linear homogeneous constant coefficient differential equations
TRANSCRIPT
logo1
Overview Examples
Linear Homogeneous ConstantCoefficient Differential Equations
Bernd Schroder
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Equation Type and Solution Method
We will focus on linear homogeneous constant coefficientdifferential equations of second order, because they areencountered most frequently.
1. The general form is ay′′+by′+ cy = 0, where a,b,c arereal constants.
2. Substituting y = eλx into the equation leads to the equationaλ 2 +bλ + c = 0 for λ .2.1 If the equation has real solutions λ1 6= λ2, then the general
solution is y = c1eλ1x + c2eλ2x.2.2 If the equation has complex solutions λ1,2 = u± iv, then
the general solution is y = c1eux cos(vx)+ c2eux sin(vx).2.3 If the equation has only one real solution λ , then the
general solution is y = c1eλx + c2xeλx.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Equation Type and Solution MethodWe will focus on linear homogeneous constant coefficientdifferential equations of second order, because they areencountered most frequently.
1. The general form is ay′′+by′+ cy = 0, where a,b,c arereal constants.
2. Substituting y = eλx into the equation leads to the equationaλ 2 +bλ + c = 0 for λ .2.1 If the equation has real solutions λ1 6= λ2, then the general
solution is y = c1eλ1x + c2eλ2x.2.2 If the equation has complex solutions λ1,2 = u± iv, then
the general solution is y = c1eux cos(vx)+ c2eux sin(vx).2.3 If the equation has only one real solution λ , then the
general solution is y = c1eλx + c2xeλx.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Equation Type and Solution MethodWe will focus on linear homogeneous constant coefficientdifferential equations of second order, because they areencountered most frequently.
1. The general form is ay′′+by′+ cy = 0, where a,b,c arereal constants.
2. Substituting y = eλx into the equation leads to the equationaλ 2 +bλ + c = 0 for λ .
2.1 If the equation has real solutions λ1 6= λ2, then the generalsolution is y = c1eλ1x + c2eλ2x.
2.2 If the equation has complex solutions λ1,2 = u± iv, thenthe general solution is y = c1eux cos(vx)+ c2eux sin(vx).
2.3 If the equation has only one real solution λ , then thegeneral solution is y = c1eλx + c2xeλx.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Equation Type and Solution MethodWe will focus on linear homogeneous constant coefficientdifferential equations of second order, because they areencountered most frequently.
1. The general form is ay′′+by′+ cy = 0, where a,b,c arereal constants.
2. Substituting y = eλx into the equation leads to the equationaλ 2 +bλ + c = 0 for λ .2.1 If the equation has real solutions λ1 6= λ2, then the general
solution is y = c1eλ1x + c2eλ2x.
2.2 If the equation has complex solutions λ1,2 = u± iv, thenthe general solution is y = c1eux cos(vx)+ c2eux sin(vx).
2.3 If the equation has only one real solution λ , then thegeneral solution is y = c1eλx + c2xeλx.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Equation Type and Solution MethodWe will focus on linear homogeneous constant coefficientdifferential equations of second order, because they areencountered most frequently.
1. The general form is ay′′+by′+ cy = 0, where a,b,c arereal constants.
2. Substituting y = eλx into the equation leads to the equationaλ 2 +bλ + c = 0 for λ .2.1 If the equation has real solutions λ1 6= λ2, then the general
solution is y = c1eλ1x + c2eλ2x.2.2 If the equation has complex solutions λ1,2 = u± iv, then
the general solution is y = c1eux cos(vx)+ c2eux sin(vx).
2.3 If the equation has only one real solution λ , then thegeneral solution is y = c1eλx + c2xeλx.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Equation Type and Solution MethodWe will focus on linear homogeneous constant coefficientdifferential equations of second order, because they areencountered most frequently.
