.6).
I Convolution of two functions.
I Properties of convolutions.
I Laplace Transform of a convolution.
I Impulse response solution.
I Solution decomposition theorem.
Laplace Transform of a convolution.
Theorem (Laplace Transform)
If f , g have well-defined Laplace Transforms L[f ], L[g ], then
L[f ∗ g ] = L[f ]L[g ].
Proof: The key step is to interchange two integrals. We start wethe product of the Laplace transforms,
L[f ]L[g ] =[∫ ∞
0
e−st f (t) dt] [∫ ∞
0
e−st̃g(t̃) dt̃],
L[f ]L[g ] =
∫ ∞
0
e−st̃g(t̃)(∫ ∞
0
e−st f (t) dt)
dt̃,
L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
0
e−s(t+t̃)f (t) dt)
dt̃.
Laplace Transform of a convolution.
Theorem (Laplace Transform)
If f , g have well-defined Laplace Transforms L[f ], L[g ], then
L[f ∗ g ] = L[f ]L[g ].
Proof: The key step is to interchange two integrals.
We start wethe product of the Laplace transforms,
L[f ]L[g ] =[∫ ∞
0
e−st f (t) dt] [∫ ∞
0
e−st̃g(t̃) dt̃],
L[f ]L[g ] =
∫ ∞
0
e−st̃g(t̃)(∫ ∞
0
e−st f (t) dt)
dt̃,
L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
0
e−s(t+t̃)f (t) dt)
dt̃.
Laplace Transform of a convolution.
Theorem (Laplace Transform)
If f , g have well-defined Laplace Transforms L[f ], L[g ], then
L[f ∗ g ] = L[f ]L[g ].
Proof: The key step is to interchange two integrals. We start wethe product of the Laplace transforms,
L[f ]L[g ] =[∫ ∞
0
e−st f (t) dt] [∫ ∞
0
e−st̃g(t̃) dt̃],
L[f ]L[g ] =
∫ ∞
0
e−st̃g(t̃)(∫ ∞
0
e−st f (t) dt)
dt̃,
L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
0
e−s(t+t̃)f (t) dt)
dt̃.
Laplace Transform of a convolution.
Theorem (Laplace Transform)
If f , g have well-defined Laplace Transforms L[f ], L[g ], then
L[f ∗ g ] = L[f ]L[g ].
Proof: The key step is to interchange two integrals. We start wethe product of the Laplace transforms,
L[f ]L[g ] =[∫ ∞
0
e−st f (t) dt] [∫ ∞
0
e−st̃g(t̃) dt̃],
L[f ]L[g ] =
∫ ∞
0
e−st̃g(t̃)(∫ ∞
0
e−st f (t) dt)
dt̃,
L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
0
e−s(t+t̃)f (t) dt)
dt̃.
Laplace Transform of a convolution.
Theorem (Laplace Transform)
If f , g have well-defined Laplace Transforms L[f ], L[g ], then
L[f ∗ g ] = L[f ]L[g ].
Proof: The key step is to interchange two integrals. We start wethe product of the Laplace transforms,
L[f ]L[g ] =[∫ ∞
0
e−st f (t) dt] [∫ ∞
0
e−st̃g(t̃) dt̃],
L[f ]L[g ] =
∫ ∞
0
e−st̃g(t̃)(∫ ∞
0
e−st f (t) dt)
dt̃,
L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
0
e−s(t+t̃)f (t) dt)
dt̃.
Laplace Transform of a convolution.
Proof: Recall: L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
0
e−s(t+t̃)f (t) dt)
dt̃.
Change variables: τ = t + t̃, hence dτ = dt;
L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
t̃e−sτ f (τ − t̃) dτ
)dt̃.
L[f ]L[g ] =
∫ ∞
0
∫ ∞
t̃e−sτ g(t̃) f (τ − t̃) dτ dt̃.
The key step: Switch the order of integration.
t = tau
tau
t
0
L[f ]L[g ] =
∫ ∞
0
∫ τ
0e−sτ g(t̃) f (τ − t̃) dt̃ dτ.
Laplace Transform of a convolution.
Proof: Recall: L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
0
e−s(t+t̃)f (t) dt)
dt̃.
Change variables: τ = t + t̃,
hence dτ = dt;
L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
t̃e−sτ f (τ − t̃) dτ
)dt̃.
L[f ]L[g ] =
∫ ∞
0
∫ ∞
t̃e−sτ g(t̃) f (τ − t̃) dτ dt̃.
The key step: Switch the order of integration.
t = tau
tau
t
0
L[f ]L[g ] =
∫ ∞
0
∫ τ
0e−sτ g(t̃) f (τ − t̃) dt̃ dτ.
Laplace Transform of a convolution.
Proof: Recall: L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
0
e−s(t+t̃)f (t) dt)
dt̃.
Change variables: τ = t + t̃, hence dτ = dt;
L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
t̃e−sτ f (τ − t̃) dτ
)dt̃.
L[f ]L[g ] =
∫ ∞
0
∫ ∞
t̃e−sτ g(t̃) f (τ − t̃) dτ dt̃.
The key step: Switch the order of integration.
t = tau
tau
t
0
L[f ]L[g ] =
∫ ∞
0
∫ τ
0e−sτ g(t̃) f (τ − t̃) dt̃ dτ.
Laplace Transform of a convolution.
Proof: Recall: L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
0
e−s(t+t̃)f (t) dt)
dt̃.
Change variables: τ = t + t̃, hence dτ = dt;
L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
t̃e−sτ f (τ − t̃) dτ
)dt̃.
L[f ]L[g ] =
∫ ∞
0
∫ ∞
t̃e−sτ g(t̃) f (τ − t̃) dτ dt̃.
The key step: Switch the order of integration.
t = tau
tau
t
0
L[f ]L[g ] =
∫ ∞
0
∫ τ
0e−sτ g(t̃) f (τ − t̃) dt̃ dτ.
Laplace Transform of a convolution.
Proof: Recall: L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
0
e−s(t+t̃)f (t) dt)
dt̃.
Change variables: τ = t + t̃, hence dτ = dt;
L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
t̃e−sτ f (τ − t̃) dτ
)dt̃.
L[f ]L[g ] =
∫ ∞
0
∫ ∞
t̃e−sτ g(t̃) f (τ − t̃) dτ dt̃.
The key step: Switch the order of integration.
t = tau
tau
t
0
L[f ]L[g ] =
∫ ∞
0
∫ τ
0e−sτ g(t̃) f (τ − t̃) dt̃ dτ.
Laplace Transform of a convolution.
Proof: Recall: L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
0
e−s(t+t̃)f (t) dt)
dt̃.
Change variables: τ = t + t̃, hence dτ = dt;
L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
t̃e−sτ f (τ − t̃) dτ
)dt̃.
L[f ]L[g ] =
∫ ∞
0
∫ ∞
t̃e−sτ g(t̃) f (τ − t̃) dτ dt̃.
The key step: Switch the order of integration.
t = tau
tau
t
0
L[f ]L[g ] =
∫ ∞
0
∫ τ
0e−sτ g(t̃) f (τ − t̃) dt̃ dτ.
Laplace Transform of a convolution.
Proof: Recall: L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
0
e−s(t+t̃)f (t) dt)
dt̃.
Change variables: τ = t + t̃, hence dτ = dt;
L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
t̃e−sτ f (τ − t̃) dτ
)dt̃.
L[f ]L[g ] =
∫ ∞
0
∫ ∞
t̃e−sτ g(t̃) f (τ − t̃) dτ dt̃.
The key step: Switch the order of integration.
t = tau
tau
t
0
L[f ]L[g ] =
∫ ∞
0
∫ τ
0e−sτ g(t̃) f (τ − t̃) dt̃ dτ.
Laplace Transform of a convolution.
Proof: Recall: L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
0
e−s(t+t̃)f (t) dt)
dt̃.
Change variables: τ = t + t̃, hence dτ = dt;
L[f ]L[g ] =
∫ ∞
0
g(t̃)(∫ ∞
t̃e−sτ f (τ − t̃) dτ
)dt̃.
L[f ]L[g ] =
∫ ∞
0
∫ ∞
t̃e−sτ g(t̃) f (τ − t̃) dτ dt̃.
The key step: Switch the order of integration.
t = tau
tau
t
0
L[f ]L[g ] =
∫ ∞
0
∫ τ
0e−sτ g(t̃) f (τ − t̃) dt̃ dτ.
Laplace Transform of a convolution.
Proof: Recall: L[f ]L[g ] =
∫ ∞
0
∫ τ
0e−sτ g(t̃) f (τ − t̃) dt̃ dτ .
Then, is straightforward to check that
L[f ]L[g ] =
∫ ∞
0
e−sτ(∫ τ
0g(t̃) f (τ − t̃) dt̃
)dτ,
L[f ]L[g ] =
∫ ∞
0
e−sτ (g ∗ f )(τ) dτ
L[f ]L[g ] = L[g ∗ f ]
We conclude: L[f ∗ g ] = L[f ]L[g ].
Laplace Transform of a convolution.
Proof: Recall: L[f ]L[g ] =
∫ ∞
0
∫ τ
0e−sτ g(t̃) f (τ − t̃) dt̃ dτ .
Then, is straightforward to check that
L[f ]L[g ] =
∫ ∞
0
e−sτ(∫ τ
0g(t̃) f (τ − t̃) dt̃
)dτ,
L[f ]L[g ] =
∫ ∞
0
e−sτ (g ∗ f )(τ) dτ
L[f ]L[g ] = L[g ∗ f ]
We conclude: L[f ∗ g ] = L[f ]L[g ].
Laplace Transform of a convolution.
Proof: Recall: L[f ]L[g ] =
∫ ∞
0
∫ τ
0e−sτ g(t̃) f (τ − t̃) dt̃ dτ .
Then, is straightforward to check that
L[f ]L[g ] =
∫ ∞
0
e−sτ(∫ τ
0g(t̃) f (τ − t̃) dt̃
)dτ,
L[f ]L[g ] =
∫ ∞
0
e−sτ (g ∗ f )(τ) dτ
L[f ]L[g ] = L[g ∗ f ]
We conclude: L[f ∗ g ] = L[f ]L[g ].
Laplace Transform of a convolution.
Proof: Recall: L[f ]L[g ] =
∫ ∞
0
∫ τ
0e−sτ g(t̃) f (τ − t̃) dt̃ dτ .
Then, is straightforward to check that
L[f ]L[g ] =
∫ ∞
0
e−sτ(∫ τ
0g(t̃) f (τ − t̃) dt̃
)dτ,
L[f ]L[g ] =
∫ ∞
0
e−sτ (g ∗ f )(τ) dτ
L[f ]L[g ] = L[g ∗ f ]
We conclude: L[f ∗ g ] = L[f ]L[g ].
Laplace Transform of a convolution.
Proof: Recall: L[f ]L[g ] =
∫ ∞
0
∫ τ
0e−sτ g(t̃) f (τ − t̃) dt̃ dτ .
Then, is straightforward to check that
L[f ]L[g ] =
∫ ∞
0
e−sτ(∫ τ
0g(t̃) f (τ − t̃) dt̃
)dτ,
L[f ]L[g ] =
∫ ∞
0
e−sτ (g ∗ f )(τ) dτ
L[f ]L[g ] = L[g ∗ f ]
We conclude: L[f ∗ g ] = L[f ]L[g ].
Convolution solutions (Sect. 6.6).
I Convolution of two functions.
I Properties of convolutions.
I Laplace Transform of a convolution.
I Impulse response solution.
I Solution decomposition theorem.
Impulse response solution.
DefinitionThe impulse response solution is the function yδ solution of the IVP
y ′′δ + a1 y ′δ + a0 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, c ∈ R.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0.
Solution: L[y ′′δ ] + 2L[y ′δ] + 2L[yδ] = L[δ(t − c)].
(s2 + 2s + 2)L[yδ] = e−cs ⇒ L[yδ] =e−cs
(s2 + 2s + 2).
Impulse response solution.
DefinitionThe impulse response solution is the function yδ solution of the IVP
y ′′δ + a1 y ′δ + a0 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, c ∈ R.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0.
Solution: L[y ′′δ ] + 2L[y ′δ] + 2L[yδ] = L[δ(t − c)].
(s2 + 2s + 2)L[yδ] = e−cs ⇒ L[yδ] =e−cs
(s2 + 2s + 2).
Impulse response solution.
DefinitionThe impulse response solution is the function yδ solution of the IVP
y ′′δ + a1 y ′δ + a0 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, c ∈ R.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0.
Solution: L[y ′′δ ] + 2L[y ′δ] + 2L[yδ] = L[δ(t − c)].
(s2 + 2s + 2)L[yδ] = e−cs ⇒ L[yδ] =e−cs
(s2 + 2s + 2).
Impulse response solution.
DefinitionThe impulse response solution is the function yδ solution of the IVP
y ′′δ + a1 y ′δ + a0 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, c ∈ R.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0.
Solution: L[y ′′δ ] + 2L[y ′δ] + 2L[yδ] = L[δ(t − c)].
(s2 + 2s + 2)L[yδ] = e−cs
⇒ L[yδ] =e−cs
(s2 + 2s + 2).
Impulse response solution.
DefinitionThe impulse response solution is the function yδ solution of the IVP
y ′′δ + a1 y ′δ + a0 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, c ∈ R.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0.
Solution: L[y ′′δ ] + 2L[y ′δ] + 2L[yδ] = L[δ(t − c)].
(s2 + 2s + 2)L[yδ] = e−cs ⇒ L[yδ] =e−cs
(s2 + 2s + 2).
Impulse response solution.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .
Solution: Recall: L[yδ] =e−cs
(s2 + 2s + 2).
Find the roots of the denominator,
s2 + 2s + 2 = 0 ⇒ s± =1
2
[−2±
√4− 8
]Complex roots. We complete the square:
s2 + 2s + 2 =[s2 + 2
(2
2
)s + 1
]− 1 + 2 = (s + 1)2 + 1.
Therefore, L[yδ] =e−cs
(s + 1)2 + 1.
Impulse response solution.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .
Solution: Recall: L[yδ] =e−cs
(s2 + 2s + 2).
Find the roots of the denominator,
s2 + 2s + 2 = 0
⇒ s± =1
2
[−2±
√4− 8
]Complex roots. We complete the square:
s2 + 2s + 2 =[s2 + 2
(2
2
)s + 1
]− 1 + 2 = (s + 1)2 + 1.
Therefore, L[yδ] =e−cs
(s + 1)2 + 1.
Impulse response solution.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .
Solution: Recall: L[yδ] =e−cs
(s2 + 2s + 2).
Find the roots of the denominator,
s2 + 2s + 2 = 0 ⇒ s± =1
2
[−2±
√4− 8
]
Complex roots. We complete the square:
s2 + 2s + 2 =[s2 + 2
(2
2
)s + 1
]− 1 + 2 = (s + 1)2 + 1.
Therefore, L[yδ] =e−cs
(s + 1)2 + 1.
Impulse response solution.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .
Solution: Recall: L[yδ] =e−cs
(s2 + 2s + 2).
Find the roots of the denominator,
s2 + 2s + 2 = 0 ⇒ s± =1
2
[−2±
√4− 8
]Complex roots.
We complete the square:
s2 + 2s + 2 =[s2 + 2
(2
2
)s + 1
]− 1 + 2 = (s + 1)2 + 1.
Therefore, L[yδ] =e−cs
(s + 1)2 + 1.
Impulse response solution.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .
Solution: Recall: L[yδ] =e−cs
(s2 + 2s + 2).
Find the roots of the denominator,
s2 + 2s + 2 = 0 ⇒ s± =1
2
[−2±
√4− 8
]Complex roots. We complete the square:
s2 + 2s + 2 =[s2 + 2
(2
2
)s + 1
]− 1 + 2 = (s + 1)2 + 1.
Therefore, L[yδ] =e−cs
(s + 1)2 + 1.
Impulse response solution.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .
Solution: Recall: L[yδ] =e−cs
(s2 + 2s + 2).
Find the roots of the denominator,
s2 + 2s + 2 = 0 ⇒ s± =1
2
[−2±
√4− 8
]Complex roots. We complete the square:
s2 + 2s + 2 =[s2 + 2
(2
2
)s + 1
]− 1 + 2
= (s + 1)2 + 1.
Therefore, L[yδ] =e−cs
(s + 1)2 + 1.
Impulse response solution.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .
Solution: Recall: L[yδ] =e−cs
(s2 + 2s + 2).
Find the roots of the denominator,
s2 + 2s + 2 = 0 ⇒ s± =1
2
[−2±
√4− 8
]Complex roots. We complete the square:
s2 + 2s + 2 =[s2 + 2
(2
2
)s + 1
]− 1 + 2 = (s + 1)2 + 1.
Therefore, L[yδ] =e−cs
(s + 1)2 + 1.
Impulse response solution.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .
Solution: Recall: L[yδ] =e−cs
(s2 + 2s + 2).
Find the roots of the denominator,
s2 + 2s + 2 = 0 ⇒ s± =1
2
[−2±
√4− 8
]Complex roots. We complete the square:
s2 + 2s + 2 =[s2 + 2
(2
2
)s + 1
]− 1 + 2 = (s + 1)2 + 1.
Therefore, L[yδ] =e−cs
(s + 1)2 + 1.
Impulse response solution.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .
Solution: Recall: L[yδ] =e−cs
(s + 1)2 + 1.
Recall: L[sin(t)] =1
s2 + 1, and L[f ](s − c) = L[ect f (t)].
1
(s + 1)2 + 1= L[e−t sin(t)] ⇒ L[yδ] = e−cs L[e−t sin(t)].
Since e−cs L[f ](s) = L[u(t − c) f (t − c)],
we conclude yδ(t) = u(t − c) e−(t−c) sin(t − c). C
Impulse response solution.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .
Solution: Recall: L[yδ] =e−cs
(s + 1)2 + 1.
Recall: L[sin(t)] =1
s2 + 1,
and L[f ](s − c) = L[ect f (t)].
1
(s + 1)2 + 1= L[e−t sin(t)] ⇒ L[yδ] = e−cs L[e−t sin(t)].
Since e−cs L[f ](s) = L[u(t − c) f (t − c)],
we conclude yδ(t) = u(t − c) e−(t−c) sin(t − c). C
Impulse response solution.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .
