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Calculus For Biologists
Linear Systems of ODE: Solving the IVP
James K. Peterson
Department of Biological Sciences and Department of Mathematical SciencesClemson University
October 6, 2013
Calculus For Biologists
Outline
1 Linear Systems of ODE
2 The Characteristic EquationExamples
3 Finding The General SolutionExamples
Calculus For Biologists
Abstract
This lecture is about linear systems of ODE and how to solve theresulting IVP.
Calculus For Biologists
Linear Systems of ODE
We are now ready to solve what are called Linear Systems of firstorder ODE. These have the form
x ′(t) = a x(t) + b y(t)
y ′(t) = c x(t) + d y(t)
x(0) = x0
y(0) = y0
for any numbers a, b, c and d and initial conditions x0 and y0.The full problem is called, as usual, an Initial Value Problem orIVP for short. The two initial conditions are just called the IC’s forthe problem to save writing.
Calculus For Biologists
Linear Systems of ODE
For example, we might be interested in the system
x ′(t) = −2 x(t) + 3 y(t)
y ′(t) = 4 x(t) + 5 y(t)
x(0) = 5
y(0) = −3
Here the IC’s are x(0) = 5 and y(0) = −3.
Another sample problem might be the one below.
x ′(t) = 14 x(t) + 5 y(t)
y ′(t) = −4 x(t) + 8 y(t)
x(0) = 2
y(0) = 7
Calculus For Biologists
The Characteristic Equation
For linear first order problems like u′ = 3u and so forth, wefind the solution has the form u(t) = A e3t for some numberA. We then determine the value of A to use by looking at theinitial condition.
To find the solutions here, we begin by rewriting the model inmatrix - vector notation.[
x ′(t)y ′(t)
]=
[a bc d
] [x(t)y(t)
].
The 2× 2 matrix is called the coefficient matrix of thismodel.
The initial conditions can then be redone in vector form as[x(0)y(0)
]=
[x0y0
].
Calculus For Biologists
The Characteristic Equation
Now it seems reasonable to believe that if a constant times ert
solves a first order linear problem like u′ = ru, perhaps a vectortimes ert will work here. Let’s make this formal. So let’s look atthe problem below
x ′(t) = 3 x(t) + 2 y(t)
y ′(t) = −4 x(t) + 5 y(t)
x(0) = 2
y(0) = −3
Assume the solution has the form V ert
Let’s denote the components of V as follows:
V =
[V1
V2
].
Calculus For Biologists
The Characteristic Equation
We assume the solution is[x(t)y(t)
]= V ert .
Then the derivative of V ert is(V ert
)′=
([V1
V2
]ert)′
=
([V1 ert
V2 ert
] )′=
[V1 (ert)′
V2 (ert)′
]=
[V1 r ert
V2 r ert
]=
[V1
V2
]r ert = r V ert
Hence,
[x ′(t)y ′(t)
]= r V ert .
Calculus For Biologists
The Characteristic Equation
When we plug these terms into the matrix - vector form of theproblem, we find
r V ert =
[3 2−4 5
]V ert
Rewrite as
r V ert −[
3 2−4 5
]V ert =
[00
].
Recall that the 2× 2 identity matrix I has the form
I =
[1 00 1
]⇒ I V = V.
Calculus For Biologists
The Characteristic Equation
So
r V ert −[
3 2−4 5
]V ert = r I V ert −
[3 2−4 5
]V ert
=
(r I −
[3 2−4 5
])V ert
=
([r 00 r
]−[
3 2−4 5
])V ert
=
[r − 3 −2−(−4) r − 5
]V ert
Plugging this into our model, we find[r − 3 −2
4 r − 5
]V ert =
[00
].
Calculus For Biologists
The Characteristic Equation
But ert is never 0, so we want r satisfying[r − 3 −2
4 r − 5
]V =
[00
].