1. The general form is ay′′+by′+ cy = 0, where a,b,c arereal constants.
2. Substituting y = eλx into the equation leads to the equationaλ 2 +bλ + c = 0 for λ .2.1 If the equation has real solutions λ1 6= λ2, then the general
solution is y = c1eλ1x + c2eλ2x.2.2 If the equation has complex solutions λ1,2 = u± iv, then
the general solution is y = c1eux cos(vx)+ c2eux sin(vx).2.3 If the equation has only one real solution λ , then the
general solution is y = c1eλx + c2xeλx.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of 2y′′+5y′+2y = 0.
2y′′+5y′+2y = 0 y = eλx
2(
eλx)′′
+5(
eλx)′
+2(
eλx)
= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of 2y′′+5y′+2y = 0.
2y′′+5y′+2y = 0
y = eλx
2(
eλx)′′
+5(
eλx)′
+2(
eλx)
= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of 2y′′+5y′+2y = 0.
2y′′+5y′+2y = 0 y = eλx
2(
eλx)′′
+5(
eλx)′
+2(
eλx)
= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of 2y′′+5y′+2y = 0.
2y′′+5y′+2y = 0 y = eλx
2(
eλx)′′
+5(
eλx)′
+2(
eλx)
= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of 2y′′+5y′+2y = 0.
2y′′+5y′+2y = 0 y = eλx
2(
eλx)′′
+5(
eλx)′
+2eλx = 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of 2y′′+5y′+2y = 0.
2y′′+5y′+2y = 0 y = eλx
2(
eλx)′′
+5λeλx +2eλx = 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of 2y′′+5y′+2y = 0.
2y′′+5y′+2y = 0 y = eλx
2λ2eλx +5λeλx +2eλx = 0
2λ2 +5λ +2 = 0
λ1,2 =−5±
√25−16
4
=−5±3
4
= −12,−2
y = c1e−12 x + c2e−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of 2y′′+5y′+2y = 0.
2y′′+5y′+2y = 0 y = eλx
2λ2eλx +5λeλx +2eλx = 0
2λ2 +5λ +2 = 0
λ1,2 =−5±
√25−16
4
=−5±3
4
= −12,−2
y = c1e−12 x + c2e−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of 2y′′+5y′+2y = 0.
2y′′+5y′+2y = 0 y = eλx
2λ2eλx +5λeλx +2eλx = 0
2λ2 +5λ +2 = 0
λ1,2 =−5±
√25−16
4
=−5±3
4
= −12,−2
y = c1e−12 x + c2e−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of 2y′′+5y′+2y = 0.
2y′′+5y′+2y = 0 y = eλx
2λ2eλx +5λeλx +2eλx = 0
2λ2 +5λ +2 = 0
λ1,2 =−5±
√25−16
4
=−5±3
4
= −12,−2
y = c1e−12 x + c2e−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of 2y′′+5y′+2y = 0.
2y′′+5y′+2y = 0 y = eλx
2λ2eλx +5λeλx +2eλx = 0
2λ2 +5λ +2 = 0
λ1,2 =−5±
√25−16
4
=−5±3
4
= −12,−2
y = c1e−12 x + c2e−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of 2y′′+5y′+2y = 0.
2y′′+5y′+2y = 0 y = eλx
2λ2eλx +5λeλx +2eλx = 0
2λ2 +5λ +2 = 0
λ1,2 =−5±
√25−16
4
=−5±3
4
= −12,−2
y = c1e−12 x + c2e−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−12x + c2e−2x Really Solve
2y′′+5y′+2y = 0?
2y′′+5y′+2y ?= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−12x + c2e−2x Really Solve
2y′′+5y′+2y = 0?
2y′′+5y′+2y ?= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−12x + c2e−2x Really Solve
2y′′+5y′+2y = 0?
2y′′+5y′+2(
c1e−12 x+ c2e−2x
)?= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−12x + c2e−2x Really Solve
2y′′+5y′+2y = 0?
2y′′+5(−1
2c1e−
12 x−2c2e−2x
)+2
(c1e−
12 x+c2e−2x
)?= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−12x + c2e−2x Really Solve
2y′′+5y′+2y = 0?