Solution: Recall: L[yδ] =e−cs
(s + 1)2 + 1.
Recall: L[sin(t)] =1
s2 + 1, and L[f ](s − c) = L[ect f (t)].
1
(s + 1)2 + 1= L[e−t sin(t)] ⇒ L[yδ] = e−cs L[e−t sin(t)].
Since e−cs L[f ](s) = L[u(t − c) f (t − c)],
we conclude yδ(t) = u(t − c) e−(t−c) sin(t − c). C
Impulse response solution.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .
Solution: Recall: L[yδ] =e−cs
(s + 1)2 + 1.
Recall: L[sin(t)] =1
s2 + 1, and L[f ](s − c) = L[ect f (t)].
1
(s + 1)2 + 1= L[e−t sin(t)]
⇒ L[yδ] = e−cs L[e−t sin(t)].
Since e−cs L[f ](s) = L[u(t − c) f (t − c)],
we conclude yδ(t) = u(t − c) e−(t−c) sin(t − c). C
Impulse response solution.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .
Solution: Recall: L[yδ] =e−cs
(s + 1)2 + 1.
Recall: L[sin(t)] =1
s2 + 1, and L[f ](s − c) = L[ect f (t)].
1
(s + 1)2 + 1= L[e−t sin(t)] ⇒ L[yδ] = e−cs L[e−t sin(t)].
Since e−cs L[f ](s) = L[u(t − c) f (t − c)],
we conclude yδ(t) = u(t − c) e−(t−c) sin(t − c). C
Impulse response solution.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .
Solution: Recall: L[yδ] =e−cs
(s + 1)2 + 1.
Recall: L[sin(t)] =1
s2 + 1, and L[f ](s − c) = L[ect f (t)].
1
(s + 1)2 + 1= L[e−t sin(t)] ⇒ L[yδ] = e−cs L[e−t sin(t)].
Since e−cs L[f ](s) = L[u(t − c) f (t − c)],
we conclude yδ(t) = u(t − c) e−(t−c) sin(t − c). C
Impulse response solution.
Example
Find the impulse response solution of the IVP
y ′′δ + 2 y ′δ + 2 yδ = δ(t − c), yδ(0) = 0, y ′δ(0) = 0, .
Solution: Recall: L[yδ] =e−cs
(s + 1)2 + 1.
Recall: L[sin(t)] =1
s2 + 1, and L[f ](s − c) = L[ect f (t)].
1
(s + 1)2 + 1= L[e−t sin(t)] ⇒ L[yδ] = e−cs L[e−t sin(t)].
Since e−cs L[f ](s) = L[u(t − c) f (t − c)],
we conclude yδ(t) = u(t − c) e−(t−c) sin(t − c). C
Convolution solutions (Sect. 6.6).
I Convolution of two functions.
I Properties of convolutions.
I Laplace Transform of a convolution.
I Impulse response solution.
I Solution decomposition theorem.
Solution decomposition theorem.
Theorem (Solution decomposition)
The solution y to the IVP
y ′′ + a1 y ′ + a0 y = g(t), y(0) = y0, y ′(0) = y1,
can be decomposed as
y(t) = yh(t) + (yδ ∗ g)(t),
where yh is the solution of the homogeneous IVP
y ′′h + a1 y ′h + a0 yh = 0, yh(0) = y0, y ′h(0) = y1,
and yδ is the impulse response solution, that is,
y ′′δ + a1 y ′δ + a0 yδ = δ(t), yδ(0) = 0, y ′δ(0) = 0.
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.
Solution: L[y ′′] + 2L[y ′] + 2L[y ] = L[sin(at)], and recall,
L[y ′′] = s2 L[y ]− s (1)− (−1), L[y ′] = s L[y ]− 1.
(s2 + 2s + 2)L[y ]− s + 1− 2 = L[sin(at)].
L[y ] =(s + 1)
(s2 + 2s + 2)+
1
(s2 + 2s + 2)L[sin(at)].
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.
Solution: L[y ′′] + 2L[y ′] + 2L[y ] = L[sin(at)],
and recall,
L[y ′′] = s2 L[y ]− s (1)− (−1), L[y ′] = s L[y ]− 1.
(s2 + 2s + 2)L[y ]− s + 1− 2 = L[sin(at)].
L[y ] =(s + 1)
(s2 + 2s + 2)+
1
(s2 + 2s + 2)L[sin(at)].
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.
Solution: L[y ′′] + 2L[y ′] + 2L[y ] = L[sin(at)], and recall,
L[y ′′] = s2 L[y ]− s (1)− (−1),
L[y ′] = s L[y ]− 1.
(s2 + 2s + 2)L[y ]− s + 1− 2 = L[sin(at)].
L[y ] =(s + 1)
(s2 + 2s + 2)+
1
(s2 + 2s + 2)L[sin(at)].
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.
Solution: L[y ′′] + 2L[y ′] + 2L[y ] = L[sin(at)], and recall,
L[y ′′] = s2 L[y ]− s (1)− (−1), L[y ′] = s L[y ]− 1.
(s2 + 2s + 2)L[y ]− s + 1− 2 = L[sin(at)].
L[y ] =(s + 1)
(s2 + 2s + 2)+
1
(s2 + 2s + 2)L[sin(at)].
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.
Solution: L[y ′′] + 2L[y ′] + 2L[y ] = L[sin(at)], and recall,
L[y ′′] = s2 L[y ]− s (1)− (−1), L[y ′] = s L[y ]− 1.
(s2 + 2s + 2)L[y ]− s + 1− 2 = L[sin(at)].
L[y ] =(s + 1)
(s2 + 2s + 2)+
1
(s2 + 2s + 2)L[sin(at)].
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.
Solution: L[y ′′] + 2L[y ′] + 2L[y ] = L[sin(at)], and recall,
L[y ′′] = s2 L[y ]− s (1)− (−1), L[y ′] = s L[y ]− 1.
(s2 + 2s + 2)L[y ]− s + 1− 2 = L[sin(at)].
L[y ] =(s + 1)
(s2 + 2s + 2)+
1
(s2 + 2s + 2)L[sin(at)].
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.
Solution: Recall: L[y ] =(s + 1)
(s2 + 2s + 2)+
1
(s2 + 2s + 2)L[sin(at)].
But: L[yh] =(s + 1)
(s2 + 2s + 2)=
(s + 1)
(s + 1)2 + 1= L[e−t cos(t)],
and: L[yδ] =1
(s2 + 2s + 2)=
1
(s + 1)2 + 1= L[e−t sin(t)]. So,
L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),
So: y(t) = e−t cos(t) +
∫ t
0e−τ sin(τ) sin[a(t − τ)] dτ . C
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.
Solution: Recall: L[y ] =(s + 1)
(s2 + 2s + 2)+
1
(s2 + 2s + 2)L[sin(at)].
But: L[yh] =(s + 1)
(s2 + 2s + 2)
=(s + 1)
(s + 1)2 + 1= L[e−t cos(t)],
and: L[yδ] =1
(s2 + 2s + 2)=
1
(s + 1)2 + 1= L[e−t sin(t)]. So,
L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),
So: y(t) = e−t cos(t) +
∫ t
0e−τ sin(τ) sin[a(t − τ)] dτ . C
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.
Solution: Recall: L[y ] =(s + 1)
(s2 + 2s + 2)+
1
(s2 + 2s + 2)L[sin(at)].
But: L[yh] =(s + 1)
(s2 + 2s + 2)=
(s + 1)
(s + 1)2 + 1
= L[e−t cos(t)],
and: L[yδ] =1
(s2 + 2s + 2)=
1
(s + 1)2 + 1= L[e−t sin(t)]. So,
L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),
So: y(t) = e−t cos(t) +
∫ t
0e−τ sin(τ) sin[a(t − τ)] dτ . C
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.
Solution: Recall: L[y ] =(s + 1)
(s2 + 2s + 2)+
1
(s2 + 2s + 2)L[sin(at)].
But: L[yh] =(s + 1)
(s2 + 2s + 2)=
(s + 1)
(s + 1)2 + 1= L[e−t cos(t)],
and: L[yδ] =1
(s2 + 2s + 2)=
1
(s + 1)2 + 1= L[e−t sin(t)]. So,
L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),
So: y(t) = e−t cos(t) +
∫ t
0e−τ sin(τ) sin[a(t − τ)] dτ . C
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.
Solution: Recall: L[y ] =(s + 1)
(s2 + 2s + 2)+
1
(s2 + 2s + 2)L[sin(at)].
But: L[yh] =(s + 1)
(s2 + 2s + 2)=
(s + 1)
(s + 1)2 + 1= L[e−t cos(t)],
and: L[yδ] =1
(s2 + 2s + 2)
=1
(s + 1)2 + 1= L[e−t sin(t)]. So,
L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),
So: y(t) = e−t cos(t) +
∫ t
0e−τ sin(τ) sin[a(t − τ)] dτ . C
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.
Solution: Recall: L[y ] =(s + 1)
(s2 + 2s + 2)+
1
(s2 + 2s + 2)L[sin(at)].
But: L[yh] =(s + 1)
(s2 + 2s + 2)=
(s + 1)
(s + 1)2 + 1= L[e−t cos(t)],
and: L[yδ] =1
(s2 + 2s + 2)=
1
(s + 1)2 + 1
= L[e−t sin(t)]. So,
L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),
So: y(t) = e−t cos(t) +
∫ t
0e−τ sin(τ) sin[a(t − τ)] dτ . C
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.
Solution: Recall: L[y ] =(s + 1)
(s2 + 2s + 2)+
1
(s2 + 2s + 2)L[sin(at)].
But: L[yh] =(s + 1)
(s2 + 2s + 2)=
(s + 1)
(s + 1)2 + 1= L[e−t cos(t)],
and: L[yδ] =1
(s2 + 2s + 2)=
1
(s + 1)2 + 1= L[e−t sin(t)].
So,
L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),
So: y(t) = e−t cos(t) +
∫ t
0e−τ sin(τ) sin[a(t − τ)] dτ . C
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.
Solution: Recall: L[y ] =(s + 1)
(s2 + 2s + 2)+
1
(s2 + 2s + 2)L[sin(at)].
But: L[yh] =(s + 1)
(s2 + 2s + 2)=
(s + 1)
(s + 1)2 + 1= L[e−t cos(t)],
and: L[yδ] =1
(s2 + 2s + 2)=
1
(s + 1)2 + 1= L[e−t sin(t)]. So,
L[y ] = L[yh] + L[yδ]L[g(t)]
⇒ y(t) = yh(t) + (yδ ∗ g)(t),
So: y(t) = e−t cos(t) +
∫ t
0e−τ sin(τ) sin[a(t − τ)] dτ . C
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.
Solution: Recall: L[y ] =(s + 1)
(s2 + 2s + 2)+
1
(s2 + 2s + 2)L[sin(at)].
But: L[yh] =(s + 1)
(s2 + 2s + 2)=
(s + 1)
(s + 1)2 + 1= L[e−t cos(t)],
and: L[yδ] =1
(s2 + 2s + 2)=
1
(s + 1)2 + 1= L[e−t sin(t)]. So,
L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),
So: y(t) = e−t cos(t) +
∫ t
0e−τ sin(τ) sin[a(t − τ)] dτ . C
Solution decomposition theorem.
Example
Use the Solution Decomposition Theorem to express the solution of
y ′′ + 2 y ′ + 2 y = sin(at), y(0) = 1, y ′(0) = −1.
Solution: Recall: L[y ] =(s + 1)
(s2 + 2s + 2)+
1
(s2 + 2s + 2)L[sin(at)].
But: L[yh] =(s + 1)
(s2 + 2s + 2)=
(s + 1)
(s + 1)2 + 1= L[e−t cos(t)],
and: L[yδ] =1
(s2 + 2s + 2)=
1
(s + 1)2 + 1= L[e−t sin(t)]. So,
L[y ] = L[yh] + L[yδ]L[g(t)] ⇒ y(t) = yh(t) + (yδ ∗ g)(t),
So: y(t) = e−t cos(t) +
∫ t
0e−τ sin(τ) sin[a(t − τ)] dτ . C
Solution decomposition theorem.
Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)],
and recall,
L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.
(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].
L[y ] =(s + a1)y0 + y1
(s2 + a1s + a0)+
1
(s2 + a1s + a0)L[g(t)].
Recall: L[yh] =(s + a1)y0 + y1
(s2 + a1s + a0), and L[yδ] =
1
(s2 + a1s + a0).
Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).
Equivalently: y(t) = yh(t) +
∫ t
0yδ(τ)g(t − τ) dτ .
Solution decomposition theorem.
Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,
L[y ′′] = s2 L[y ]− sy0 − y1,
L[y ′] = s L[y ]− y0.
(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].
L[y ] =(s + a1)y0 + y1
(s2 + a1s + a0)+
1
(s2 + a1s + a0)L[g(t)].
Recall: L[yh] =(s + a1)y0 + y1
(s2 + a1s + a0), and L[yδ] =
1
(s2 + a1s + a0).
Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).
Equivalently: y(t) = yh(t) +
∫ t
0yδ(τ)g(t − τ) dτ .
Solution decomposition theorem.
Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,
L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.
(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].
L[y ] =(s + a1)y0 + y1
(s2 + a1s + a0)+
1
(s2 + a1s + a0)L[g(t)].
Recall: L[yh] =(s + a1)y0 + y1
(s2 + a1s + a0), and L[yδ] =
1
(s2 + a1s + a0).
Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).
Equivalently: y(t) = yh(t) +
∫ t
0yδ(τ)g(t − τ) dτ .
Solution decomposition theorem.
Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,
L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.
(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].
L[y ] =(s + a1)y0 + y1
(s2 + a1s + a0)+
1
(s2 + a1s + a0)L[g(t)].
Recall: L[yh] =(s + a1)y0 + y1
(s2 + a1s + a0), and L[yδ] =
1
(s2 + a1s + a0).
Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).
Equivalently: y(t) = yh(t) +
∫ t
0yδ(τ)g(t − τ) dτ .
Solution decomposition theorem.
Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,
L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.
(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].
L[y ] =(s + a1)y0 + y1
(s2 + a1s + a0)+
1
(s2 + a1s + a0)L[g(t)].
Recall: L[yh] =(s + a1)y0 + y1
(s2 + a1s + a0), and L[yδ] =
1
(s2 + a1s + a0).
Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).
Equivalently: y(t) = yh(t) +
∫ t
0yδ(τ)g(t − τ) dτ .
Solution decomposition theorem.
Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,
L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.
(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].
L[y ] =(s + a1)y0 + y1
(s2 + a1s + a0)+
1
(s2 + a1s + a0)L[g(t)].
Recall: L[yh] =(s + a1)y0 + y1
(s2 + a1s + a0),
and L[yδ] =1
(s2 + a1s + a0).
Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).
Equivalently: y(t) = yh(t) +
∫ t
0yδ(τ)g(t − τ) dτ .
Solution decomposition theorem.
Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,
L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.
(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].
L[y ] =(s + a1)y0 + y1
(s2 + a1s + a0)+
1
(s2 + a1s + a0)L[g(t)].
Recall: L[yh] =(s + a1)y0 + y1
(s2 + a1s + a0), and L[yδ] =
1
(s2 + a1s + a0).
Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).
Equivalently: y(t) = yh(t) +
∫ t
0yδ(τ)g(t − τ) dτ .
Solution decomposition theorem.
Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,
L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.
(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].
L[y ] =(s + a1)y0 + y1
(s2 + a1s + a0)+
1
(s2 + a1s + a0)L[g(t)].
Recall: L[yh] =(s + a1)y0 + y1
(s2 + a1s + a0), and L[yδ] =
1
(s2 + a1s + a0).
Since, L[y ] = L[yh] + L[yδ]L[g(t)],
so y(t) = yh(t) + (yδ ∗ g)(t).
Equivalently: y(t) = yh(t) +
∫ t
0yδ(τ)g(t − τ) dτ .
Solution decomposition theorem.
Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,
L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.
(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].
L[y ] =(s + a1)y0 + y1
(s2 + a1s + a0)+
1
(s2 + a1s + a0)L[g(t)].
Recall: L[yh] =(s + a1)y0 + y1
(s2 + a1s + a0), and L[yδ] =
1
(s2 + a1s + a0).
Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).
Equivalently: y(t) = yh(t) +
∫ t
0yδ(τ)g(t − τ) dτ .
Solution decomposition theorem.
Proof: Compute: L[y ′′] + a1 L[y ′] + a0 L[y ] = L[g(t)], and recall,
L[y ′′] = s2 L[y ]− sy0 − y1, L[y ′] = s L[y ]− y0.
(s2 + a1s + a0)L[y ]− sy0 − y1 − a1y0 = L[g(t)].
L[y ] =(s + a1)y0 + y1
(s2 + a1s + a0)+
1
(s2 + a1s + a0)L[g(t)].
Recall: L[yh] =(s + a1)y0 + y1
(s2 + a1s + a0), and L[yδ] =
1
(s2 + a1s + a0).
Since, L[y ] = L[yh] + L[yδ]L[g(t)], so y(t) = yh(t) + (yδ ∗ g)(t).
Equivalently: y(t) = yh(t) +
∫ t
0yδ(τ)g(t − τ) dτ .
Systems of linear differential equations (Sect. 7.1).
I n × n systems of linear differential equations.
I Second order equations and first order systems.
I Main concepts from Linear Algebra.
n × n systems of linear differential equations.
Remark: Many physical systems must be described with morethan one differential equation.
Example
Newton’s law of motion for a particle of mass m moving in space.The unknown and the force are vector-valued functions,
x(t) =
x1(t)x2(t)x3(t)
, F(t) =
F1(t, x)F2(t, x)F3(t, x)
.
The equation of motion are: md2x
dt2= F
(t, x(t)
).
These are three differential equations,
md2x1
dt2= F1
(t, x(t)
), m
d2x2
dt2= F2
(t, x(t)
), m
d2x3
dt2= F3
(t, x(t)
).
C
n × n systems of linear differential equations.
Remark: Many physical systems must be described with morethan one differential equation.
Example
Newton’s law of motion for a particle of mass m moving in space.
The unknown and the force are vector-valued functions,
x(t) =
x1(t)x2(t)x3(t)
, F(t) =
F1(t, x)F2(t, x)F3(t, x)
.