For each r , we get two equations in V1 and V2:
(r − 3)V1 − 2V2 = 0, 4V1 + (r − 5)V2 = 0.
Let Ar be this matrix.Any r for which the det Ar 6= 0 tells us these two lines havedifferent slopes and so cross at the origin implying V1 = 0 andV2 = 0. Thus [
xy
]=
[00
]ert =
[00
].
which will not satisfy nonzero initial conditions. So rejectthese r .Any value of r for which det Ar = 0 gives an infinite number ofsolutions which allows us to pick one that matches the initialconditions we have.
Calculus For Biologists
The Characteristic Equation
The equation
det(r I− A) = det
[r − 3 −2
4 r − 5
]= 0.
is called the characteristic equation of this linear system.
The characteristic equation is a quadratic, so there arethree possibilities:
two distinct roots – this is the only case we handle in this class.the real roots are the same – not done in this class. The thirdCalculus for Biology class does this one.the roots are complex – not done in this class. The thirdCalculus for Biology class does this one.
Calculus For Biologists
The Characteristic Equation
Examples
Example
Derive the characteristic equation for the system below
x ′(t) = 8 x(t) + 9 y(t)
y ′(t) = 3 x(t) − 2 y(t)
x(0) = 12
y(0) = 4
Solution
The matrix - vector form is[x ′(t)y ′(t)
]=
[8 93 −2
] [x(t)y(t)
].[
x(0)y(0)
]=
[12
4
].
Calculus For Biologists
The Characteristic Equation
Examples
Solution
The coefficient matrix A is thus
A =
[8 93 −2
].
Assume the solution has the form V ert
Plug this into the system.
r V ert −[
8 93 −2
]V ert =
[00
].
Rewrite using the identity matrix I and factor(r I −
[8 93 −2
])V ert =
[00
].
Calculus For Biologists
The Characteristic Equation
Examples
Solution
Since ert 6= 0 ever, we find r and V satisfy(r I −
[8 93 −2
])V =
[00
].
If r is chosen so that det (rI − A) 6= 0, the only solution tothis system of two linear equations in the two unknowns V1
and V2 is V1 = 0 and V2 = 0. This leads to x(t) = 0 andy(t) = 0 always and this solution does not satisfy the initialconditions. Hence, we must find r which givedet (rI − A) = 0. The characteristic equation is thus
det
[r − 8 −9−3 r + 2
]= (r − 8)(r + 2)− 27
= r2 − 6r − 43 = 0.
Calculus For Biologists
The Characteristic Equation
Examples
Example
Derive the characteristic equation for the system below
x ′(t) = −10 x(t) − 7 y(t)
y ′(t) = 8 x(t) + 5 y(t)
x(0) = −1
y(0) = −4
Solution
The matrix - vector form is[x ′(t)y ′(t)
]=
[−10 −7
8 5
] [x(t)y(t)
].[
x(0)y(0)
]=
[−1−4
].
Calculus For Biologists
The Characteristic Equation
Examples
Solution
The coefficient matrix A is thus
A =
[−10 −7
8 5
].
Assume the solution has the form V ert
Plug this into the system.
r V ert −[−10 −7
8 5
]V ert =
[00
].
Rewrite using the identity matrix I and factor(r I −
[−10 −7
8 5
])V ert =
[00
].
Calculus For Biologists
The Characteristic Equation
Examples
Solution
Since ert 6= 0 ever, we find r and V satisfy(r I −
[−10 −7
8 5
])V =
[00
].
If r is chosen so that det (rI − A) 6= 0, the only solution tothis system of two linear equations in the two unknowns V1
and V2 is V1 = 0 and V2 = 0. This leads to x(t) = 0 andy(t) = 0 always and this solution does not satisfy the initialconditions. Hence, we must find r which givedet (rI − A) = 0. The characteristic equation is thus
det
[r + 10 7−8 r − 5
]= (r + 10)(r − 5) + 56
= r2 + 5r + 6 = 0.