2(
14
c1e−12 x+4c2e−2x
)+5
(−1
2c1e−
12 x−2c2e−2x
)+2
(c1e−
12 x+c2e−2x
)?= 0
12
c1e−12 x+8c2e−2x−5
2c1e−
12 x−10c2e−2x+2c1e−
12 x+2c2e−2x ?= 0(
12− 5
2+2
)c1e−
12 x +(8−10+2)c2e−2x ?= 0
0 = 0√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−12x + c2e−2x Really Solve
2y′′+5y′+2y = 0?
2(
14
c1e−12 x+4c2e−2x
)+5
(−1
2c1e−
12 x−2c2e−2x
)+2
(c1e−
12 x+c2e−2x
)?= 0
12
c1e−12 x+8c2e−2x−5
2c1e−
12 x−10c2e−2x+2c1e−
12 x+2c2e−2x ?= 0
(12− 5
2+2
)c1e−
12 x +(8−10+2)c2e−2x ?= 0
0 = 0√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−12x + c2e−2x Really Solve
2y′′+5y′+2y = 0?
2(
14
c1e−12 x+4c2e−2x
)+5
(−1
2c1e−
12 x−2c2e−2x
)+2
(c1e−
12 x+c2e−2x
)?= 0
12
c1e−12 x+8c2e−2x−5
2c1e−
12 x−10c2e−2x+2c1e−
12 x+2c2e−2x ?= 0(
12− 5
2+2
)c1e−
12 x
+(8−10+2)c2e−2x ?= 0
0 = 0√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−12x + c2e−2x Really Solve
2y′′+5y′+2y = 0?
2(
14
c1e−12 x+4c2e−2x
)+5
(−1
2c1e−
12 x−2c2e−2x
)+2
(c1e−
12 x+c2e−2x
)?= 0
12
c1e−12 x+8c2e−2x−5
2c1e−
12 x−10c2e−2x+2c1e−
12 x+2c2e−2x ?= 0(
12− 5
2+2
)c1e−
12 x +(8−10+2)c2e−2x
?= 0
0 = 0√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−12x + c2e−2x Really Solve
2y′′+5y′+2y = 0?
2(
14
c1e−12 x+4c2e−2x
)+5
(−1
2c1e−
12 x−2c2e−2x
)+2
(c1e−
12 x+c2e−2x
)?= 0
12
c1e−12 x+8c2e−2x−5
2c1e−
12 x−10c2e−2x+2c1e−
12 x+2c2e−2x ?= 0(
12− 5
2+2
)c1e−
12 x +(8−10+2)c2e−2x ?= 0
0 = 0√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−12x + c2e−2x Really Solve
2y′′+5y′+2y = 0?
2(
14
c1e−12 x+4c2e−2x
)+5
(−1
2c1e−
12 x−2c2e−2x
)+2
(c1e−
12 x+c2e−2x
)?= 0
12
c1e−12 x+8c2e−2x−5
2c1e−
12 x−10c2e−2x+2c1e−
12 x+2c2e−2x ?= 0(
12− 5
2+2
)c1e−
12 x +(8−10+2)c2e−2x ?= 0
0 = 0
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−12x + c2e−2x Really Solve
2y′′+5y′+2y = 0?
2(
14
c1e−12 x+4c2e−2x
)+5
(−1
2c1e−
12 x−2c2e−2x
)+2
(c1e−
12 x+c2e−2x
)?= 0
12
c1e−12 x+8c2e−2x−5
2c1e−
12 x−10c2e−2x+2c1e−
12 x+2c2e−2x ?= 0(
12− 5
2+2
)c1e−
12 x +(8−10+2)c2e−2x ?= 0
0 = 0√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of y′′+4y′+5y = 0.
y′′+4y′+5y = 0λ
2eλx +4λeλx +5eλx = 0λ
2 +4λ +5 = 0
λ1,2 =−4±
√16−20
2= −2± i
y = c1e−2x cos(x)+ c2e−2x sin(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of y′′+4y′+5y = 0.