The equation of motion are: md2x
dt2= F
(t, x(t)
).
These are three differential equations,
md2x1
dt2= F1
(t, x(t)
), m
d2x2
dt2= F2
(t, x(t)
), m
d2x3
dt2= F3
(t, x(t)
).
C
n × n systems of linear differential equations.
Remark: Many physical systems must be described with morethan one differential equation.
Example
Newton’s law of motion for a particle of mass m moving in space.The unknown and the force are vector-valued functions,
x(t) =
x1(t)x2(t)x3(t)
, F(t) =
F1(t, x)F2(t, x)F3(t, x)
.
The equation of motion are: md2x
dt2= F
(t, x(t)
).
These are three differential equations,
md2x1
dt2= F1
(t, x(t)
), m
d2x2
dt2= F2
(t, x(t)
), m
d2x3
dt2= F3
(t, x(t)
).
C
n × n systems of linear differential equations.
Remark: Many physical systems must be described with morethan one differential equation.
Example
Newton’s law of motion for a particle of mass m moving in space.The unknown and the force are vector-valued functions,
x(t) =
x1(t)x2(t)x3(t)
,
F(t) =
F1(t, x)F2(t, x)F3(t, x)
.
The equation of motion are: md2x
dt2= F
(t, x(t)
).
These are three differential equations,
md2x1
dt2= F1
(t, x(t)
), m
d2x2
dt2= F2
(t, x(t)
), m
d2x3
dt2= F3
(t, x(t)
).
C
n × n systems of linear differential equations.
Remark: Many physical systems must be described with morethan one differential equation.
Example
Newton’s law of motion for a particle of mass m moving in space.The unknown and the force are vector-valued functions,
x(t) =
x1(t)x2(t)x3(t)
, F(t) =
F1(t, x)F2(t, x)F3(t, x)
.
The equation of motion are: md2x
dt2= F
(t, x(t)
).
These are three differential equations,
md2x1
dt2= F1
(t, x(t)
), m
d2x2
dt2= F2
(t, x(t)
), m
d2x3
dt2= F3
(t, x(t)
).
C
n × n systems of linear differential equations.
Remark: Many physical systems must be described with morethan one differential equation.
Example
Newton’s law of motion for a particle of mass m moving in space.The unknown and the force are vector-valued functions,
x(t) =
x1(t)x2(t)x3(t)
, F(t) =
F1(t, x)F2(t, x)F3(t, x)
.
The equation of motion are: md2x
dt2= F
(t, x(t)
).
These are three differential equations,
md2x1
dt2= F1
(t, x(t)
), m
d2x2
dt2= F2
(t, x(t)
), m
d2x3
dt2= F3
(t, x(t)
).
C
n × n systems of linear differential equations.
Remark: Many physical systems must be described with morethan one differential equation.
Example
Newton’s law of motion for a particle of mass m moving in space.The unknown and the force are vector-valued functions,
x(t) =
x1(t)x2(t)x3(t)
, F(t) =
F1(t, x)F2(t, x)F3(t, x)
.
The equation of motion are: md2x
dt2= F
(t, x(t)
).
These are three differential equations,
md2x1
dt2= F1
(t, x(t)
), m
d2x2
dt2= F2
(t, x(t)
), m
d2x3
dt2= F3
(t, x(t)
).
C
n × n systems of linear differential equations.
DefinitionAn n × n system of linear first order differential equations is thefollowing: Given the functions aij , gi : [a, b] → R, wherei , j = 1, · · · , n, find n functions xj : [a, b] → R solutions of the nlinear differential equations
x ′1 = a11(t) x1 + · · ·+ a1n(t) xn + g1(t)
...
x ′n = an1(t) x1 + · · ·+ ann(t) xn + gn(t).
The system is called homogeneous iff the source functions satisfythat g1 = · · · = gn = 0.
n × n systems of linear differential equations.
Example
n = 1: Single differential equation: Find x1(t) solution of
x ′1 = a11(t) x1 + g1(t).
Example
n = 2: 2× 2 linear system: Find x1(t) and x2(t) solutions of
x ′1 = a11(t) x1 + a12(t) x2 + g1(t),
x ′2 = a21(t) x1 + a22(t) x2 + g2(t).
Example
n = 2: 2× 2 homogeneous linear system: Find x1(t) and x2(t),
x ′1 = a11(t) x1 + a12(t) x2
x ′2 = a21(t) x1 + a22(t) x2.
n × n systems of linear differential equations.
Example
n = 1: Single differential equation: Find x1(t) solution of
x ′1 = a11(t) x1 + g1(t).
Example
n = 2: 2× 2 linear system: Find x1(t) and x2(t) solutions of
x ′1 = a11(t) x1 + a12(t) x2 + g1(t),
x ′2 = a21(t) x1 + a22(t) x2 + g2(t).
Example
n = 2: 2× 2 homogeneous linear system: Find x1(t) and x2(t),
x ′1 = a11(t) x1 + a12(t) x2
x ′2 = a21(t) x1 + a22(t) x2.
n × n systems of linear differential equations.
Example
n = 1: Single differential equation: Find x1(t) solution of
x ′1 = a11(t) x1 + g1(t).
Example
n = 2: 2× 2 linear system: Find x1(t) and x2(t) solutions of
x ′1 = a11(t) x1 + a12(t) x2 + g1(t),
x ′2 = a21(t) x1 + a22(t) x2 + g2(t).
Example
n = 2: 2× 2 homogeneous linear system: Find x1(t) and x2(t),
x ′1 = a11(t) x1 + a12(t) x2
x ′2 = a21(t) x1 + a22(t) x2.
n × n systems of linear differential equations.
Example
Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system
x ′1 = x1 − x2,
x ′2 = −x1 + x2.
Solution: Add up the equations, and subtract the equations,
(x1 + x2)′ = 0, (x1 − x2)
′ = 2(x1 − x2).
Introduce the unknowns v = x1 + x2, w = x1 − x2, then
v ′ = 0 ⇒ v = c1,
w ′ = 2w ⇒ w = c2e2t .
Back to x1 and x2: x1 =1
2(v + w), x2 =
1
2(v − w).
We conclude: x1(t) =1
2(c1 + c2e
2t), x2(t) =1
2(c1 − c2e
2t).C
n × n systems of linear differential equations.
Example
Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system
x ′1 = x1 − x2,
x ′2 = −x1 + x2.
Solution: Add up the equations, and subtract the equations,
(x1 + x2)′ = 0, (x1 − x2)
′ = 2(x1 − x2).
Introduce the unknowns v = x1 + x2, w = x1 − x2, then
v ′ = 0 ⇒ v = c1,
w ′ = 2w ⇒ w = c2e2t .
Back to x1 and x2: x1 =1
2(v + w), x2 =
1
2(v − w).
We conclude: x1(t) =1
2(c1 + c2e
2t), x2(t) =1
2(c1 − c2e
2t).C
n × n systems of linear differential equations.
Example
Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system
x ′1 = x1 − x2,
x ′2 = −x1 + x2.
Solution: Add up the equations, and subtract the equations,
(x1 + x2)′ = 0,
(x1 − x2)′ = 2(x1 − x2).
Introduce the unknowns v = x1 + x2, w = x1 − x2, then
v ′ = 0 ⇒ v = c1,
w ′ = 2w ⇒ w = c2e2t .
Back to x1 and x2: x1 =1
2(v + w), x2 =
1
2(v − w).
We conclude: x1(t) =1
2(c1 + c2e
2t), x2(t) =1
2(c1 − c2e
2t).C
n × n systems of linear differential equations.
Example
Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system
x ′1 = x1 − x2,
x ′2 = −x1 + x2.
Solution: Add up the equations, and subtract the equations,
(x1 + x2)′ = 0, (x1 − x2)
′ = 2(x1 − x2).
Introduce the unknowns v = x1 + x2, w = x1 − x2, then
v ′ = 0 ⇒ v = c1,
w ′ = 2w ⇒ w = c2e2t .
Back to x1 and x2: x1 =1
2(v + w), x2 =
1
2(v − w).
We conclude: x1(t) =1
2(c1 + c2e
2t), x2(t) =1
2(c1 − c2e
2t).C
n × n systems of linear differential equations.
Example
Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system
x ′1 = x1 − x2,
x ′2 = −x1 + x2.
Solution: Add up the equations, and subtract the equations,
(x1 + x2)′ = 0, (x1 − x2)
′ = 2(x1 − x2).
Introduce the unknowns v = x1 + x2,
w = x1 − x2, then
v ′ = 0 ⇒ v = c1,
w ′ = 2w ⇒ w = c2e2t .
Back to x1 and x2: x1 =1
2(v + w), x2 =
1
2(v − w).
We conclude: x1(t) =1
2(c1 + c2e
2t), x2(t) =1
2(c1 − c2e
2t).C
n × n systems of linear differential equations.
Example
Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system
x ′1 = x1 − x2,
x ′2 = −x1 + x2.
Solution: Add up the equations, and subtract the equations,
(x1 + x2)′ = 0, (x1 − x2)
′ = 2(x1 − x2).
Introduce the unknowns v = x1 + x2, w = x1 − x2,
then
v ′ = 0 ⇒ v = c1,
w ′ = 2w ⇒ w = c2e2t .
Back to x1 and x2: x1 =1
2(v + w), x2 =
1
2(v − w).
We conclude: x1(t) =1
2(c1 + c2e
2t), x2(t) =1
2(c1 − c2e
2t).C
n × n systems of linear differential equations.
Example
Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system
x ′1 = x1 − x2,
x ′2 = −x1 + x2.
Solution: Add up the equations, and subtract the equations,
(x1 + x2)′ = 0, (x1 − x2)
′ = 2(x1 − x2).
Introduce the unknowns v = x1 + x2, w = x1 − x2, then
v ′ = 0
⇒ v = c1,
w ′ = 2w ⇒ w = c2e2t .
Back to x1 and x2: x1 =1
2(v + w), x2 =
1
2(v − w).
We conclude: x1(t) =1
2(c1 + c2e
2t), x2(t) =1
2(c1 − c2e
2t).C
n × n systems of linear differential equations.
Example
Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system
x ′1 = x1 − x2,
x ′2 = −x1 + x2.
Solution: Add up the equations, and subtract the equations,
(x1 + x2)′ = 0, (x1 − x2)
′ = 2(x1 − x2).
Introduce the unknowns v = x1 + x2, w = x1 − x2, then
v ′ = 0 ⇒ v = c1,
w ′ = 2w ⇒ w = c2e2t .
Back to x1 and x2: x1 =1
2(v + w), x2 =
1
2(v − w).
We conclude: x1(t) =1
2(c1 + c2e
2t), x2(t) =1
2(c1 − c2e
2t).C
n × n systems of linear differential equations.
Example
Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system
x ′1 = x1 − x2,
x ′2 = −x1 + x2.
Solution: Add up the equations, and subtract the equations,
(x1 + x2)′ = 0, (x1 − x2)
′ = 2(x1 − x2).
Introduce the unknowns v = x1 + x2, w = x1 − x2, then
v ′ = 0 ⇒ v = c1,
w ′ = 2w
⇒ w = c2e2t .
Back to x1 and x2: x1 =1
2(v + w), x2 =
1
2(v − w).
We conclude: x1(t) =1
2(c1 + c2e
2t), x2(t) =1
2(c1 − c2e
2t).C
n × n systems of linear differential equations.
Example
Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system
x ′1 = x1 − x2,
x ′2 = −x1 + x2.
Solution: Add up the equations, and subtract the equations,
(x1 + x2)′ = 0, (x1 − x2)
′ = 2(x1 − x2).
Introduce the unknowns v = x1 + x2, w = x1 − x2, then
v ′ = 0 ⇒ v = c1,
w ′ = 2w ⇒ w = c2e2t .
Back to x1 and x2: x1 =1
2(v + w), x2 =
1
2(v − w).
We conclude: x1(t) =1
2(c1 + c2e
2t), x2(t) =1
2(c1 − c2e
2t).C
n × n systems of linear differential equations.
Example
Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system
x ′1 = x1 − x2,
x ′2 = −x1 + x2.
Solution: Add up the equations, and subtract the equations,
(x1 + x2)′ = 0, (x1 − x2)
′ = 2(x1 − x2).
Introduce the unknowns v = x1 + x2, w = x1 − x2, then
v ′ = 0 ⇒ v = c1,
w ′ = 2w ⇒ w = c2e2t .
Back to x1 and x2:
x1 =1
2(v + w), x2 =
1
2(v − w).
We conclude: x1(t) =1
2(c1 + c2e
2t), x2(t) =1
2(c1 − c2e
2t).C
n × n systems of linear differential equations.
Example
Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system
x ′1 = x1 − x2,
x ′2 = −x1 + x2.
Solution: Add up the equations, and subtract the equations,
(x1 + x2)′ = 0, (x1 − x2)
′ = 2(x1 − x2).
Introduce the unknowns v = x1 + x2, w = x1 − x2, then
v ′ = 0 ⇒ v = c1,
w ′ = 2w ⇒ w = c2e2t .
Back to x1 and x2: x1 =1
2(v + w),
x2 =1
2(v − w).
We conclude: x1(t) =1
2(c1 + c2e
2t), x2(t) =1
2(c1 − c2e
2t).C
n × n systems of linear differential equations.
Example
Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system
x ′1 = x1 − x2,
x ′2 = −x1 + x2.
Solution: Add up the equations, and subtract the equations,
(x1 + x2)′ = 0, (x1 − x2)
′ = 2(x1 − x2).
Introduce the unknowns v = x1 + x2, w = x1 − x2, then
v ′ = 0 ⇒ v = c1,
w ′ = 2w ⇒ w = c2e2t .
Back to x1 and x2: x1 =1
2(v + w), x2 =
1
2(v − w).
We conclude: x1(t) =1
2(c1 + c2e
2t), x2(t) =1
2(c1 − c2e
2t).C
n × n systems of linear differential equations.
Example
Find x1(t), x2(t) solutions of the 2× 2,constant coefficients, homogeneous system
x ′1 = x1 − x2,
x ′2 = −x1 + x2.
Solution: Add up the equations, and subtract the equations,
(x1 + x2)′ = 0, (x1 − x2)
′ = 2(x1 − x2).
Introduce the unknowns v = x1 + x2, w = x1 − x2, then
v ′ = 0 ⇒ v = c1,
w ′ = 2w ⇒ w = c2e2t .
Back to x1 and x2: x1 =1
2(v + w), x2 =
1
2(v − w).
We conclude: x1(t) =1
2(c1 + c2e
2t), x2(t) =1
2(c1 − c2e
2t).C
Systems of linear differential equations (Sect. 7.1).
I n × n systems of linear differential equations.
I Second order equations and first order systems.
I Main concepts from Linear Algebra.
Second order equations and first order systems.
Theorem (Reduction to first order)
Every solution y to the second order linear equation
y ′′ + p(t) y ′ + q(t) y = g(t), (1)
defines a solution x1 = y and x2 = y ′ of the 2× 2 first order lineardifferential system
x ′1 = x2, (2)
x ′2 = −q(t) x1 − p(t) x2 + g(t). (3)
Conversely, every solution x1, x2 of the 2× 2 first order linearsystem in Eqs. (2)-(3) defines a solution y = x1 of the second orderdifferential equation in (1).
Second order equations and first order systems.
Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),
introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2, that is,
x ′1 = x2.
Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t). That is,
x ′2 = −q(t) x1 − p(t) x2 + g(t).
(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).
x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),
that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).
Second order equations and first order systems.
Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),
introduce x1 = y and x2 = y ′,
hence x ′1 = y ′ = x2, that is,
x ′1 = x2.
Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t). That is,
x ′2 = −q(t) x1 − p(t) x2 + g(t).
(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).
x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),
that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).
Second order equations and first order systems.
Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),
introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2,
that is,
x ′1 = x2.
Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t). That is,
x ′2 = −q(t) x1 − p(t) x2 + g(t).
(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).
x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),
that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).
Second order equations and first order systems.
Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),
introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2, that is,
x ′1 = x2.
Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t). That is,
x ′2 = −q(t) x1 − p(t) x2 + g(t).
(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).
x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),
that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).
Second order equations and first order systems.
Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),
introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2, that is,
x ′1 = x2.
Then, x ′2 = y ′′
= −q(t) y − p(t) y ′ + g(t). That is,
x ′2 = −q(t) x1 − p(t) x2 + g(t).
(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).
x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),
that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).
Second order equations and first order systems.
Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),
introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2, that is,
x ′1 = x2.
Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t).
That is,
x ′2 = −q(t) x1 − p(t) x2 + g(t).
(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).
x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),
that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).
Second order equations and first order systems.
Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),
introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2, that is,
x ′1 = x2.
Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t). That is,
x ′2 = −q(t) x1 − p(t) x2 + g(t).
(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).
x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),
that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).
Second order equations and first order systems.
Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),
introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2, that is,
x ′1 = x2.
Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t). That is,
x ′2 = −q(t) x1 − p(t) x2 + g(t).
(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).
x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),
that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).
Second order equations and first order systems.
Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),
introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2, that is,
x ′1 = x2.
Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t). That is,
x ′2 = −q(t) x1 − p(t) x2 + g(t).
(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).
x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),
that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).
Second order equations and first order systems.
Proof:(⇒) Given y solution of y ′′ + p(t) y ′ + q(t) y = g(t),
introduce x1 = y and x2 = y ′, hence x ′1 = y ′ = x2, that is,
x ′1 = x2.
Then, x ′2 = y ′′ = −q(t) y − p(t) y ′ + g(t). That is,
x ′2 = −q(t) x1 − p(t) x2 + g(t).
(⇐) Introduce x2 = x ′1 into x ′2 = −q(t) x1 − p(t) x2 + g(t).
x ′′1 = −q(t) x1 − p(t) x ′1 + g(t),
that isx ′′1 + p(t) x ′1 + q(t) x1 = g(t).
Second order equations and first order systems.
Example
Express as a first order system the equation
y ′′ + 2y ′ + 2y = sin(at).
Solution: Introduce the new unknowns
x1 = y , x2 = y ′ ⇒ x ′1 = x2.
Then, the differential equation can be written as
x ′2 + 2x2 + 2x1 = sin(at).
We conclude thatx ′1 = x2.
x ′2 = −2x1 − 2x2 + sin(at). C
Second order equations and first order systems.