Calculus For Biologists
The Characteristic Equation
Examples
Homework 50
For each of these problems,
Write the matrix - vector form.Derive the characteristic equation but you don’t have to findthe roots.
50.1
x ′ = 2 x + 3 y
y ′ = 8 x − 2 y
x(0) = 3
y(0) = 5.
50.2
x ′ = −4 x + 6 y
y ′ = 9 x + 2 y
x(0) = 4
y(0) = −6.
Calculus For Biologists
Finding The General Solution
The roots to the characteristic equation are calledeigenvalues.
Organize the eigenvalues as r1 < r2.
Example: The eigenvalues are −2 and −1. So r1 = −2 andr2 = −1. Since e−2t decays faster than e−t , we say the rootr2 = −1 is the dominant part of the solution.Example: The eigenvalues are −2 and 3. So r1 = −2 andr2 = 3. Since e−2t decays and e3t grows, we say the rootr2 = 3 is the dominant part of the solution.Example: The eigenvalues are 2 and 3. So r1 = 2 and r2 = 3.Since e2t grows slower than e3t , we say the root r2 = 3 is thedominant part of the solution.
For each eigenvalue r we want to find nonzero vectors V sothat (r I − A) V = 0 where to help with our writing we let 0be the two dimensional zero vector.
These nonzero V are called the eigenvectors for eigenvaluer and satisfy AV = rV.
Calculus For Biologists
Finding The General Solution
The roots to the characteristic equation are calledeigenvalues.
Organize the eigenvalues as r1 < r2.
Example: The eigenvalues are −2 and −1. So r1 = −2 andr2 = −1. Since e−2t decays faster than e−t , we say the rootr2 = −1 is the dominant part of the solution.Example: The eigenvalues are −2 and 3. So r1 = −2 andr2 = 3. Since e−2t decays and e3t grows, we say the rootr2 = 3 is the dominant part of the solution.Example: The eigenvalues are 2 and 3. So r1 = 2 and r2 = 3.Since e2t grows slower than e3t , we say the root r2 = 3 is thedominant part of the solution.
For each eigenvalue r we want to find nonzero vectors V sothat (r I − A) V = 0 where to help with our writing we let 0be the two dimensional zero vector.
These nonzero V are called the eigenvectors for eigenvaluer and satisfy AV = rV.
Calculus For Biologists
Finding The General Solution
The roots to the characteristic equation are calledeigenvalues.
Organize the eigenvalues as r1 < r2.
Example: The eigenvalues are −2 and −1. So r1 = −2 andr2 = −1. Since e−2t decays faster than e−t , we say the rootr2 = −1 is the dominant part of the solution.Example: The eigenvalues are −2 and 3. So r1 = −2 andr2 = 3. Since e−2t decays and e3t grows, we say the rootr2 = 3 is the dominant part of the solution.Example: The eigenvalues are 2 and 3. So r1 = 2 and r2 = 3.Since e2t grows slower than e3t , we say the root r2 = 3 is thedominant part of the solution.
For each eigenvalue r we want to find nonzero vectors V sothat (r I − A) V = 0 where to help with our writing we let 0be the two dimensional zero vector.
These nonzero V are called the eigenvectors for eigenvaluer and satisfy AV = rV.
Calculus For Biologists
Finding The General Solution
For eigenvalue r1Find V so that (r1 I − A) V = 0There will be an infinite number of V’s that solve this; we pickone and call it eigenvector E1.
For eigenvalue r2Find V so that (r2 I − A) V = 0There will again be an infinite number of V’s that solve this;we pick one and call it eigenvector E2.
The general solution to our model will be[x(t)y(t)
]= AE1 er1t + BE2 er2t .
where A and B are arbitrary. We use the ICs to find A and B.
Best to show all this with some examples.