y′′+4y′+5y = 0
λ2eλx +4λeλx +5eλx = 0
λ2 +4λ +5 = 0
λ1,2 =−4±
√16−20
2= −2± i
y = c1e−2x cos(x)+ c2e−2x sin(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of y′′+4y′+5y = 0.
y′′+4y′+5y = 0λ
2eλx +4λeλx +5eλx = 0
λ2 +4λ +5 = 0
λ1,2 =−4±
√16−20
2= −2± i
y = c1e−2x cos(x)+ c2e−2x sin(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of y′′+4y′+5y = 0.
y′′+4y′+5y = 0λ
2eλx +4λeλx +5eλx = 0λ
2 +4λ +5 = 0
λ1,2 =−4±
√16−20
2= −2± i
y = c1e−2x cos(x)+ c2e−2x sin(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of y′′+4y′+5y = 0.
y′′+4y′+5y = 0λ
2eλx +4λeλx +5eλx = 0λ
2 +4λ +5 = 0
λ1,2 =−4±
√16−20
2
= −2± i
y = c1e−2x cos(x)+ c2e−2x sin(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of y′′+4y′+5y = 0.
y′′+4y′+5y = 0λ
2eλx +4λeλx +5eλx = 0λ
2 +4λ +5 = 0
λ1,2 =−4±
√16−20
2= −2± i
y = c1e−2x cos(x)+ c2e−2x sin(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Find the general solution of y′′+4y′+5y = 0.
y′′+4y′+5y = 0λ
2eλx +4λeλx +5eλx = 0λ
2 +4λ +5 = 0
λ1,2 =−4±
√16−20
2= −2± i
y = c1e−2x cos(x)+ c2e−2x sin(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−2x cos(x)+ c2e−2x sin(x) ReallySolve y′′+4y′+5y = 0?
y′′+4y′+5y ?= 0
(e−2x cos(x)
)′′+4
(e−2x cos(x)
)′+5
(e−2x cos(x)
)?= 0(
3e−2x cos(x)+4e−2x sin(x))
+4(−2e−2x cos(x)− e−2x sin(x)
)+5
(e−2x cos(x)
)?= 0
(3−8+5)e−2x cos(x)+(4−4)e−2x sin(x) ?= 00 = 0
√
Check y = e−2x sin(x) yourself.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−2x cos(x)+ c2e−2x sin(x) ReallySolve y′′+4y′+5y = 0?
y′′+4y′+5y ?= 0(e−2x cos(x)
)′′+4
(e−2x cos(x)
)′+5
(e−2x cos(x)
)?= 0
(3e−2x cos(x)+4e−2x sin(x)
)+4
(−2e−2x cos(x)− e−2x sin(x)
)+5
(e−2x cos(x)
)?= 0
(3−8+5)e−2x cos(x)+(4−4)e−2x sin(x) ?= 00 = 0
√
Check y = e−2x sin(x) yourself.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−2x cos(x)+ c2e−2x sin(x) ReallySolve y′′+4y′+5y = 0?
y′′+4y′+5y ?= 0(e−2x cos(x)
)′′+4
(e−2x cos(x)
)′+5
(e−2x cos(x)
)?= 0(
3e−2x cos(x)+4e−2x sin(x))
+4(−2e−2x cos(x)− e−2x sin(x)
)+5
(e−2x cos(x)
)?= 0
(3−8+5)e−2x cos(x)+(4−4)e−2x sin(x) ?= 00 = 0
√
Check y = e−2x sin(x) yourself.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−2x cos(x)+ c2e−2x sin(x) ReallySolve y′′+4y′+5y = 0?