Example
Express as a first order system the equation
y ′′ + 2y ′ + 2y = sin(at).
Solution: Introduce the new unknowns
x1 = y , x2 = y ′
⇒ x ′1 = x2.
Then, the differential equation can be written as
x ′2 + 2x2 + 2x1 = sin(at).
We conclude thatx ′1 = x2.
x ′2 = −2x1 − 2x2 + sin(at). C
Second order equations and first order systems.
Example
Express as a first order system the equation
y ′′ + 2y ′ + 2y = sin(at).
Solution: Introduce the new unknowns
x1 = y , x2 = y ′ ⇒ x ′1 = x2.
Then, the differential equation can be written as
x ′2 + 2x2 + 2x1 = sin(at).
We conclude thatx ′1 = x2.
x ′2 = −2x1 − 2x2 + sin(at). C
Second order equations and first order systems.
Example
Express as a first order system the equation
y ′′ + 2y ′ + 2y = sin(at).
Solution: Introduce the new unknowns
x1 = y , x2 = y ′ ⇒ x ′1 = x2.
Then, the differential equation can be written as
x ′2 + 2x2 + 2x1 = sin(at).
We conclude thatx ′1 = x2.
x ′2 = −2x1 − 2x2 + sin(at). C
Second order equations and first order systems.
Example
Express as a first order system the equation
y ′′ + 2y ′ + 2y = sin(at).
Solution: Introduce the new unknowns
x1 = y , x2 = y ′ ⇒ x ′1 = x2.
Then, the differential equation can be written as
x ′2 + 2x2 + 2x1 = sin(at).
We conclude thatx ′1 = x2.
x ′2 = −2x1 − 2x2 + sin(at). C
Second order equations and first order systems.
Remark: Systems of first order equations can, sometimes, betransformed into a second order single equation.
Example
Express as a single second order equationthe 2× 2 system and solve it,
x ′1 = −x1 + 3x2,
x ′2 = x1 − x2.
Solution: Compute x1 from the second equation: x1 = x ′2 + x2.Introduce this expression into the first equation,
(x ′2 + x2)′ = −(x ′2 + x2) + 3x2,
x ′′2 + x ′2 = −x ′2 − x2 + 3x2,
x ′′2 + 2x ′2 − 2x2 = 0.
Second order equations and first order systems.
Remark: Systems of first order equations can, sometimes, betransformed into a second order single equation.
Example
Express as a single second order equationthe 2× 2 system and solve it,
x ′1 = −x1 + 3x2,
x ′2 = x1 − x2.
Solution: Compute x1 from the second equation: x1 = x ′2 + x2.Introduce this expression into the first equation,
(x ′2 + x2)′ = −(x ′2 + x2) + 3x2,
x ′′2 + x ′2 = −x ′2 − x2 + 3x2,
x ′′2 + 2x ′2 − 2x2 = 0.
Second order equations and first order systems.
Remark: Systems of first order equations can, sometimes, betransformed into a second order single equation.
Example
Express as a single second order equationthe 2× 2 system and solve it,
x ′1 = −x1 + 3x2,
x ′2 = x1 − x2.
Solution: Compute x1 from the second equation:
x1 = x ′2 + x2.Introduce this expression into the first equation,
(x ′2 + x2)′ = −(x ′2 + x2) + 3x2,
x ′′2 + x ′2 = −x ′2 − x2 + 3x2,
x ′′2 + 2x ′2 − 2x2 = 0.
Second order equations and first order systems.
Remark: Systems of first order equations can, sometimes, betransformed into a second order single equation.
Example
Express as a single second order equationthe 2× 2 system and solve it,
x ′1 = −x1 + 3x2,
x ′2 = x1 − x2.
Solution: Compute x1 from the second equation: x1 = x ′2 + x2.
Introduce this expression into the first equation,
(x ′2 + x2)′ = −(x ′2 + x2) + 3x2,
x ′′2 + x ′2 = −x ′2 − x2 + 3x2,
x ′′2 + 2x ′2 − 2x2 = 0.
Second order equations and first order systems.
Remark: Systems of first order equations can, sometimes, betransformed into a second order single equation.
Example
Express as a single second order equationthe 2× 2 system and solve it,
x ′1 = −x1 + 3x2,
x ′2 = x1 − x2.
Solution: Compute x1 from the second equation: x1 = x ′2 + x2.Introduce this expression into the first equation,
(x ′2 + x2)′ = −(x ′2 + x2) + 3x2,
x ′′2 + x ′2 = −x ′2 − x2 + 3x2,
x ′′2 + 2x ′2 − 2x2 = 0.
Second order equations and first order systems.
Remark: Systems of first order equations can, sometimes, betransformed into a second order single equation.
Example
Express as a single second order equationthe 2× 2 system and solve it,
x ′1 = −x1 + 3x2,
x ′2 = x1 − x2.
Solution: Compute x1 from the second equation: x1 = x ′2 + x2.Introduce this expression into the first equation,
(x ′2 + x2)′ = −(x ′2 + x2) + 3x2,
x ′′2 + x ′2 = −x ′2 − x2 + 3x2,
x ′′2 + 2x ′2 − 2x2 = 0.
Second order equations and first order systems.
Remark: Systems of first order equations can, sometimes, betransformed into a second order single equation.
Example
Express as a single second order equationthe 2× 2 system and solve it,
x ′1 = −x1 + 3x2,
x ′2 = x1 − x2.
Solution: Compute x1 from the second equation: x1 = x ′2 + x2.Introduce this expression into the first equation,
(x ′2 + x2)′ = −(x ′2 + x2) + 3x2,
x ′′2 + x ′2 = −x ′2 − x2 + 3x2,
x ′′2 + 2x ′2 − 2x2 = 0.
Second order equations and first order systems.
Remark: Systems of first order equations can, sometimes, betransformed into a second order single equation.
Example
Express as a single second order equationthe 2× 2 system and solve it,
x ′1 = −x1 + 3x2,
x ′2 = x1 − x2.
Solution: Compute x1 from the second equation: x1 = x ′2 + x2.Introduce this expression into the first equation,
(x ′2 + x2)′ = −(x ′2 + x2) + 3x2,
x ′′2 + x ′2 = −x ′2 − x2 + 3x2,
x ′′2 + 2x ′2 − 2x2 = 0.
Second order equations and first order systems.
Example
Express as a single second order equationthe 2× 2 system and solve it,
x ′1 = −x1 + 3x2,
x ′2 = x1 − x2.
Solution: Recall: x ′′2 + 2x ′2 − 2x2 = 0.
r2+2r−2 = 0 ⇒ r± =1
2
[−2±
√4 + 8
]⇒ r± = −1±
√3.
Therefore, x2 = c1 er+ t + c2 er− t . Since x1 = x ′2 + x2,
x1 =(c1r+ er+ t + c2r− er− t
)+
(c1 er+ t + c2 er− t
),
We conclude: x1 = c1(1 + r+) er+ t + c2(1 + r−) er− t . C
Second order equations and first order systems.
Example
Express as a single second order equationthe 2× 2 system and solve it,
x ′1 = −x1 + 3x2,
x ′2 = x1 − x2.
Solution: Recall: x ′′2 + 2x ′2 − 2x2 = 0.
r2+2r−2 = 0
⇒ r± =1
2
[−2±
√4 + 8
]⇒ r± = −1±
√3.
Therefore, x2 = c1 er+ t + c2 er− t . Since x1 = x ′2 + x2,
x1 =(c1r+ er+ t + c2r− er− t
)+
(c1 er+ t + c2 er− t
),
We conclude: x1 = c1(1 + r+) er+ t + c2(1 + r−) er− t . C
Second order equations and first order systems.
Example
Express as a single second order equationthe 2× 2 system and solve it,
x ′1 = −x1 + 3x2,
x ′2 = x1 − x2.
Solution: Recall: x ′′2 + 2x ′2 − 2x2 = 0.
r2+2r−2 = 0 ⇒ r± =1
2
[−2±
√4 + 8
]
⇒ r± = −1±√
3.
Therefore, x2 = c1 er+ t + c2 er− t . Since x1 = x ′2 + x2,
x1 =(c1r+ er+ t + c2r− er− t
)+
(c1 er+ t + c2 er− t
),
We conclude: x1 = c1(1 + r+) er+ t + c2(1 + r−) er− t . C
Second order equations and first order systems.
Example
Express as a single second order equationthe 2× 2 system and solve it,
x ′1 = −x1 + 3x2,
x ′2 = x1 − x2.
Solution: Recall: x ′′2 + 2x ′2 − 2x2 = 0.
r2+2r−2 = 0 ⇒ r± =1
2
[−2±
√4 + 8
]⇒ r± = −1±
√3.
Therefore, x2 = c1 er+ t + c2 er− t . Since x1 = x ′2 + x2,
x1 =(c1r+ er+ t + c2r− er− t
)+
(c1 er+ t + c2 er− t
),
We conclude: x1 = c1(1 + r+) er+ t + c2(1 + r−) er− t . C
Second order equations and first order systems.
Example
Express as a single second order equationthe 2× 2 system and solve it,
x ′1 = −x1 + 3x2,
x ′2 = x1 − x2.
Solution: Recall: x ′′2 + 2x ′2 − 2x2 = 0.
r2+2r−2 = 0 ⇒ r± =1
2
[−2±
√4 + 8
]⇒ r± = −1±
√3.
Therefore, x2 = c1 er+ t + c2 er− t .
Since x1 = x ′2 + x2,
x1 =(c1r+ er+ t + c2r− er− t
)+
(c1 er+ t + c2 er− t
),
We conclude: x1 = c1(1 + r+) er+ t + c2(1 + r−) er− t . C
Second order equations and first order systems.
Example
Express as a single second order equationthe 2× 2 system and solve it,
x ′1 = −x1 + 3x2,
x ′2 = x1 − x2.
Solution: Recall: x ′′2 + 2x ′2 − 2x2 = 0.
r2+2r−2 = 0 ⇒ r± =1
2
[−2±
√4 + 8
]⇒ r± = −1±
√3.
Therefore, x2 = c1 er+ t + c2 er− t . Since x1 = x ′2 + x2,
x1 =(c1r+ er+ t + c2r− er− t
)+
(c1 er+ t + c2 er− t
),
We conclude: x1 = c1(1 + r+) er+ t + c2(1 + r−) er− t . C
Second order equations and first order systems.
Example
Express as a single second order equationthe 2× 2 system and solve it,
x ′1 = −x1 + 3x2,
x ′2 = x1 − x2.
Solution: Recall: x ′′2 + 2x ′2 − 2x2 = 0.
r2+2r−2 = 0 ⇒ r± =1
2
[−2±
√4 + 8
]⇒ r± = −1±
√3.
Therefore, x2 = c1 er+ t + c2 er− t . Since x1 = x ′2 + x2,
x1 =(c1r+ er+ t + c2r− er− t
)+
(c1 er+ t + c2 er− t
),
We conclude: x1 = c1(1 + r+) er+ t + c2(1 + r−) er− t . C
Second order equations and first order systems.
Example
Express as a single second order equationthe 2× 2 system and solve it,
x ′1 = −x1 + 3x2,
x ′2 = x1 − x2.
Solution: Recall: x ′′2 + 2x ′2 − 2x2 = 0.
r2+2r−2 = 0 ⇒ r± =1
2
[−2±
√4 + 8
]⇒ r± = −1±
√3.
Therefore, x2 = c1 er+ t + c2 er− t . Since x1 = x ′2 + x2,
x1 =(c1r+ er+ t + c2r− er− t
)+
(c1 er+ t + c2 er− t
),
We conclude: x1 = c1(1 + r+) er+ t + c2(1 + r−) er− t . C
Systems of linear differential equations (Sect. 7.1).
I n × n systems of linear differential equations.
I Second order equations and first order systems.
I Main concepts from Linear Algebra.
Main concepts from Linear Algebra.
Remark: Ideas from LinearAlgebra are useful to studysystems of linear differentialequations.
We review:
I Matrices m × n.
I Matrix operations.
I n-vectors, dot product.
I matrix-vector product.
DefinitionAn m × n matrix, A, is an array of numbers
A =
a11 · · · a1n...
...am1 · · · amn
,m rows,
n columns.
where aij ∈ C and i = 1, · · · ,m, and j = 1, · · · , n. An n × nmatrix is called a square matrix.
Main concepts from Linear Algebra.
Remark: Ideas from LinearAlgebra are useful to studysystems of linear differentialequations.
We review:
I Matrices m × n.
I Matrix operations.
I n-vectors, dot product.
I matrix-vector product.
DefinitionAn m × n matrix, A, is an array of numbers
A =
a11 · · · a1n...
...am1 · · · amn
,m rows,
n columns.
where aij ∈ C and i = 1, · · · ,m, and j = 1, · · · , n. An n × nmatrix is called a square matrix.
Main concepts from Linear Algebra.
Remark: Ideas from LinearAlgebra are useful to studysystems of linear differentialequations.
We review:
I Matrices m × n.
I Matrix operations.
I n-vectors, dot product.
I matrix-vector product.
DefinitionAn m × n matrix, A, is an array of numbers
A =
a11 · · · a1n...
...am1 · · · amn
,m rows,
n columns.
where aij ∈ C and i = 1, · · · ,m, and j = 1, · · · , n. An n × nmatrix is called a square matrix.
Main concepts from Linear Algebra.
Example
(a) 2× 2 matrix: A =
[1 23 4
].
(b) 2× 3 matrix: A =
[1 2 34 5 6
].
(c) 3× 2 matrix: A =
1 23 45 6
.
(d) 2× 2 complex-valued matrix: A =
[1 + i 2− i
3 4i
].
(e) The coefficients of a linear system can be grouped in a matrix,
x ′1 = −x1 + 3x2
x ′2 = x1 − x2
}⇒ A =
[−1 31 −1
].
Main concepts from Linear Algebra.
Example
(a) 2× 2 matrix: A =
[1 23 4
].
(b) 2× 3 matrix: A =
[1 2 34 5 6
].
(c) 3× 2 matrix: A =
1 23 45 6
.
(d) 2× 2 complex-valued matrix: A =
[1 + i 2− i
3 4i
].
(e) The coefficients of a linear system can be grouped in a matrix,
x ′1 = −x1 + 3x2
x ′2 = x1 − x2
}⇒ A =
[−1 31 −1
].
Main concepts from Linear Algebra.
Example
(a) 2× 2 matrix: A =
[1 23 4
].
(b) 2× 3 matrix: A =
[1 2 34 5 6
].
(c) 3× 2 matrix: A =
1 23 45 6
.
(d) 2× 2 complex-valued matrix: A =
[1 + i 2− i
3 4i
].
(e) The coefficients of a linear system can be grouped in a matrix,
x ′1 = −x1 + 3x2
x ′2 = x1 − x2
}⇒ A =
[−1 31 −1
].
Main concepts from Linear Algebra.
Example
(a) 2× 2 matrix: A =
[1 23 4
].
(b) 2× 3 matrix: A =
[1 2 34 5 6
].
(c) 3× 2 matrix: A =
1 23 45 6
.
(d) 2× 2 complex-valued matrix: A =
[1 + i 2− i
3 4i
].
(e) The coefficients of a linear system can be grouped in a matrix,
x ′1 = −x1 + 3x2
x ′2 = x1 − x2
}⇒ A =
[−1 31 −1
].
Main concepts from Linear Algebra.
Example
(a) 2× 2 matrix: A =
[1 23 4
].
(b) 2× 3 matrix: A =
[1 2 34 5 6
].
(c) 3× 2 matrix: A =
1 23 45 6
.
(d) 2× 2 complex-valued matrix: A =
[1 + i 2− i
3 4i
].
(e) The coefficients of a linear system can be grouped in a matrix,
x ′1 = −x1 + 3x2
x ′2 = x1 − x2
}⇒ A =
[−1 31 −1
].
Main concepts from Linear Algebra.
Example
(a) 2× 2 matrix: A =
[1 23 4
].
(b) 2× 3 matrix: A =
[1 2 34 5 6
].
(c) 3× 2 matrix: A =
1 23 45 6
.
(d) 2× 2 complex-valued matrix: A =
[1 + i 2− i
3 4i
].
(e) The coefficients of a linear system can be grouped in a matrix,
x ′1 = −x1 + 3x2
x ′2 = x1 − x2
}⇒ A =
[−1 31 −1
].
Main concepts from Linear Algebra.
Remark: An m × 1 matrix is called an m-vector.
Definition
An m-vector, v, is the array of numbers v =
v1...
vm
, where the
vector components vi ∈ C, with i = 1, · · · ,m.
Example
The unknowns of a 2× 2 linear system can be grouped in a2-vector, for example,
x ′1 = −x1 + 3x2
x ′2 = x1 − x2
}⇒ x =
[x1
x2
].
Main concepts from Linear Algebra.
Remark: An m × 1 matrix is called an m-vector.
Definition
An m-vector, v, is the array of numbers v =
v1...
vm
, where the
vector components vi ∈ C, with i = 1, · · · ,m.
Example
The unknowns of a 2× 2 linear system can be grouped in a2-vector, for example,
x ′1 = −x1 + 3x2
x ′2 = x1 − x2
}⇒ x =
[x1
x2
].
Main concepts from Linear Algebra.
Remark: An m × 1 matrix is called an m-vector.
Definition
An m-vector, v, is the array of numbers v =
v1...
vm
, where the
vector components vi ∈ C, with i = 1, · · · ,m.
Example
The unknowns of a 2× 2 linear system can be grouped in a2-vector,
for example,
x ′1 = −x1 + 3x2
x ′2 = x1 − x2
}⇒ x =
[x1
x2
].
Main concepts from Linear Algebra.
Remark: An m × 1 matrix is called an m-vector.
Definition
An m-vector, v, is the array of numbers v =
v1...
vm
, where the
vector components vi ∈ C, with i = 1, · · · ,m.
Example
The unknowns of a 2× 2 linear system can be grouped in a2-vector, for example,
x ′1 = −x1 + 3x2
x ′2 = x1 − x2
}⇒ x =
[x1
x2
].
Main concepts from Linear Algebra.
Remark: An m × 1 matrix is called an m-vector.
Definition
An m-vector, v, is the array of numbers v =
v1...
vm
, where the
vector components vi ∈ C, with i = 1, · · · ,m.