Calculus For Biologists
Finding The General Solution
Examples
Example
For the system below[x ′(t)y ′(t)
]=
[−20 12−13 5
] [x(t)y(t)
][
x(0)y(0)
]=
[−1
2
]
Find the characteristic equation
Find the general solution
Solve the IVP
Calculus For Biologists
Finding The General Solution
Examples
Solution
The characteristic equation is
det
(r
[1 00 1
]−[−20 12−13 5
])= 0
Thus
0 = det
([r + 20 −12
13 r − 5
])= (r + 20)(r − 5) + 156
= r2 + 15r + 56
= (r + 8)(r + 7)
Hence, eigenvalues or roots of the characteristic equationare r1 = −8 and r2 = −7.
Calculus For Biologists
Finding The General Solution
Examples
First, for eigenvalue r1 = −8, let’s find our eigenvector.
Solution
For eigenvalue r1 = −8, substitute the value into[r + 20 −12
13 r − 5
] [V1
V2
]⇒[
12 −1213 −13
] [V1
V2
]This system of equations should be collinear: i.e. the rowsshould be multiples; i.e. both give rise to the same line. Ourrows are multiples, so we can pick any row to find V2 in termsof V1. Picking the top row, we get 12V1 − 12V2 = 0implying V2 = V1.
Letting V1 = a, we find V1 = a and V2 = a: so[V1
V2
]= a
[11
]
Calculus For Biologists
Finding The General Solution
Examples
Solution
Choose E1:
The vector
E1 =
[11
]is our choice for an eigenvector corresponding to eigenvaluer1 = −8.
So one of the solutions is[x1(t)y1(t)
]= E1 e−8t =
[11
]e−8t .
Calculus For Biologists
Finding The General Solution
Examples
First, for eigenvalue r2 = −7, let’s find our eigenvector.
Solution
For eigenvalue r2 = −7, substitute the value into[r + 20 −12
13 r − 5
] [V1
V2
]⇒[
13 −1213 −12
] [V1
V2
]This system of equations should be collinear: i.e. the rowsshould be multiples; i.e. both give rise to the same line. Ourrows are multiples, so we can pick any row to find V2 in termsof V1. Picking the top row, we get 13V1 − 12V2 = 0implying V2 = (13/12)V1.
Letting V1 = b, we find V1 = b and V2 = (13/12)b: so[V1
V2
]= b
[1
1312
]
Calculus For Biologists
Finding The General Solution
Examples
Solution
Choose E2:
The vector
E2 =
[1
1312
]is our choice for an eigenvector corresponding to eigenvaluer2 = −7.
So one of the solutions is[x2(t)y2(t)
]= E2 e−7t =
[1
1312
]e−7t .
Calculus For Biologists
Finding The General Solution
Examples
Solution
The general solution:
[x(t)y(t)
]= A E1 e−8t + B E2 e−7t
= A
[11
]e−8t + B
[1
1312
]e−7t .
Find A and B: use the ICs.[x(0)y(0)
]=
[−1
2
]= A
[11
]e0 + B
[1
1312
]e0
= A
[11
]+ B
[1
1312
].
Calculus For Biologists
Finding The General Solution
Examples
Solution
So
A + B = −1
A +13
12B = 2
Subtracting the bottom equation from the top equation, weget − 1
12B = −3 or B = 36. Thus, A = −1− B = −37. So[x(t)y(t)
]= −37
[11
]e−8t + 36
[1
1312
]e−7t
Calculus For Biologists
Finding The General Solution
Examples
Example
For the system below[x ′(t)y ′(t)
]=
[4 9−1 −6
] [x(t)y(t)
][
x(0)y(0)
]=
[4−2
]
Find the characteristic equation
Find the general solution
Solve the IVP
Calculus For Biologists
Finding The General Solution
Examples
Solution
The characteristic equation is
det
(r
[1 00 1
]−[
4 9−1 −6
])= 0
Thus
0 = det
([r − 4 −9
1 r + 6
])= (r − 4)(r + 6) + 9
= r2 + 2r − 15
= (r + 5)(r − 3)
Hence, eigenvalues or roots of the characteristic equationare r1 = −5 and r2 = 3.