y′′+4y′+5y ?= 0(e−2x cos(x)
)′′+4
(e−2x cos(x)
)′+5
(e−2x cos(x)
)?= 0(
3e−2x cos(x)+4e−2x sin(x))
+4(−2e−2x cos(x)− e−2x sin(x)
)+5
(e−2x cos(x)
)?= 0
(3−8+5)e−2x cos(x)
+(4−4)e−2x sin(x) ?= 00 = 0
√
Check y = e−2x sin(x) yourself.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−2x cos(x)+ c2e−2x sin(x) ReallySolve y′′+4y′+5y = 0?
y′′+4y′+5y ?= 0(e−2x cos(x)
)′′+4
(e−2x cos(x)
)′+5
(e−2x cos(x)
)?= 0(
3e−2x cos(x)+4e−2x sin(x))
+4(−2e−2x cos(x)− e−2x sin(x)
)+5
(e−2x cos(x)
)?= 0
(3−8+5)e−2x cos(x)+(4−4)e−2x sin(x) ?= 0
0 = 0√
Check y = e−2x sin(x) yourself.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−2x cos(x)+ c2e−2x sin(x) ReallySolve y′′+4y′+5y = 0?
y′′+4y′+5y ?= 0(e−2x cos(x)
)′′+4
(e−2x cos(x)
)′+5
(e−2x cos(x)
)?= 0(
3e−2x cos(x)+4e−2x sin(x))
+4(−2e−2x cos(x)− e−2x sin(x)
)+5
(e−2x cos(x)
)?= 0
(3−8+5)e−2x cos(x)+(4−4)e−2x sin(x) ?= 00 = 0
√
Check y = e−2x sin(x) yourself.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−2x cos(x)+ c2e−2x sin(x) ReallySolve y′′+4y′+5y = 0?
y′′+4y′+5y ?= 0(e−2x cos(x)
)′′+4
(e−2x cos(x)
)′+5
(e−2x cos(x)
)?= 0(
3e−2x cos(x)+4e−2x sin(x))
+4(−2e−2x cos(x)− e−2x sin(x)
)+5
(e−2x cos(x)
)?= 0
(3−8+5)e−2x cos(x)+(4−4)e−2x sin(x) ?= 00 = 0
√
Check y = e−2x sin(x) yourself.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = c1e−2x cos(x)+ c2e−2x sin(x) ReallySolve y′′+4y′+5y = 0?
y′′+4y′+5y ?= 0(e−2x cos(x)
)′′+4
(e−2x cos(x)
)′+5
(e−2x cos(x)
)?= 0(
3e−2x cos(x)+4e−2x sin(x))
+4(−2e−2x cos(x)− e−2x sin(x)
)+5
(e−2x cos(x)
)?= 0
(3−8+5)e−2x cos(x)+(4−4)e−2x sin(x) ?= 00 = 0
√
Check y = e−2x sin(x) yourself.Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
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Overview Examples
Solve the Initial Value Problem y′′+4y′+4y = 0,y(0) = 1, y′(0) = 0.
y′′+4y′+4y = 0
λ2eλx +4λeλx +4eλx = 0
λ2 +4λ +4 = 0
λ1,2 =−4±
√16−16
2= −2
y = c1e−2x + c2xe−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
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Overview Examples
Solve the Initial Value Problem y′′+4y′+4y = 0,y(0) = 1, y′(0) = 0.
y′′+4y′+4y = 0λ
2eλx +4λeλx +4eλx = 0
λ2 +4λ +4 = 0
λ1,2 =−4±
√16−16
2= −2
y = c1e−2x + c2xe−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
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Overview Examples
Solve the Initial Value Problem y′′+4y′+4y = 0,y(0) = 1, y′(0) = 0.
y′′+4y′+4y = 0λ
2eλx +4λeλx +4eλx = 0λ
2 +4λ +4 = 0
λ1,2 =−4±
√16−16
2= −2
y = c1e−2x + c2xe−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Solve the Initial Value Problem y′′+4y′+4y = 0,y(0) = 1, y′(0) = 0.
y′′+4y′+4y = 0λ
2eλx +4λeλx +4eλx = 0λ
2 +4λ +4 = 0
λ1,2 =−4±
√16−16
2
= −2
y = c1e−2x + c2xe−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Solve the Initial Value Problem y′′+4y′+4y = 0,y(0) = 1, y′(0) = 0.