Example
The unknowns of a 2× 2 linear system can be grouped in a2-vector, for example,
x ′1 = −x1 + 3x2
x ′2 = x1 − x2
}⇒ x =
[x1
x2
].
Main concepts from Linear Algebra.
Remark: We present only examples of matrix operations.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-transpose: Interchange rows with columns:
AT =
1 3i2 + i 2−1 + 2i 1
. Notice that:(AT
)T= A.
(b) A-conjugate: Conjugate every matrix coefficient:
A =
[1 2− i −1− 2i−3i 2 1
]. Notice that:
(A
)= A.
Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.
Main concepts from Linear Algebra.
Remark: We present only examples of matrix operations.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-transpose: Interchange rows with columns:
AT =
1 3i2 + i 2−1 + 2i 1
. Notice that:(AT
)T= A.
(b) A-conjugate: Conjugate every matrix coefficient:
A =
[1 2− i −1− 2i−3i 2 1
]. Notice that:
(A
)= A.
Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.
Main concepts from Linear Algebra.
Remark: We present only examples of matrix operations.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-transpose: Interchange rows with columns:
AT =
1 3i2 + i 2−1 + 2i 1
. Notice that:(AT
)T= A.
(b) A-conjugate: Conjugate every matrix coefficient:
A =
[1 2− i −1− 2i−3i 2 1
]. Notice that:
(A
)= A.
Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.
Main concepts from Linear Algebra.
Remark: We present only examples of matrix operations.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-transpose: Interchange rows with columns:
AT =
1 3i2 + i 2−1 + 2i 1
.
Notice that:(AT
)T= A.
(b) A-conjugate: Conjugate every matrix coefficient:
A =
[1 2− i −1− 2i−3i 2 1
]. Notice that:
(A
)= A.
Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.
Main concepts from Linear Algebra.
Remark: We present only examples of matrix operations.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-transpose: Interchange rows with columns:
AT =
1 3i2 + i 2−1 + 2i 1
. Notice that:(AT
)T= A.
(b) A-conjugate: Conjugate every matrix coefficient:
A =
[1 2− i −1− 2i−3i 2 1
]. Notice that:
(A
)= A.
Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.
Main concepts from Linear Algebra.
Remark: We present only examples of matrix operations.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-transpose: Interchange rows with columns:
AT =
1 3i2 + i 2−1 + 2i 1
. Notice that:(AT
)T= A.
(b) A-conjugate: Conjugate every matrix coefficient:
A =
[1 2− i −1− 2i−3i 2 1
]. Notice that:
(A
)= A.
Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.
Main concepts from Linear Algebra.
Remark: We present only examples of matrix operations.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-transpose: Interchange rows with columns:
AT =
1 3i2 + i 2−1 + 2i 1
. Notice that:(AT
)T= A.
(b) A-conjugate: Conjugate every matrix coefficient:
A =
[1 2− i −1− 2i−3i 2 1
].
Notice that:(A
)= A.
Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.
Main concepts from Linear Algebra.
Remark: We present only examples of matrix operations.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-transpose: Interchange rows with columns:
AT =
1 3i2 + i 2−1 + 2i 1
. Notice that:(AT
)T= A.
(b) A-conjugate: Conjugate every matrix coefficient:
A =
[1 2− i −1− 2i−3i 2 1
]. Notice that:
(A
)= A.
Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.
Main concepts from Linear Algebra.
Remark: We present only examples of matrix operations.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-transpose: Interchange rows with columns:
AT =
1 3i2 + i 2−1 + 2i 1
. Notice that:(AT
)T= A.
(b) A-conjugate: Conjugate every matrix coefficient:
A =
[1 2− i −1− 2i−3i 2 1
]. Notice that:
(A
)= A.
Matrix A is real iff A = A.
Matrix A is imaginary iff A = −A.
Main concepts from Linear Algebra.
Remark: We present only examples of matrix operations.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-transpose: Interchange rows with columns:
AT =
1 3i2 + i 2−1 + 2i 1
. Notice that:(AT
)T= A.
(b) A-conjugate: Conjugate every matrix coefficient:
A =
[1 2− i −1− 2i−3i 2 1
]. Notice that:
(A
)= A.
Matrix A is real iff A = A. Matrix A is imaginary iff A = −A.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-adjoint: Conjugate and transpose:
A∗ =
1 −3i2− i 2−1− 2i 1
. Notice that:(A∗
)∗= A.
(b) Addition of two m × n matrices is performed component-wise:[1 23 4
]+
[2 35 1
]=
[(1 + 2) (2 + 3)(3 + 5) (4 + 1)
]=
[3 58 5
].
The addition
[1 23 4
]+
[1 2 34 5 6
]is not defined.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-adjoint: Conjugate and transpose:
A∗ =
1 −3i2− i 2−1− 2i 1
. Notice that:(A∗
)∗= A.
(b) Addition of two m × n matrices is performed component-wise:[1 23 4
]+
[2 35 1
]=
[(1 + 2) (2 + 3)(3 + 5) (4 + 1)
]=
[3 58 5
].
The addition
[1 23 4
]+
[1 2 34 5 6
]is not defined.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-adjoint: Conjugate and transpose:
A∗ =
1 −3i2− i 2−1− 2i 1
.
Notice that:(A∗
)∗= A.
(b) Addition of two m × n matrices is performed component-wise:[1 23 4
]+
[2 35 1
]=
[(1 + 2) (2 + 3)(3 + 5) (4 + 1)
]=
[3 58 5
].
The addition
[1 23 4
]+
[1 2 34 5 6
]is not defined.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-adjoint: Conjugate and transpose:
A∗ =
1 −3i2− i 2−1− 2i 1
. Notice that:(A∗
)∗= A.
(b) Addition of two m × n matrices is performed component-wise:[1 23 4
]+
[2 35 1
]=
[(1 + 2) (2 + 3)(3 + 5) (4 + 1)
]=
[3 58 5
].
The addition
[1 23 4
]+
[1 2 34 5 6
]is not defined.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-adjoint: Conjugate and transpose:
A∗ =
1 −3i2− i 2−1− 2i 1
. Notice that:(A∗
)∗= A.
(b) Addition of two m × n matrices is performed component-wise:
[1 23 4
]+
[2 35 1
]=
[(1 + 2) (2 + 3)(3 + 5) (4 + 1)
]=
[3 58 5
].
The addition
[1 23 4
]+
[1 2 34 5 6
]is not defined.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-adjoint: Conjugate and transpose:
A∗ =
1 −3i2− i 2−1− 2i 1
. Notice that:(A∗
)∗= A.
(b) Addition of two m × n matrices is performed component-wise:[1 23 4
]+
[2 35 1
]
=
[(1 + 2) (2 + 3)(3 + 5) (4 + 1)
]=
[3 58 5
].
The addition
[1 23 4
]+
[1 2 34 5 6
]is not defined.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-adjoint: Conjugate and transpose:
A∗ =
1 −3i2− i 2−1− 2i 1
. Notice that:(A∗
)∗= A.
(b) Addition of two m × n matrices is performed component-wise:[1 23 4
]+
[2 35 1
]=
[(1 + 2) (2 + 3)(3 + 5) (4 + 1)
]
=
[3 58 5
].
The addition
[1 23 4
]+
[1 2 34 5 6
]is not defined.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-adjoint: Conjugate and transpose:
A∗ =
1 −3i2− i 2−1− 2i 1
. Notice that:(A∗
)∗= A.
(b) Addition of two m × n matrices is performed component-wise:[1 23 4
]+
[2 35 1
]=
[(1 + 2) (2 + 3)(3 + 5) (4 + 1)
]=
[3 58 5
].
The addition
[1 23 4
]+
[1 2 34 5 6
]is not defined.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 2 + i −1 + 2i3i 2 1
].
(a) A-adjoint: Conjugate and transpose:
A∗ =
1 −3i2− i 2−1− 2i 1
. Notice that:(A∗
)∗= A.
(b) Addition of two m × n matrices is performed component-wise:[1 23 4
]+
[2 35 1
]=
[(1 + 2) (2 + 3)(3 + 5) (4 + 1)
]=
[3 58 5
].
The addition
[1 23 4
]+
[1 2 34 5 6
]is not defined.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 3 52 4 6
].
(a) Multiplication of a matrix by a number is performedcomponent-wise:
2A = 2
[1 3 52 4 6
]=
[2 6 104 8 12
],
[8 1216 20
]= 4
[2 34 5
].
Also:
A
3=
1
3
[1 3 52 4 6
]=
1
31
5
3
2
3
4
32
.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 3 52 4 6
].
(a) Multiplication of a matrix by a number is performedcomponent-wise:
2A = 2
[1 3 52 4 6
]=
[2 6 104 8 12
],
[8 1216 20
]= 4
[2 34 5
].
Also:
A
3=
1
3
[1 3 52 4 6
]=
1
31
5
3
2
3
4
32
.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 3 52 4 6
].
(a) Multiplication of a matrix by a number is performedcomponent-wise:
2A = 2
[1 3 52 4 6
]
=
[2 6 104 8 12
],
[8 1216 20
]= 4
[2 34 5
].
Also:
A
3=
1
3
[1 3 52 4 6
]=
1
31
5
3
2
3
4
32
.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 3 52 4 6
].
(a) Multiplication of a matrix by a number is performedcomponent-wise:
2A = 2
[1 3 52 4 6
]=
[2 6 104 8 12
],
[8 1216 20
]= 4
[2 34 5
].
Also:
A
3=
1
3
[1 3 52 4 6
]=
1
31
5
3
2
3
4
32
.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 3 52 4 6
].
(a) Multiplication of a matrix by a number is performedcomponent-wise:
2A = 2
[1 3 52 4 6
]=
[2 6 104 8 12
],
[8 1216 20
]
= 4
[2 34 5
].
Also:
A
3=
1
3
[1 3 52 4 6
]=
1
31
5
3
2
3
4
32
.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 3 52 4 6
].
(a) Multiplication of a matrix by a number is performedcomponent-wise:
2A = 2
[1 3 52 4 6
]=
[2 6 104 8 12
],
[8 1216 20
]= 4
[2 34 5
].
Also:
A
3=
1
3
[1 3 52 4 6
]=
1
31
5
3
2
3
4
32
.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 3 52 4 6
].
(a) Multiplication of a matrix by a number is performedcomponent-wise:
2A = 2
[1 3 52 4 6
]=
[2 6 104 8 12
],
[8 1216 20
]= 4
[2 34 5
].
Also:
A
3
=1
3
[1 3 52 4 6
]=
1
31
5
3
2
3
4
32
.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 3 52 4 6
].
(a) Multiplication of a matrix by a number is performedcomponent-wise:
2A = 2
[1 3 52 4 6
]=
[2 6 104 8 12
],
[8 1216 20
]= 4
[2 34 5
].
Also:
A
3=
1
3
[1 3 52 4 6
]
=
1
31
5
3
2
3
4
32
.
Main concepts from Linear Algebra.
Example
Consider a 2× 3 matrix A =
[1 3 52 4 6
].
(a) Multiplication of a matrix by a number is performedcomponent-wise:
2A = 2
[1 3 52 4 6
]=
[2 6 104 8 12
],
[8 1216 20
]= 4
[2 34 5
].
Also:
A
3=
1
3
[1 3 52 4 6
]=
1
31
5
3
2
3
4
32
.
Main concepts from Linear Algebra.
Example
(a) Matrix multiplication.
The matrix sizes is important:
Am × n
times Bn × `
defines ABm × `
Example: A is 2× 2, B is 2× 3, so AB is 2× 3:
AB =
[4 32 1
] [1 2 34 5 6
]=
[16 23 306 9 12
].
Notice B is 2× 3, A is 2× 2, so BA is not defined.
BA =
[1 2 34 5 6
] [4 32 1
]not defined.
Main concepts from Linear Algebra.
Example
(a) Matrix multiplication. The matrix sizes is important:
Am × n
times Bn × `
defines ABm × `
Example: A is 2× 2, B is 2× 3, so AB is 2× 3:
AB =
[4 32 1
] [1 2 34 5 6
]=
[16 23 306 9 12
].
Notice B is 2× 3, A is 2× 2, so BA is not defined.
BA =
[1 2 34 5 6
] [4 32 1
]not defined.
Main concepts from Linear Algebra.
Example
(a) Matrix multiplication. The matrix sizes is important:
Am × n
times Bn × `
defines ABm × `
Example: A is 2× 2, B is 2× 3, so AB is 2× 3:
AB =
[4 32 1
] [1 2 34 5 6
]=
[16 23 306 9 12
].
Notice B is 2× 3, A is 2× 2, so BA is not defined.
BA =
[1 2 34 5 6
] [4 32 1
]not defined.
Main concepts from Linear Algebra.
Example
(a) Matrix multiplication. The matrix sizes is important:
Am × n
times Bn × `
defines ABm × `
Example: A is 2× 2, B is 2× 3, so AB is 2× 3:
AB =
[4 32 1
] [1 2 34 5 6
]=
[16 23 306 9 12
].
Notice B is 2× 3, A is 2× 2, so BA is not defined.
BA =
[1 2 34 5 6
] [4 32 1
]not defined.
Main concepts from Linear Algebra.
Example
(a) Matrix multiplication. The matrix sizes is important:
Am × n
times Bn × `
defines ABm × `
Example: A is 2× 2, B is 2× 3, so AB is 2× 3:
AB =
[4 32 1
] [1 2 34 5 6
]=
[16 23 306 9 12
].
Notice B is 2× 3, A is 2× 2, so BA is not defined.
BA =
[1 2 34 5 6
] [4 32 1
]not defined.
Main concepts from Linear Algebra.
Example
(a) Matrix multiplication. The matrix sizes is important:
Am × n
times Bn × `
defines ABm × `
Example: A is 2× 2, B is 2× 3, so AB is 2× 3:
AB =
[4 32 1
] [1 2 34 5 6
]=
[16 23 306 9 12
].
Notice B is 2× 3, A is 2× 2, so BA is not defined.
BA =
[1 2 34 5 6
] [4 32 1
]not defined.
Main concepts from Linear Algebra.
Example
(a) Matrix multiplication. The matrix sizes is important:
Am × n
times Bn × `
defines ABm × `
Example: A is 2× 2, B is 2× 3, so AB is 2× 3:
AB =
[4 32 1
] [1 2 34 5 6
]=
[16 23 306 9 12
].
Notice B is 2× 3, A is 2× 2, so BA is not defined.
BA =
[1 2 34 5 6
] [4 32 1
]not defined.
Main concepts from Linear Algebra.
Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.
Example
Find AB and BA for A =
[2 −1−1 2
]and B =
[3 02 −1
].
Solution:
AB =
[2 −1−1 2
] [3 02 −1
]=
[(6− 2) (0 + 1)
(−3 + 4) (0− 2)
]=
[4 11 −2
].
BA =
[3 02 −1
] [2 −1−1 2
]=
[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)
]=
[6 −35 −4
].
So AB 6= BA. C
Main concepts from Linear Algebra.
Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.
Example
Find AB and BA for A =
[2 −1−1 2
]and B =
[3 02 −1
].
Solution:
AB =
[2 −1−1 2
] [3 02 −1
]=
[(6− 2) (0 + 1)
(−3 + 4) (0− 2)
]=
[4 11 −2
].
BA =
[3 02 −1
] [2 −1−1 2
]=
[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)
]=
[6 −35 −4
].
So AB 6= BA. C
Main concepts from Linear Algebra.
Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.
Example
Find AB and BA for A =
[2 −1−1 2
]and B =
[3 02 −1
].
Solution:
AB =
[2 −1−1 2
] [3 02 −1
]
=
[(6− 2) (0 + 1)
(−3 + 4) (0− 2)
]=
[4 11 −2
].
BA =
[3 02 −1
] [2 −1−1 2
]=
[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)
]=
[6 −35 −4
].
So AB 6= BA. C
Main concepts from Linear Algebra.
Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.
Example
Find AB and BA for A =
[2 −1−1 2
]and B =
[3 02 −1
].
Solution:
AB =
[2 −1−1 2
] [3 02 −1
]=
[(6− 2) (0 + 1)
(−3 + 4) (0− 2)
]
=
[4 11 −2
].
BA =
[3 02 −1
] [2 −1−1 2
]=
[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)
]=
[6 −35 −4
].
So AB 6= BA. C
Main concepts from Linear Algebra.
Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.
Example
Find AB and BA for A =
[2 −1−1 2
]and B =
[3 02 −1
].
Solution:
AB =
[2 −1−1 2
] [3 02 −1
]=
[(6− 2) (0 + 1)
(−3 + 4) (0− 2)
]=
[4 11 −2
].
BA =
[3 02 −1
] [2 −1−1 2
]=
[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)
]=
[6 −35 −4
].
So AB 6= BA. C
Main concepts from Linear Algebra.
Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.
Example
Find AB and BA for A =
[2 −1−1 2
]and B =
[3 02 −1
].
Solution:
AB =
[2 −1−1 2
] [3 02 −1
]=
[(6− 2) (0 + 1)
(−3 + 4) (0− 2)
]=
[4 11 −2
].
BA =
[3 02 −1
] [2 −1−1 2
]
=
[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)
]=
[6 −35 −4
].
So AB 6= BA. C
Main concepts from Linear Algebra.
Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.
Example
Find AB and BA for A =
[2 −1−1 2
]and B =
[3 02 −1
].
Solution:
AB =
[2 −1−1 2
] [3 02 −1
]=
[(6− 2) (0 + 1)
(−3 + 4) (0− 2)
]=
[4 11 −2
].
BA =
[3 02 −1
] [2 −1−1 2
]=
[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)
]
=
[6 −35 −4
].
So AB 6= BA. C
Main concepts from Linear Algebra.
Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.
Example
Find AB and BA for A =
[2 −1−1 2
]and B =
[3 02 −1
].
Solution:
AB =
[2 −1−1 2
] [3 02 −1
]=
[(6− 2) (0 + 1)
(−3 + 4) (0− 2)
]=
[4 11 −2
].
BA =
[3 02 −1
] [2 −1−1 2
]=
[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)
]=
[6 −35 −4
].
So AB 6= BA. C
Main concepts from Linear Algebra.
Remark: The matrix product is not commutative, that is, ingeneral holds AB 6= BA.
Example
Find AB and BA for A =
[2 −1−1 2
]and B =
[3 02 −1
].