Calculus For Biologists
Finding The General Solution
Examples
First, for eigenvalue r1 = −5, let’s find our eigenvector.
Solution
For eigenvalue r1 = −5, substitute the value into[r − 4 −9
1 r + 6
] [V1
V2
]⇒[−9 −9
1 1
] [V1
V2
]This system of equations should be collinear: i.e. the rowsshould be multiples; i.e. both give rise to the same line. Ourrows are multiples, so we can pick any row to find V2 in termsof V1. Picking the bottom row, we get V1 + V2 = 0 implyingV2 = −V1.
Letting V1 = a, we find V1 = a and V2 = −a: so[V1
V2
]= a
[1−1
]
Calculus For Biologists
Finding The General Solution
Examples
Solution
Choose E1:
The vector
E1 =
[1−1
]is our choice for an eigenvector corresponding to eigenvaluer1 = −5.
So one of the solutions is[x1(t)y1(t)
]= E1 e−5t =
[1−1
]e−5t .
Calculus For Biologists
Finding The General Solution
Examples
First, for eigenvalue r2 = 3, let’s find our eigenvector.
Solution
For eigenvalue r2 = 3, substitute the value into[r − 4 −9
1 r + 6
] [V1
V2
]⇒[−1 −9
1 9
] [V1
V2
]This system of equations should be collinear: i.e. the rowsshould be multiples; i.e. both give rise to the same line. Ourrows are multiples, so we can pick any row to find V2 in termsof V1. Picking the bottom row, we get V1 + 9V2 = 0implying V2 = −(1/9)V1.
Letting V1 = b, we find V1 = b and V2 = −(1/9)b: so[V1
V2
]= b
[1−1
9
]
Calculus For Biologists
Finding The General Solution
Examples
Solution
Choose E2:
The vector
E2 =
[1−1
9
]is our choice for an eigenvector corresponding to eigenvaluer2 = 3.
So one of the solutions is[x2(t)y2(t)
]= E2 e3t =
[1−1
9
]e3t .
Calculus For Biologists
Finding The General Solution
Examples
Solution
The general solution:
[x(t)y(t)
]= A E1 e−5t + B E2 e3t
= A
[1−1
]e−5t + B
[1−1
9
]e3t .
Find A and B: use the ICs.[x(0)y(0)
]=
[4−2
]= A
[1−1
]e0 + B
[1−1
9
]e0
= A
[1−1
]+ B
[1−1
9
].
Calculus For Biologists
Finding The General Solution
Examples
Solution
So
A + B = 4
−A− 1
9B = −2
Adding the bottom equation and the top equation, we get89B = 2 or B = 9/4. Thus, A = 4− B = 7/4. So[
x(t)y(t)
]=
7
4
[1−1
]e−5t +
9
4
[1−1
9
]e3t
Calculus For Biologists
Finding The General Solution
Examples
Homework 51
For these problems:
write matrix, vector form.
find characteristic equation. No derivation needed this time.
find the two eigenvalues.
find the two associated eigenvectors in glorious detail.
Write general solution.
solve the IVP.
51.1
x ′ = 3 x + y
y ′ = 5 x − y
x(0) = 4
y(0) = −6.
The eigenvalues should be −2 and 4.
Calculus For Biologists
Finding The General Solution
Examples
Homework 51 Continued
51.2
x ′ = x + 4 y
y ′ = 5 x + 2 y
x(0) = 4
y(0) = −5.
The eigenvalues should be −3 and 6.
51.3
x ′ = −3 x + y
y ′ = −4 x + 2 y
x(0) = 1
y(0) = 6.
The eigenvalues should be −2 and 1.