y′′+4y′+4y = 0λ
2eλx +4λeλx +4eλx = 0λ
2 +4λ +4 = 0
λ1,2 =−4±
√16−16
2= −2
y = c1e−2x + c2xe−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Solve the Initial Value Problem y′′+4y′+4y = 0,y(0) = 1, y′(0) = 0.
y′′+4y′+4y = 0λ
2eλx +4λeλx +4eλx = 0λ
2 +4λ +4 = 0
λ1,2 =−4±
√16−16
2= −2
y = c1e−2x + c2xe−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Solve the Initial Value Problem y′′+4y′+4y = 0,y(0) = 1, y′(0) = 0.
y = c1e−2x + c2xe−2x
y′ = −2c1e−2x + c2
(e−2x−2xe−2x
)1 = c1
0 = −2c1 + c2
c2 = 2c1 = 2
y = e−2x +2xe−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Solve the Initial Value Problem y′′+4y′+4y = 0,y(0) = 1, y′(0) = 0.
y = c1e−2x + c2xe−2x
y′ = −2c1e−2x + c2
(e−2x−2xe−2x
)
1 = c1
0 = −2c1 + c2
c2 = 2c1 = 2
y = e−2x +2xe−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Solve the Initial Value Problem y′′+4y′+4y = 0,y(0) = 1, y′(0) = 0.
y = c1e−2x + c2xe−2x
y′ = −2c1e−2x + c2
(e−2x−2xe−2x
)1 = c1
0 = −2c1 + c2
c2 = 2c1 = 2
y = e−2x +2xe−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Solve the Initial Value Problem y′′+4y′+4y = 0,y(0) = 1, y′(0) = 0.
y = c1e−2x + c2xe−2x
y′ = −2c1e−2x + c2
(e−2x−2xe−2x
)1 = c1
0 = −2c1 + c2
c2 = 2c1 = 2
y = e−2x +2xe−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Solve the Initial Value Problem y′′+4y′+4y = 0,y(0) = 1, y′(0) = 0.
y = c1e−2x + c2xe−2x
y′ = −2c1e−2x + c2
(e−2x−2xe−2x
)1 = c1
0 = −2c1 + c2
c2 = 2c1 = 2
y = e−2x +2xe−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Solve the Initial Value Problem y′′+4y′+4y = 0,y(0) = 1, y′(0) = 0.
y = c1e−2x + c2xe−2x
y′ = −2c1e−2x + c2
(e−2x−2xe−2x
)1 = c1
0 = −2c1 + c2
c2 = 2c1 = 2
y = e−2x +2xe−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Solve the Initial Value Problem y′′+4y′+4y = 0,y(0) = 1, y′(0) = 0.
y = e−2x +2xe−2x
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = e−2x +2xe−2x Solve the Initial ValueProblem y′′+4y′+4y = 0, y(0) = 1, y′(0) = 0?
y′′+4y′+4y ?= 0
(−4e−2x +8xe−2x
)+4
(−2e−2x +2e−2x−4xe−2x
)+4
(e−2x +2xe−2x
)?= 0(
−4e−2x +8xe−2x)
+4(−4xe−2x
)+4
(e−2x +2xe−2x
)?= 0
(−4+4)e−2x +(8−16+8)xe−2x ?= 00 = 0
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = e−2x +2xe−2x Solve the Initial ValueProblem y′′+4y′+4y = 0, y(0) = 1, y′(0) = 0?
y′′+4y′+4y ?= 0(−4e−2x +8xe−2x
)+4
(−2e−2x +2e−2x−4xe−2x
)+4
(e−2x +2xe−2x
)?= 0
(−4e−2x +8xe−2x
)+4
(−4xe−2x
)+4
(e−2x +2xe−2x
)?= 0
(−4+4)e−2x +(8−16+8)xe−2x ?= 00 = 0
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = e−2x +2xe−2x Solve the Initial ValueProblem y′′+4y′+4y = 0, y(0) = 1, y′(0) = 0?