Solution:
AB =
[2 −1−1 2
] [3 02 −1
]=
[(6− 2) (0 + 1)
(−3 + 4) (0− 2)
]=
[4 11 −2
].
BA =
[3 02 −1
] [2 −1−1 2
]=
[(6 + 0) (−3 + 0)(4 + 1) (−2− 2)
]=
[6 −35 −4
].
So AB 6= BA. C
Main concepts from Linear Algebra.
Remark: There exist matrices A 6= 0 and B 6= 0 with AB = 0.
Example
Find AB for A =
[1 −1−1 1
]and B =
[1 −11 −1
].
Solution:
AB =
[1 −1−1 1
] [1 −11 −1
]=
[(1− 1) (−1 + 1)
(−1 + 1) (1− 1)
]=
[0 00 0
].
C
Recall: If a, b ∈ R and ab = 0, then either a = 0 or b = 0.
We have just shown that this statement is not true for matrices.
Main concepts from Linear Algebra.
Remark: There exist matrices A 6= 0 and B 6= 0 with AB = 0.
Example
Find AB for A =
[1 −1−1 1
]and B =
[1 −11 −1
].
Solution:
AB =
[1 −1−1 1
] [1 −11 −1
]=
[(1− 1) (−1 + 1)
(−1 + 1) (1− 1)
]=
[0 00 0
].
C
Recall: If a, b ∈ R and ab = 0, then either a = 0 or b = 0.
We have just shown that this statement is not true for matrices.
Main concepts from Linear Algebra.
Remark: There exist matrices A 6= 0 and B 6= 0 with AB = 0.
Example
Find AB for A =
[1 −1−1 1
]and B =
[1 −11 −1
].
Solution:
AB =
[1 −1−1 1
] [1 −11 −1
]
=
[(1− 1) (−1 + 1)
(−1 + 1) (1− 1)
]=
[0 00 0
].
C
Recall: If a, b ∈ R and ab = 0, then either a = 0 or b = 0.
We have just shown that this statement is not true for matrices.
Main concepts from Linear Algebra.
Remark: There exist matrices A 6= 0 and B 6= 0 with AB = 0.
Example
Find AB for A =
[1 −1−1 1
]and B =
[1 −11 −1
].
Solution:
AB =
[1 −1−1 1
] [1 −11 −1
]=
[(1− 1) (−1 + 1)
(−1 + 1) (1− 1)
]
=
[0 00 0
].
C
Recall: If a, b ∈ R and ab = 0, then either a = 0 or b = 0.
We have just shown that this statement is not true for matrices.
Main concepts from Linear Algebra.
Remark: There exist matrices A 6= 0 and B 6= 0 with AB = 0.
Example
Find AB for A =
[1 −1−1 1
]and B =
[1 −11 −1
].
Solution:
AB =
[1 −1−1 1
] [1 −11 −1
]=
[(1− 1) (−1 + 1)
(−1 + 1) (1− 1)
]=
[0 00 0
].
C
Recall: If a, b ∈ R and ab = 0, then either a = 0 or b = 0.
We have just shown that this statement is not true for matrices.
Main concepts from Linear Algebra.
Remark: There exist matrices A 6= 0 and B 6= 0 with AB = 0.
Example
Find AB for A =
[1 −1−1 1
]and B =
[1 −11 −1
].
Solution:
AB =
[1 −1−1 1
] [1 −11 −1
]=
[(1− 1) (−1 + 1)
(−1 + 1) (1− 1)
]=
[0 00 0
].
C
Recall: If a, b ∈ R and ab = 0, then either a = 0 or b = 0.
We have just shown that this statement is not true for matrices.
Main concepts from Linear Algebra.
Remark: There exist matrices A 6= 0 and B 6= 0 with AB = 0.
Example
Find AB for A =
[1 −1−1 1
]and B =
[1 −11 −1
].
Solution:
AB =
[1 −1−1 1
] [1 −11 −1
]=
[(1− 1) (−1 + 1)
(−1 + 1) (1− 1)
]=
[0 00 0
].
C
Recall: If a, b ∈ R and ab = 0, then either a = 0 or b = 0.
We have just shown that this statement is not true for matrices.
Review Exam 3.
I Sections 6.1-6.6.
I 5 or 6 problems.
I 50 minutes.
I Laplace Transform table included.
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Compute the LT of the equation,
L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)] = e−2s
L[y ′′] = s2 L[y ]− s y(0)− y ′(0), L[y ′] = s L[y ]− y(0).
(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s
(s2 − 2s + 2)L[y ]− s − 1 = e−2s
L[y ] =(s + 1)
(s2 − 2s + 2)+
1
(s2 − 2s + 2)e−2s .
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Compute the LT of the equation,
L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)] = e−2s
L[y ′′] = s2 L[y ]− s y(0)− y ′(0), L[y ′] = s L[y ]− y(0).
(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s
(s2 − 2s + 2)L[y ]− s − 1 = e−2s
L[y ] =(s + 1)
(s2 − 2s + 2)+
1
(s2 − 2s + 2)e−2s .
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Compute the LT of the equation,
L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)]
= e−2s
L[y ′′] = s2 L[y ]− s y(0)− y ′(0), L[y ′] = s L[y ]− y(0).
(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s
(s2 − 2s + 2)L[y ]− s − 1 = e−2s
L[y ] =(s + 1)
(s2 − 2s + 2)+
1
(s2 − 2s + 2)e−2s .
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Compute the LT of the equation,
L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)] = e−2s
L[y ′′] = s2 L[y ]− s y(0)− y ′(0), L[y ′] = s L[y ]− y(0).
(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s
(s2 − 2s + 2)L[y ]− s − 1 = e−2s
L[y ] =(s + 1)
(s2 − 2s + 2)+
1
(s2 − 2s + 2)e−2s .
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Compute the LT of the equation,
L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)] = e−2s
L[y ′′] = s2 L[y ]− s y(0)− y ′(0),
L[y ′] = s L[y ]− y(0).
(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s
(s2 − 2s + 2)L[y ]− s − 1 = e−2s
L[y ] =(s + 1)
(s2 − 2s + 2)+
1
(s2 − 2s + 2)e−2s .
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Compute the LT of the equation,
L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)] = e−2s
L[y ′′] = s2 L[y ]− s y(0)− y ′(0), L[y ′] = s L[y ]− y(0).
(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s
(s2 − 2s + 2)L[y ]− s − 1 = e−2s
L[y ] =(s + 1)
(s2 − 2s + 2)+
1
(s2 − 2s + 2)e−2s .
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Compute the LT of the equation,
L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)] = e−2s
L[y ′′] = s2 L[y ]− s y(0)− y ′(0), L[y ′] = s L[y ]− y(0).
(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s
(s2 − 2s + 2)L[y ]− s − 1 = e−2s
L[y ] =(s + 1)
(s2 − 2s + 2)+
1
(s2 − 2s + 2)e−2s .
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Compute the LT of the equation,
L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)] = e−2s
L[y ′′] = s2 L[y ]− s y(0)− y ′(0), L[y ′] = s L[y ]− y(0).
(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s
(s2 − 2s + 2)L[y ]− s − 1 = e−2s
L[y ] =(s + 1)
(s2 − 2s + 2)+
1
(s2 − 2s + 2)e−2s .
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Compute the LT of the equation,
L[y ′′]− 2L[y ′] + 2L[y ] = L[δ(t − 2)] = e−2s
L[y ′′] = s2 L[y ]− s y(0)− y ′(0), L[y ′] = s L[y ]− y(0).
(s2 − 2s + 2)L[y ]− s y(0)− y ′(0) + 2 y(0) = e−2s
(s2 − 2s + 2)L[y ]− s − 1 = e−2s
L[y ] =(s + 1)
(s2 − 2s + 2)+
1
(s2 − 2s + 2)e−2s .
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall: L[y ] =(s + 1)
(s2 − 2s + 2)+
1
(s2 − 2s + 2)e−2s .
s2 − 2s + 2 = 0 ⇒ s± =1
2
[2±
√4− 8
], complex roots.
s2 − 2s + 2 = (s2 − 2s + 1)− 1 + 2 = (s − 1)2 + 1.
L[y ] =s + 1
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s
L[y ] =(s − 1 + 1) + 1
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall: L[y ] =(s + 1)
(s2 − 2s + 2)+
1
(s2 − 2s + 2)e−2s .
s2 − 2s + 2 = 0
⇒ s± =1
2
[2±
√4− 8
], complex roots.
s2 − 2s + 2 = (s2 − 2s + 1)− 1 + 2 = (s − 1)2 + 1.
L[y ] =s + 1
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s
L[y ] =(s − 1 + 1) + 1
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall: L[y ] =(s + 1)
(s2 − 2s + 2)+
1
(s2 − 2s + 2)e−2s .
s2 − 2s + 2 = 0 ⇒ s± =1
2
[2±
√4− 8
],
complex roots.
s2 − 2s + 2 = (s2 − 2s + 1)− 1 + 2 = (s − 1)2 + 1.
L[y ] =s + 1
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s
L[y ] =(s − 1 + 1) + 1
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall: L[y ] =(s + 1)
(s2 − 2s + 2)+
1
(s2 − 2s + 2)e−2s .
s2 − 2s + 2 = 0 ⇒ s± =1
2
[2±
√4− 8
], complex roots.
s2 − 2s + 2 = (s2 − 2s + 1)− 1 + 2 = (s − 1)2 + 1.
L[y ] =s + 1
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s
L[y ] =(s − 1 + 1) + 1
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall: L[y ] =(s + 1)
(s2 − 2s + 2)+
1
(s2 − 2s + 2)e−2s .
s2 − 2s + 2 = 0 ⇒ s± =1
2
[2±
√4− 8
], complex roots.
s2 − 2s + 2 = (s2 − 2s + 1)− 1 + 2
= (s − 1)2 + 1.
L[y ] =s + 1
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s
L[y ] =(s − 1 + 1) + 1
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall: L[y ] =(s + 1)
(s2 − 2s + 2)+
1
(s2 − 2s + 2)e−2s .
s2 − 2s + 2 = 0 ⇒ s± =1
2
[2±
√4− 8
], complex roots.
s2 − 2s + 2 = (s2 − 2s + 1)− 1 + 2 = (s − 1)2 + 1.
L[y ] =s + 1
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s
L[y ] =(s − 1 + 1) + 1
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall: L[y ] =(s + 1)
(s2 − 2s + 2)+
1
(s2 − 2s + 2)e−2s .
s2 − 2s + 2 = 0 ⇒ s± =1
2
[2±
√4− 8
], complex roots.
s2 − 2s + 2 = (s2 − 2s + 1)− 1 + 2 = (s − 1)2 + 1.
L[y ] =s + 1
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s
L[y ] =(s − 1 + 1) + 1
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall: L[y ] =(s + 1)
(s2 − 2s + 2)+
1
(s2 − 2s + 2)e−2s .
s2 − 2s + 2 = 0 ⇒ s± =1
2
[2±
√4− 8
], complex roots.
s2 − 2s + 2 = (s2 − 2s + 1)− 1 + 2 = (s − 1)2 + 1.
L[y ] =s + 1
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s
L[y ] =(s − 1 + 1) + 1
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall: L[y ] =(s − 1) + 2
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s ,
L[y ] =(s − 1)
(s − 1)2 + 1+ 2
1
(s − 1)2 + 1+ e−2s 1
(s − 1)2 + 1,
L[cos(at)] =s
s2 + a2, L[sin(at)] =
a
s2 + a2,
L[y ] = L[cos(t)]∣∣(s−1)
+ 2L[sin(t)]∣∣(s−1)
+ e−2s L[sin(t)]∣∣(s−1)
.
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall: L[y ] =(s − 1) + 2
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s ,
L[y ] =(s − 1)
(s − 1)2 + 1+ 2
1
(s − 1)2 + 1+ e−2s 1
(s − 1)2 + 1,
L[cos(at)] =s
s2 + a2, L[sin(at)] =
a
s2 + a2,
L[y ] = L[cos(t)]∣∣(s−1)
+ 2L[sin(t)]∣∣(s−1)
+ e−2s L[sin(t)]∣∣(s−1)
.
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall: L[y ] =(s − 1) + 2
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s ,
L[y ] =(s − 1)
(s − 1)2 + 1+ 2
1
(s − 1)2 + 1+ e−2s 1
(s − 1)2 + 1,
L[cos(at)] =s
s2 + a2,
L[sin(at)] =a
s2 + a2,
L[y ] = L[cos(t)]∣∣(s−1)
+ 2L[sin(t)]∣∣(s−1)
+ e−2s L[sin(t)]∣∣(s−1)
.
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall: L[y ] =(s − 1) + 2
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s ,
L[y ] =(s − 1)
(s − 1)2 + 1+ 2
1
(s − 1)2 + 1+ e−2s 1
(s − 1)2 + 1,
L[cos(at)] =s
s2 + a2, L[sin(at)] =
a
s2 + a2,
L[y ] = L[cos(t)]∣∣(s−1)
+ 2L[sin(t)]∣∣(s−1)
+ e−2s L[sin(t)]∣∣(s−1)
.
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall: L[y ] =(s − 1) + 2
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s ,
L[y ] =(s − 1)
(s − 1)2 + 1+ 2
1
(s − 1)2 + 1+ e−2s 1
(s − 1)2 + 1,
L[cos(at)] =s
s2 + a2, L[sin(at)] =
a
s2 + a2,
L[y ] = L[cos(t)]∣∣(s−1)
+ 2L[sin(t)]∣∣(s−1)
+ e−2s L[sin(t)]∣∣(s−1)
.
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall: L[y ] =(s − 1) + 2
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s ,
L[y ] =(s − 1)
(s − 1)2 + 1+ 2
1
(s − 1)2 + 1+ e−2s 1
(s − 1)2 + 1,
L[cos(at)] =s
s2 + a2, L[sin(at)] =
a
s2 + a2,
L[y ] = L[cos(t)]∣∣(s−1)
+ 2L[sin(t)]∣∣(s−1)
+ e−2s L[sin(t)]∣∣(s−1)
.
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall: L[y ] =(s − 1) + 2
(s − 1)2 + 1+
1
(s − 1)2 + 1e−2s ,
L[y ] =(s − 1)
(s − 1)2 + 1+ 2
1
(s − 1)2 + 1+ e−2s 1
(s − 1)2 + 1,
L[cos(at)] =s
s2 + a2, L[sin(at)] =
a
s2 + a2,
L[y ] = L[cos(t)]∣∣(s−1)
+ 2L[sin(t)]∣∣(s−1)
+ e−2s L[sin(t)]∣∣(s−1)
.
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall:
L[y ] = L[cos(t)]∣∣(s−1)
+ 2L[sin(t)]∣∣(s−1)
+ e−2s L[sin(t)]∣∣(s−1)
and L[f (t)]∣∣(s−c)
= L[ect f (t)]. Therefore,
L[y ] = L[et cos(t)] + 2L[et sin(t)] + e−2s L[et sin(t)].
Also recall: e−cs L[f (t)] = L[uc(t) f (t − c)]. Therefore,
L[y ] = L[et cos(t)] + 2L[et sin(t)] + L[u2(t) e(t−2) sin(t − 2)].
y(t) =[cos(t) + 2 sin(t)
]et + u2(t) sin(t − 2) e(t−2). C
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall:
L[y ] = L[cos(t)]∣∣(s−1)
+ 2L[sin(t)]∣∣(s−1)
+ e−2s L[sin(t)]∣∣(s−1)
and L[f (t)]∣∣(s−c)
= L[ect f (t)].
Therefore,
L[y ] = L[et cos(t)] + 2L[et sin(t)] + e−2s L[et sin(t)].
Also recall: e−cs L[f (t)] = L[uc(t) f (t − c)]. Therefore,
L[y ] = L[et cos(t)] + 2L[et sin(t)] + L[u2(t) e(t−2) sin(t − 2)].
y(t) =[cos(t) + 2 sin(t)
]et + u2(t) sin(t − 2) e(t−2). C
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall:
L[y ] = L[cos(t)]∣∣(s−1)
+ 2L[sin(t)]∣∣(s−1)
+ e−2s L[sin(t)]∣∣(s−1)
and L[f (t)]∣∣(s−c)
= L[ect f (t)]. Therefore,
L[y ] = L[et cos(t)]
+ 2L[et sin(t)] + e−2s L[et sin(t)].
Also recall: e−cs L[f (t)] = L[uc(t) f (t − c)]. Therefore,
L[y ] = L[et cos(t)] + 2L[et sin(t)] + L[u2(t) e(t−2) sin(t − 2)].
y(t) =[cos(t) + 2 sin(t)
]et + u2(t) sin(t − 2) e(t−2). C
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall:
L[y ] = L[cos(t)]∣∣(s−1)
+ 2L[sin(t)]∣∣(s−1)
+ e−2s L[sin(t)]∣∣(s−1)
and L[f (t)]∣∣(s−c)
= L[ect f (t)]. Therefore,
L[y ] = L[et cos(t)] + 2L[et sin(t)]
+ e−2s L[et sin(t)].
Also recall: e−cs L[f (t)] = L[uc(t) f (t − c)]. Therefore,
L[y ] = L[et cos(t)] + 2L[et sin(t)] + L[u2(t) e(t−2) sin(t − 2)].
y(t) =[cos(t) + 2 sin(t)
]et + u2(t) sin(t − 2) e(t−2). C
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall:
L[y ] = L[cos(t)]∣∣(s−1)
+ 2L[sin(t)]∣∣(s−1)
+ e−2s L[sin(t)]∣∣(s−1)
and L[f (t)]∣∣(s−c)
= L[ect f (t)]. Therefore,
L[y ] = L[et cos(t)] + 2L[et sin(t)] + e−2s L[et sin(t)].
Also recall: e−cs L[f (t)] = L[uc(t) f (t − c)]. Therefore,
L[y ] = L[et cos(t)] + 2L[et sin(t)] + L[u2(t) e(t−2) sin(t − 2)].
y(t) =[cos(t) + 2 sin(t)
]et + u2(t) sin(t − 2) e(t−2). C
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall:
L[y ] = L[cos(t)]∣∣(s−1)
+ 2L[sin(t)]∣∣(s−1)
+ e−2s L[sin(t)]∣∣(s−1)
and L[f (t)]∣∣(s−c)
= L[ect f (t)]. Therefore,
L[y ] = L[et cos(t)] + 2L[et sin(t)] + e−2s L[et sin(t)].
Also recall: e−cs L[f (t)] = L[uc(t) f (t − c)].
Therefore,
L[y ] = L[et cos(t)] + 2L[et sin(t)] + L[u2(t) e(t−2) sin(t − 2)].
y(t) =[cos(t) + 2 sin(t)
]et + u2(t) sin(t − 2) e(t−2). C
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall:
L[y ] = L[cos(t)]∣∣(s−1)
+ 2L[sin(t)]∣∣(s−1)
+ e−2s L[sin(t)]∣∣(s−1)
and L[f (t)]∣∣(s−c)
= L[ect f (t)]. Therefore,
L[y ] = L[et cos(t)] + 2L[et sin(t)] + e−2s L[et sin(t)].
Also recall: e−cs L[f (t)] = L[uc(t) f (t − c)]. Therefore,
L[y ] = L[et cos(t)] + 2L[et sin(t)] + L[u2(t) e(t−2) sin(t − 2)].
y(t) =[cos(t) + 2 sin(t)
]et + u2(t) sin(t − 2) e(t−2). C
Exam: November 11, 2008. Problem 2
Example
Use Laplace Transform to find y solution of
y ′′ − 2y ′ + 2y = δ(t − 2), y(0) = 1, y ′(0) = 3.
Solution: Recall:
L[y ] = L[cos(t)]∣∣(s−1)
+ 2L[sin(t)]∣∣(s−1)
+ e−2s L[sin(t)]∣∣(s−1)
and L[f (t)]∣∣(s−c)
= L[ect f (t)]. Therefore,
L[y ] = L[et cos(t)] + 2L[et sin(t)] + e−2s L[et sin(t)].
Also recall: e−cs L[f (t)] = L[uc(t) f (t − c)]. Therefore,
L[y ] = L[et cos(t)] + 2L[et sin(t)] + L[u2(t) e(t−2) sin(t − 2)].
y(t) =[cos(t) + 2 sin(t)
]et + u2(t) sin(t − 2) e(t−2). C
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution:
t
u ( t − 2 )1
t2
g ( t )
e
Express g using step functions,
g(t) = u2(t) e(t−2).
L[uc(t) f (t − c)] = e−cs L[f (t)].
Therefore,
L[g(t)] = e−2sL[et ].
We obtain: L[g(t)] =e−2s
(s − 1).
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution:
t
u ( t − 2 )1
t2
g ( t )
e
Express g using step functions,
g(t) = u2(t) e(t−2).
L[uc(t) f (t − c)] = e−cs L[f (t)].
Therefore,
L[g(t)] = e−2sL[et ].
We obtain: L[g(t)] =e−2s
(s − 1).
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution:
t
u ( t − 2 )1
t2
g ( t )
e
Express g using step functions,
g(t) = u2(t) e(t−2).
L[uc(t) f (t − c)] = e−cs L[f (t)].
Therefore,
L[g(t)] = e−2sL[et ].
We obtain: L[g(t)] =e−2s
(s − 1).
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution:
t
u ( t − 2 )1
t2
g ( t )
e
Express g using step functions,
g(t) = u2(t) e(t−2).
L[uc(t) f (t − c)] = e−cs L[f (t)].
Therefore,
L[g(t)] = e−2sL[et ].
We obtain: L[g(t)] =e−2s
(s − 1).
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution:
t
u ( t − 2 )1
t2
g ( t )
e
Express g using step functions,
g(t) = u2(t) e(t−2).
L[uc(t) f (t − c)] = e−cs L[f (t)].
Therefore,
L[g(t)] = e−2sL[et ].
We obtain: L[g(t)] =e−2s
(s − 1).
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution:
t
u ( t − 2 )1
t2
g ( t )
e
Express g using step functions,
g(t) = u2(t) e(t−2).
L[uc(t) f (t − c)] = e−cs L[f (t)].
Therefore,
L[g(t)] = e−2sL[et ].
We obtain: L[g(t)] =e−2s
(s − 1).
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution:
t
u ( t − 2 )1
t2
g ( t )
e
Express g using step functions,
g(t) = u2(t) e(t−2).
L[uc(t) f (t − c)] = e−cs L[f (t)].
Therefore,
L[g(t)] = e−2sL[et ].
We obtain: L[g(t)] =e−2s
(s − 1).
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: L[g(t)] =e−2s
(s − 1).
L[y ′′] + 3L[y ] = L[g(t)] =e−2s
(s − 1).
(s2 + 3)L[y ] =e−2s
(s − 1)⇒ L[y ] = e−2s 1
(s − 1)(s2 + 3).
H(s) =1
(s − 1)(s2 + 3)=
a
(s − 1)+
(bs + c)
(s2 + 3)
1 = a(s2 + 3) + (bs + c)(s − 1)
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: L[g(t)] =e−2s
(s − 1).
L[y ′′] + 3L[y ] = L[g(t)]
=e−2s
(s − 1).
(s2 + 3)L[y ] =e−2s
(s − 1)⇒ L[y ] = e−2s 1
(s − 1)(s2 + 3).
H(s) =1
(s − 1)(s2 + 3)=
a
(s − 1)+
(bs + c)
(s2 + 3)
1 = a(s2 + 3) + (bs + c)(s − 1)
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: L[g(t)] =e−2s
(s − 1).
L[y ′′] + 3L[y ] = L[g(t)] =e−2s
(s − 1).
(s2 + 3)L[y ] =e−2s
(s − 1)⇒ L[y ] = e−2s 1
(s − 1)(s2 + 3).
H(s) =1
(s − 1)(s2 + 3)=
a
(s − 1)+
(bs + c)
(s2 + 3)
1 = a(s2 + 3) + (bs + c)(s − 1)
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: L[g(t)] =e−2s
(s − 1).
L[y ′′] + 3L[y ] = L[g(t)] =e−2s
(s − 1).
(s2 + 3)L[y ] =e−2s
(s − 1)
⇒ L[y ] = e−2s 1
(s − 1)(s2 + 3).
H(s) =1
(s − 1)(s2 + 3)=
a
(s − 1)+
(bs + c)
(s2 + 3)
1 = a(s2 + 3) + (bs + c)(s − 1)
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: L[g(t)] =e−2s
(s − 1).
L[y ′′] + 3L[y ] = L[g(t)] =e−2s
(s − 1).
(s2 + 3)L[y ] =e−2s
(s − 1)⇒ L[y ] = e−2s 1
(s − 1)(s2 + 3).
H(s) =1
(s − 1)(s2 + 3)=
a
(s − 1)+
(bs + c)
(s2 + 3)
1 = a(s2 + 3) + (bs + c)(s − 1)
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: L[g(t)] =e−2s
(s − 1).
L[y ′′] + 3L[y ] = L[g(t)] =e−2s
(s − 1).
(s2 + 3)L[y ] =e−2s
(s − 1)⇒ L[y ] = e−2s 1
(s − 1)(s2 + 3).
H(s) =1
(s − 1)(s2 + 3)
=a
(s − 1)+
(bs + c)
(s2 + 3)
1 = a(s2 + 3) + (bs + c)(s − 1)
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: L[g(t)] =e−2s
(s − 1).
L[y ′′] + 3L[y ] = L[g(t)] =e−2s
(s − 1).
(s2 + 3)L[y ] =e−2s
(s − 1)⇒ L[y ] = e−2s 1
(s − 1)(s2 + 3).
H(s) =1
(s − 1)(s2 + 3)=
a
(s − 1)+
(bs + c)
(s2 + 3)
1 = a(s2 + 3) + (bs + c)(s − 1)
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: L[g(t)] =e−2s
(s − 1).
L[y ′′] + 3L[y ] = L[g(t)] =e−2s
(s − 1).
(s2 + 3)L[y ] =e−2s
(s − 1)⇒ L[y ] = e−2s 1
(s − 1)(s2 + 3).
H(s) =1
(s − 1)(s2 + 3)=
a
(s − 1)+
(bs + c)
(s2 + 3)
1 = a(s2 + 3) + (bs + c)(s − 1)
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).
1 = as2 + 3a + bs2 + cs − bs − c
1 = (a + b) s2 + (c − b) s + (3a− c)
a + b = 0, c − b = 0, 3a− c = 1.
a = −b, c = b, −3b−b = 1 ⇒ b = −1
4, a =
1
4, c = −1
4.
H(s) =1
4
[ 1
s − 1− s + 1
s2 + 3
].
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).
1 = as2 + 3a + bs2 + cs − bs − c
1 = (a + b) s2 + (c − b) s + (3a− c)
a + b = 0, c − b = 0, 3a− c = 1.
a = −b, c = b, −3b−b = 1 ⇒ b = −1
4, a =
1
4, c = −1
4.
H(s) =1
4
[ 1
s − 1− s + 1
s2 + 3
].
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).
1 = as2 + 3a + bs2 + cs − bs − c
1 = (a + b) s2 + (c − b) s + (3a− c)
a + b = 0, c − b = 0, 3a− c = 1.
a = −b, c = b, −3b−b = 1 ⇒ b = −1
4, a =
1
4, c = −1
4.
H(s) =1
4
[ 1
s − 1− s + 1
s2 + 3
].
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).
1 = as2 + 3a + bs2 + cs − bs − c
1 = (a + b) s2 + (c − b) s + (3a− c)
a + b = 0, c − b = 0, 3a− c = 1.
a = −b, c = b, −3b−b = 1 ⇒ b = −1
4, a =
1
4, c = −1
4.
H(s) =1
4
[ 1
s − 1− s + 1
s2 + 3
].
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).
1 = as2 + 3a + bs2 + cs − bs − c
1 = (a + b) s2 + (c − b) s + (3a− c)
a + b = 0, c − b = 0, 3a− c = 1.
a = −b,
c = b, −3b−b = 1 ⇒ b = −1
4, a =
1
4, c = −1
4.
H(s) =1
4
[ 1
s − 1− s + 1
s2 + 3
].
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).
1 = as2 + 3a + bs2 + cs − bs − c
1 = (a + b) s2 + (c − b) s + (3a− c)
a + b = 0, c − b = 0, 3a− c = 1.
a = −b, c = b,
−3b−b = 1 ⇒ b = −1
4, a =
1
4, c = −1
4.
H(s) =1
4
[ 1
s − 1− s + 1
s2 + 3
].
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).
1 = as2 + 3a + bs2 + cs − bs − c
1 = (a + b) s2 + (c − b) s + (3a− c)
a + b = 0, c − b = 0, 3a− c = 1.
a = −b, c = b, −3b−b = 1
⇒ b = −1
4, a =
1
4, c = −1
4.
H(s) =1
4
[ 1
s − 1− s + 1
s2 + 3
].
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).
1 = as2 + 3a + bs2 + cs − bs − c
1 = (a + b) s2 + (c − b) s + (3a− c)
a + b = 0, c − b = 0, 3a− c = 1.
a = −b, c = b, −3b−b = 1 ⇒ b = −1
4,
a =1
4, c = −1
4.
H(s) =1
4
[ 1
s − 1− s + 1
s2 + 3
].
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).
1 = as2 + 3a + bs2 + cs − bs − c
1 = (a + b) s2 + (c − b) s + (3a− c)
a + b = 0, c − b = 0, 3a− c = 1.
a = −b, c = b, −3b−b = 1 ⇒ b = −1
4, a =
1
4,
c = −1
4.
H(s) =1
4
[ 1
s − 1− s + 1
s2 + 3
].
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).
1 = as2 + 3a + bs2 + cs − bs − c
1 = (a + b) s2 + (c − b) s + (3a− c)
a + b = 0, c − b = 0, 3a− c = 1.
a = −b, c = b, −3b−b = 1 ⇒ b = −1
4, a =
1
4, c = −1
4.
H(s) =1
4
[ 1
s − 1− s + 1
s2 + 3
].
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: 1 = a(s2 + 3) + (bs + c)(s − 1).
1 = as2 + 3a + bs2 + cs − bs − c
1 = (a + b) s2 + (c − b) s + (3a− c)
a + b = 0, c − b = 0, 3a− c = 1.
a = −b, c = b, −3b−b = 1 ⇒ b = −1
4, a =
1
4, c = −1
4.
H(s) =1
4
[ 1
s − 1− s + 1
s2 + 3
].
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: H(s) =1
4
[ 1
s − 1− s + 1
s2 + 3
], L[y ] = e−2s H(s).
H(s) =1
4
[ 1
s − 1− s
s2 + 3− 1√
3
√3
s2 + 3
],
H(s) =1
4
[L[et ]− L
[cos
(√3 t
)]− 1√
3L
[sin
(√3 t
)]].
H(s) = L[1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
))].
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: H(s) =1
4
[ 1
s − 1− s + 1
s2 + 3
], L[y ] = e−2s H(s).
H(s) =1
4
[ 1
s − 1− s
s2 + 3− 1√
3
√3
s2 + 3
],
H(s) =1
4
[L[et ]− L
[cos
(√3 t
)]− 1√
3L
[sin
(√3 t
)]].
H(s) = L[1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
))].
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: H(s) =1
4
[ 1
s − 1− s + 1
s2 + 3
], L[y ] = e−2s H(s).
H(s) =1
4
[ 1
s − 1− s
s2 + 3− 1√
3
√3
s2 + 3
],
H(s) =1
4
[L[et ]
− L[cos
(√3 t
)]− 1√
3L
[sin
(√3 t
)]].
H(s) = L[1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
))].
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: H(s) =1
4
[ 1
s − 1− s + 1
s2 + 3
], L[y ] = e−2s H(s).
H(s) =1
4
[ 1
s − 1− s
s2 + 3− 1√
3
√3
s2 + 3
],
H(s) =1
4
[L[et ]− L
[cos
(√3 t
)]
− 1√3L
[sin
(√3 t
)]].
H(s) = L[1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
))].
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: H(s) =1
4
[ 1
s − 1− s + 1
s2 + 3
], L[y ] = e−2s H(s).
H(s) =1
4
[ 1
s − 1− s
s2 + 3− 1√
3
√3
s2 + 3
],
H(s) =1
4
[L[et ]− L
[cos
(√3 t
)]− 1√
3L
[sin
(√3 t
)]].
H(s) = L[1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
))].
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: H(s) =1
4
[ 1
s − 1− s + 1
s2 + 3
], L[y ] = e−2s H(s).
H(s) =1
4
[ 1
s − 1− s
s2 + 3− 1√
3
√3
s2 + 3
],
H(s) =1
4
[L[et ]− L
[cos
(√3 t
)]− 1√
3L
[sin
(√3 t
)]].
H(s) = L[1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
))].
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: H(s) = L[1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
))].
h(t) =1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
)), H(s) = L[h(t)].
L[y ] = e−2s H(s) = e−2s L[h(t)] = L[u2(t) h(t − 2)].
We conclude: y(t) = u2(t) h(t − 2). Equivalently,
y(t) =u2(t)
4
[e(t−2) − cos
(√3 (t − 2)
)− 1√
3sin
(√3 (t − 2)
)].C
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: H(s) = L[1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
))].
h(t) =1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
)),
H(s) = L[h(t)].
L[y ] = e−2s H(s) = e−2s L[h(t)] = L[u2(t) h(t − 2)].
We conclude: y(t) = u2(t) h(t − 2). Equivalently,
y(t) =u2(t)
4
[e(t−2) − cos
(√3 (t − 2)
)− 1√
3sin
(√3 (t − 2)
)].C
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: H(s) = L[1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
))].
h(t) =1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
)), H(s) = L[h(t)].
L[y ] = e−2s H(s) = e−2s L[h(t)] = L[u2(t) h(t − 2)].
We conclude: y(t) = u2(t) h(t − 2). Equivalently,
y(t) =u2(t)
4
[e(t−2) − cos
(√3 (t − 2)
)− 1√
3sin
(√3 (t − 2)
)].C
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: H(s) = L[1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
))].
h(t) =1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
)), H(s) = L[h(t)].
L[y ] = e−2s H(s)
= e−2s L[h(t)] = L[u2(t) h(t − 2)].
We conclude: y(t) = u2(t) h(t − 2). Equivalently,
y(t) =u2(t)
4
[e(t−2) − cos
(√3 (t − 2)
)− 1√
3sin
(√3 (t − 2)
)].C
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: H(s) = L[1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
))].
h(t) =1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
)), H(s) = L[h(t)].
L[y ] = e−2s H(s) = e−2s L[h(t)]
= L[u2(t) h(t − 2)].
We conclude: y(t) = u2(t) h(t − 2). Equivalently,
y(t) =u2(t)
4
[e(t−2) − cos
(√3 (t − 2)
)− 1√
3sin
(√3 (t − 2)
)].C
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: H(s) = L[1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
))].
h(t) =1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
)), H(s) = L[h(t)].
L[y ] = e−2s H(s) = e−2s L[h(t)] = L[u2(t) h(t − 2)].
We conclude: y(t) = u2(t) h(t − 2). Equivalently,
y(t) =u2(t)
4
[e(t−2) − cos
(√3 (t − 2)
)− 1√
3sin
(√3 (t − 2)
)].C
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: H(s) = L[1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
))].
h(t) =1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
)), H(s) = L[h(t)].
L[y ] = e−2s H(s) = e−2s L[h(t)] = L[u2(t) h(t − 2)].
We conclude: y(t) = u2(t) h(t − 2).
Equivalently,
y(t) =u2(t)
4
[e(t−2) − cos
(√3 (t − 2)
)− 1√
3sin
(√3 (t − 2)
)].C
Exam: November 11, 2008. Problem 3
Example
Sketch the graph of g and use LT to find y solution of
y ′′ + 3y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < 2,
e(t−2), t > 2.
Solution: Recall: H(s) = L[1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
))].
h(t) =1
4
(et − cos
(√3 t
)− 1√
3sin
(√3 t
)), H(s) = L[h(t)].
L[y ] = e−2s H(s) = e−2s L[h(t)] = L[u2(t) h(t − 2)].
We conclude: y(t) = u2(t) h(t − 2). Equivalently,
y(t) =u2(t)
4
[e(t−2) − cos
(√3 (t − 2)
)− 1√
3sin
(√3 (t − 2)
)].C
Extra problem
Example
Use convolutions to find f satisfying L[f (t)] =e−2s
(s − 1)(s2 + 3).
Solution: One way to solve this is with the splitting
L[f (t)] = e−2s 1
(s2 + 3)
1
(s − 1)= e−2s 1√
3
√3
(s2 + 3)
1
(s − 1),
L[f (t)] = e−2s 1√3L[sin
(√3 t
)]L[et ]
L[f (t)] =1√3L[u2(t) sin
(√3 (t − 2)
)]L[et ].
f (t) =1√3
∫ t
0u2(τ) sin
(√3 (τ − 2)
)e(t−τ) dτ. C
Extra problem
Example
Use convolutions to find f satisfying L[f (t)] =e−2s
(s − 1)(s2 + 3).
Solution: One way to solve this is with the splitting
L[f (t)] = e−2s 1
(s2 + 3)
1
(s − 1)
= e−2s 1√3
√3
(s2 + 3)
1
(s − 1),
L[f (t)] = e−2s 1√3L[sin
(√3 t
)]L[et ]
L[f (t)] =1√3L[u2(t) sin
(√3 (t − 2)
)]L[et ].
f (t) =1√3
∫ t
0u2(τ) sin
(√3 (τ − 2)
)e(t−τ) dτ. C
Extra problem
Example
Use convolutions to find f satisfying L[f (t)] =e−2s
(s − 1)(s2 + 3).
Solution: One way to solve this is with the splitting
L[f (t)] = e−2s 1
(s2 + 3)
1
(s − 1)= e−2s 1√
3
√3
(s2 + 3)
1
(s − 1),
L[f (t)] = e−2s 1√3L[sin
(√3 t
)]L[et ]
L[f (t)] =1√3L[u2(t) sin
(√3 (t − 2)
)]L[et ].
f (t) =1√3
∫ t
0u2(τ) sin
(√3 (τ − 2)
)e(t−τ) dτ. C
Extra problem
Example
Use convolutions to find f satisfying L[f (t)] =e−2s
(s − 1)(s2 + 3).
Solution: One way to solve this is with the splitting
L[f (t)] = e−2s 1
(s2 + 3)
1
(s − 1)= e−2s 1√
3
√3
(s2 + 3)
1
(s − 1),
L[f (t)] = e−2s 1√3L[sin
(√3 t
)]L[et ]
L[f (t)] =1√3L[u2(t) sin
(√3 (t − 2)
)]L[et ].
f (t) =1√3
∫ t
0u2(τ) sin
(√3 (τ − 2)
)e(t−τ) dτ. C
Extra problem
Example
Use convolutions to find f satisfying L[f (t)] =e−2s
(s − 1)(s2 + 3).
Solution: One way to solve this is with the splitting
L[f (t)] = e−2s 1
(s2 + 3)
1
(s − 1)= e−2s 1√
3
√3
(s2 + 3)
1
(s − 1),
L[f (t)] = e−2s 1√3L[sin
(√3 t
)]L[et ]
L[f (t)] =1√3L[u2(t) sin
(√3 (t − 2)
)]L[et ].
f (t) =1√3
∫ t
0u2(τ) sin
(√3 (τ − 2)
)e(t−τ) dτ. C
Extra problem
Example
Use convolutions to find f satisfying L[f (t)] =e−2s
(s − 1)(s2 + 3).
Solution: One way to solve this is with the splitting
L[f (t)] = e−2s 1
(s2 + 3)
1
(s − 1)= e−2s 1√
3
√3
(s2 + 3)
1
(s − 1),
L[f (t)] = e−2s 1√3L[sin
(√3 t
)]L[et ]
L[f (t)] =1√3L[u2(t) sin
(√3 (t − 2)
)]L[et ].
f (t) =1√3
∫ t
0u2(τ) sin
(√3 (τ − 2)
)e(t−τ) dτ. C
Exam: November 12, 2008. Problem 3
ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution:
1
g ( t )
u ( t − pi )sin ( t )
pit
Express g using step functions,
g(t) = uπ(t) sin(t − π).
L[uc(t) f (t − c)] = e−cs L[f (t)].
Therefore,
L[g(t)] = e−πsL[sin(t)].
We obtain: L[g(t)] =e−πs
s2 + 1.
Exam: November 12, 2008. Problem 3
ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution:
1
g ( t )
u ( t − pi )sin ( t )
pit
Express g using step functions,
g(t) = uπ(t) sin(t − π).
L[uc(t) f (t − c)] = e−cs L[f (t)].
Therefore,
L[g(t)] = e−πsL[sin(t)].
We obtain: L[g(t)] =e−πs
s2 + 1.
Exam: November 12, 2008. Problem 3
ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution:
1
g ( t )
u ( t − pi )sin ( t )
pit
Express g using step functions,
g(t) = uπ(t) sin(t − π).
L[uc(t) f (t − c)] = e−cs L[f (t)].
Therefore,
L[g(t)] = e−πsL[sin(t)].
We obtain: L[g(t)] =e−πs
s2 + 1.
Exam: November 12, 2008. Problem 3
ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution:
1
g ( t )
u ( t − pi )sin ( t )
pit
Express g using step functions,
g(t) = uπ(t) sin(t − π).
L[uc(t) f (t − c)] = e−cs L[f (t)].
Therefore,
L[g(t)] = e−πsL[sin(t)].
We obtain: L[g(t)] =e−πs
s2 + 1.
Exam: November 12, 2008. Problem 3
ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution:
1
g ( t )
u ( t − pi )sin ( t )
pit
Express g using step functions,
g(t) = uπ(t) sin(t − π).
L[uc(t) f (t − c)] = e−cs L[f (t)].
Therefore,
L[g(t)] = e−πsL[sin(t)].
We obtain: L[g(t)] =e−πs
s2 + 1.
Exam: November 12, 2008. Problem 3
ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution:
1
g ( t )
u ( t − pi )sin ( t )
pit
Express g using step functions,
g(t) = uπ(t) sin(t − π).
L[uc(t) f (t − c)] = e−cs L[f (t)].
Therefore,
L[g(t)] = e−πsL[sin(t)].
We obtain: L[g(t)] =e−πs
s2 + 1.
Exam: November 12, 2008. Problem 3
ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution:
1
g ( t )
u ( t − pi )sin ( t )
pit
Express g using step functions,
g(t) = uπ(t) sin(t − π).
L[uc(t) f (t − c)] = e−cs L[f (t)].
Therefore,
L[g(t)] = e−πsL[sin(t)].
We obtain: L[g(t)] =e−πs
s2 + 1.
Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: L[g(t)] =e−πs
s2 + 1.
L[y ′′]− 6L[y ] = L[g(t)] =e−πs
s2 + 1.
(s2 − 6)L[y ] =e−πs
s2 + 1⇒ L[y ] = e−πs 1
(s2 + 1)(s2 − 6).
H(s) =1
(s2 + 1)(s2 − 6)=
1
(s2 + 1)(s +√
6)(s −√
6)
H(s) =a
(s +√
6)+
b
(s −√
6)+
(cs + d)
(s2 + 1).
Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: L[g(t)] =e−πs
s2 + 1.
L[y ′′]− 6L[y ] = L[g(t)]
=e−πs
s2 + 1.
(s2 − 6)L[y ] =e−πs
s2 + 1⇒ L[y ] = e−πs 1
(s2 + 1)(s2 − 6).
H(s) =1
(s2 + 1)(s2 − 6)=
1
(s2 + 1)(s +√
6)(s −√
6)
H(s) =a
(s +√
6)+
b
(s −√
6)+
(cs + d)
(s2 + 1).
Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: L[g(t)] =e−πs
s2 + 1.
L[y ′′]− 6L[y ] = L[g(t)] =e−πs
s2 + 1.
(s2 − 6)L[y ] =e−πs
s2 + 1⇒ L[y ] = e−πs 1
(s2 + 1)(s2 − 6).
H(s) =1
(s2 + 1)(s2 − 6)=
1
(s2 + 1)(s +√
6)(s −√
6)
H(s) =a
(s +√
6)+
b
(s −√
6)+
(cs + d)
(s2 + 1).
Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: L[g(t)] =e−πs
s2 + 1.
L[y ′′]− 6L[y ] = L[g(t)] =e−πs
s2 + 1.
(s2 − 6)L[y ] =e−πs
s2 + 1
⇒ L[y ] = e−πs 1
(s2 + 1)(s2 − 6).
H(s) =1
(s2 + 1)(s2 − 6)=
1
(s2 + 1)(s +√
6)(s −√
6)
H(s) =a
(s +√
6)+
b
(s −√
6)+
(cs + d)
(s2 + 1).
Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: L[g(t)] =e−πs
s2 + 1.
L[y ′′]− 6L[y ] = L[g(t)] =e−πs
s2 + 1.
(s2 − 6)L[y ] =e−πs
s2 + 1⇒ L[y ] = e−πs 1
(s2 + 1)(s2 − 6).
H(s) =1
(s2 + 1)(s2 − 6)=
1
(s2 + 1)(s +√
6)(s −√
6)
H(s) =a
(s +√
6)+
b
(s −√
6)+
(cs + d)
(s2 + 1).
Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: L[g(t)] =e−πs
s2 + 1.
L[y ′′]− 6L[y ] = L[g(t)] =e−πs
s2 + 1.
(s2 − 6)L[y ] =e−πs
s2 + 1⇒ L[y ] = e−πs 1
(s2 + 1)(s2 − 6).
H(s) =1
(s2 + 1)(s2 − 6)
=1
(s2 + 1)(s +√
6)(s −√
6)
H(s) =a
(s +√
6)+
b
(s −√
6)+
(cs + d)
(s2 + 1).
Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: L[g(t)] =e−πs
s2 + 1.
L[y ′′]− 6L[y ] = L[g(t)] =e−πs
s2 + 1.
(s2 − 6)L[y ] =e−πs
s2 + 1⇒ L[y ] = e−πs 1
(s2 + 1)(s2 − 6).
H(s) =1
(s2 + 1)(s2 − 6)=
1
(s2 + 1)(s +√
6)(s −√
6)
H(s) =a
(s +√
6)+
b
(s −√
6)+
(cs + d)
(s2 + 1).
Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: L[g(t)] =e−πs
s2 + 1.
L[y ′′]− 6L[y ] = L[g(t)] =e−πs
s2 + 1.
(s2 − 6)L[y ] =e−πs
s2 + 1⇒ L[y ] = e−πs 1
(s2 + 1)(s2 − 6).
H(s) =1
(s2 + 1)(s2 − 6)=
1
(s2 + 1)(s +√
6)(s −√
6)
H(s) =a
(s +√
6)+
b
(s −√
6)+
(cs + d)
(s2 + 1).
Exam: November 12, 2008. Problem 3
ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: H(s) =a
(s +√
6)+
b
(s −√
6)+
(cs + d)
(s2 + 1).
1
(s2 + 1)(s +√
6)(s −√
6)=
a
(s +√
6)+
b
(s −√
6)+
(cs + d)
(s2 + 1)
1 = a(s −√
6)(s2 + 1) + b(s +√
6)(s2 + 1) + (cs + d)(s2 − 6).
The solution is: a = − 1
14√
6, b =
1
14√
6, c = 0, d = −1
7.
Exam: November 12, 2008. Problem 3
ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: H(s) =a
(s +√
6)+
b
(s −√
6)+
(cs + d)
(s2 + 1).
1
(s2 + 1)(s +√
6)(s −√
6)=
a
(s +√
6)+
b
(s −√
6)+
(cs + d)
(s2 + 1)
1 = a(s −√
6)(s2 + 1) + b(s +√
6)(s2 + 1) + (cs + d)(s2 − 6).
The solution is: a = − 1
14√
6, b =
1
14√
6, c = 0, d = −1
7.
Exam: November 12, 2008. Problem 3
ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: H(s) =a
(s +√
6)+
b
(s −√
6)+
(cs + d)
(s2 + 1).
1
(s2 + 1)(s +√
6)(s −√
6)=
a
(s +√
6)+
b
(s −√
6)+
(cs + d)
(s2 + 1)
1 = a(s −√
6)(s2 + 1) + b(s +√
6)(s2 + 1) + (cs + d)(s2 − 6).
The solution is: a = − 1
14√
6, b =
1
14√
6, c = 0, d = −1
7.
Exam: November 12, 2008. Problem 3
ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: H(s) =a
(s +√
6)+
b
(s −√
6)+
(cs + d)
(s2 + 1).
1
(s2 + 1)(s +√
6)(s −√
6)=
a
(s +√
6)+
b
(s −√
6)+
(cs + d)
(s2 + 1)
1 = a(s −√
6)(s2 + 1) + b(s +√
6)(s2 + 1) + (cs + d)(s2 − 6).
The solution is: a = − 1
14√
6, b =
1
14√
6, c = 0, d = −1
7.
Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: H(s) =1
14√
6
[− 1
(s +√
6)+
1
(s −√
6)− 2
√6
(s2 + 1)
].
H(s) =1
14√
6
[−L
[e−
√6 t
]+ L
[e√
6 t]− 2√
6L[sin(t)]]
H(s) = L[ 1
14√
6
(−e−
√6 t + e
√6 t − 2
√6 sin(t)
)].
h(t) =1
14√
6
[−e−
√6 t + e
√6 t − 2
√6 sin(t)
]⇒ H(s) = L[h(t)].
Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: H(s) =1
14√
6
[− 1
(s +√
6)+
1
(s −√
6)− 2
√6
(s2 + 1)
].
H(s) =1
14√
6
[−L
[e−
√6 t
]+ L
[e√
6 t]− 2√
6L[sin(t)]]
H(s) = L[ 1
14√
6
(−e−
√6 t + e
√6 t − 2
√6 sin(t)
)].
h(t) =1
14√
6
[−e−
√6 t + e
√6 t − 2
√6 sin(t)
]⇒ H(s) = L[h(t)].
Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: H(s) =1
14√
6
[− 1
(s +√
6)+
1
(s −√
6)− 2
√6
(s2 + 1)
].
H(s) =1
14√
6
[−L
[e−
√6 t
]+ L
[e√
6 t]− 2√
6L[sin(t)]]
H(s) = L[ 1
14√
6
(−e−
√6 t + e
√6 t − 2
√6 sin(t)
)].
h(t) =1
14√
6
[−e−
√6 t + e
√6 t − 2
√6 sin(t)
]⇒ H(s) = L[h(t)].
Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: H(s) =1
14√
6
[− 1
(s +√
6)+
1
(s −√
6)− 2
√6
(s2 + 1)
].
H(s) =1
14√
6
[−L
[e−
√6 t
]+ L
[e√
6 t]− 2√
6L[sin(t)]]
H(s) = L[ 1
14√
6
(−e−
√6 t + e
√6 t − 2
√6 sin(t)
)].
h(t) =1
14√
6
[−e−
√6 t + e
√6 t − 2
√6 sin(t)
]
⇒ H(s) = L[h(t)].
Exam: November 12, 2008. Problem 3ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: H(s) =1
14√
6
[− 1
(s +√
6)+
1
(s −√
6)− 2
√6
(s2 + 1)
].
H(s) =1
14√
6
[−L
[e−
√6 t
]+ L
[e√
6 t]− 2√
6L[sin(t)]]
H(s) = L[ 1
14√
6
(−e−
√6 t + e
√6 t − 2
√6 sin(t)
)].
h(t) =1
14√
6
[−e−
√6 t + e
√6 t − 2
√6 sin(t)
]⇒ H(s) = L[h(t)].
Exam: November 12, 2008. Problem 3
ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: Recall: L[y ] = e−πs H(s), where H(s) = L[h(t)], and
h(t) =1
14√
6
[−e−
√6 t + e
√6 t − 2
√6 sin(t)
].
L[y ] = e−πs L[h(t)] = L[uπ(t) h(t−π)] ⇒ y(t) = uπ(t) h(t−π).
Equivalently:
y(t) =uπ(t)
14√
6
[−e−
√6 (t−π) + e
√6 (t−π) − 2
√6 sin(t − π)
]. C
Exam: November 12, 2008. Problem 3
ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: Recall: L[y ] = e−πs H(s), where H(s) = L[h(t)], and
h(t) =1
14√
6
[−e−
√6 t + e
√6 t − 2
√6 sin(t)
].
L[y ] = e−πs L[h(t)]
= L[uπ(t) h(t−π)] ⇒ y(t) = uπ(t) h(t−π).
Equivalently:
y(t) =uπ(t)
14√
6
[−e−
√6 (t−π) + e
√6 (t−π) − 2
√6 sin(t − π)
]. C
Exam: November 12, 2008. Problem 3
ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: Recall: L[y ] = e−πs H(s), where H(s) = L[h(t)], and
h(t) =1
14√
6
[−e−
√6 t + e
√6 t − 2
√6 sin(t)
].
L[y ] = e−πs L[h(t)] = L[uπ(t) h(t−π)]
⇒ y(t) = uπ(t) h(t−π).
Equivalently:
y(t) =uπ(t)
14√
6
[−e−
√6 (t−π) + e
√6 (t−π) − 2
√6 sin(t − π)
]. C
Exam: November 12, 2008. Problem 3
ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: Recall: L[y ] = e−πs H(s), where H(s) = L[h(t)], and
h(t) =1
14√
6
[−e−
√6 t + e
√6 t − 2
√6 sin(t)
].
L[y ] = e−πs L[h(t)] = L[uπ(t) h(t−π)] ⇒ y(t) = uπ(t) h(t−π).
Equivalently:
y(t) =uπ(t)
14√
6
[−e−
√6 (t−π) + e
√6 (t−π) − 2
√6 sin(t − π)
]. C
Exam: November 12, 2008. Problem 3
ExampleSketch the graph of g and use LT to find y solution of
y ′′ − 6y = g(t), y(0) = y ′(0) = 0, g(t) =
{0, t < π,
sin(t − π), t > π.
Solution: Recall: L[y ] = e−πs H(s), where H(s) = L[h(t)], and
h(t) =1
14√
6
[−e−
√6 t + e
√6 t − 2
√6 sin(t)
].
L[y ] = e−πs L[h(t)] = L[uπ(t) h(t−π)] ⇒ y(t) = uπ(t) h(t−π).
Equivalently:
y(t) =uπ(t)
14√
6
[−e−
√6 (t−π) + e
√6 (t−π) − 2
√6 sin(t − π)
]. C