y′′+4y′+4y ?= 0(−4e−2x +8xe−2x
)+4
(−2e−2x +2e−2x−4xe−2x
)+4
(e−2x +2xe−2x
)?= 0(
−4e−2x +8xe−2x)
+4(−4xe−2x
)+4
(e−2x +2xe−2x
)?= 0
(−4+4)e−2x +(8−16+8)xe−2x ?= 00 = 0
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = e−2x +2xe−2x Solve the Initial ValueProblem y′′+4y′+4y = 0, y(0) = 1, y′(0) = 0?
y′′+4y′+4y ?= 0(−4e−2x +8xe−2x
)+4
(−2e−2x +2e−2x−4xe−2x
)+4
(e−2x +2xe−2x
)?= 0(
−4e−2x +8xe−2x)
+4(−4xe−2x
)+4
(e−2x +2xe−2x
)?= 0
(−4+4)e−2x +(8−16+8)xe−2x ?= 0
0 = 0√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = e−2x +2xe−2x Solve the Initial ValueProblem y′′+4y′+4y = 0, y(0) = 1, y′(0) = 0?
y′′+4y′+4y ?= 0(−4e−2x +8xe−2x
)+4
(−2e−2x +2e−2x−4xe−2x
)+4
(e−2x +2xe−2x
)?= 0(
−4e−2x +8xe−2x)
+4(−4xe−2x
)+4
(e−2x +2xe−2x
)?= 0
(−4+4)e−2x +(8−16+8)xe−2x ?= 00 = 0
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = e−2x +2xe−2x Solve the Initial ValueProblem y′′+4y′+4y = 0, y(0) = 1, y′(0) = 0?
y′′+4y′+4y ?= 0(−4e−2x +8xe−2x
)+4
(−2e−2x +2e−2x−4xe−2x
)+4
(e−2x +2xe−2x
)?= 0(
−4e−2x +8xe−2x)
+4(−4xe−2x
)+4
(e−2x +2xe−2x
)?= 0
(−4+4)e−2x +(8−16+8)xe−2x ?= 00 = 0
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = e−2x +2xe−2x Solve the Initial ValueProblem y′′+4y′+4y = 0, y(0) = 1, y′(0) = 0?
y = e−2x +2xe−2x
y(0) = 1√
y′ = −2e−2x +2(
e−2x−2xe−2x)
y′(0) = −2+2 = 0√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = e−2x +2xe−2x Solve the Initial ValueProblem y′′+4y′+4y = 0, y(0) = 1, y′(0) = 0?
y = e−2x +2xe−2x
y(0) = 1
√
y′ = −2e−2x +2(
e−2x−2xe−2x)
y′(0) = −2+2 = 0√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = e−2x +2xe−2x Solve the Initial ValueProblem y′′+4y′+4y = 0, y(0) = 1, y′(0) = 0?
y = e−2x +2xe−2x
y(0) = 1√
y′ = −2e−2x +2(
e−2x−2xe−2x)
y′(0) = −2+2 = 0√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = e−2x +2xe−2x Solve the Initial ValueProblem y′′+4y′+4y = 0, y(0) = 1, y′(0) = 0?
y = e−2x +2xe−2x
y(0) = 1√
y′ = −2e−2x +2(
e−2x−2xe−2x)
y′(0) = −2+2 = 0√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = e−2x +2xe−2x Solve the Initial ValueProblem y′′+4y′+4y = 0, y(0) = 1, y′(0) = 0?
y = e−2x +2xe−2x
y(0) = 1√
y′ = −2e−2x +2(
e−2x−2xe−2x)
y′(0) = −2+2 = 0
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations
logo1
Overview Examples
Does y = e−2x +2xe−2x Solve the Initial ValueProblem y′′+4y′+4y = 0, y(0) = 1, y′(0) = 0?
y = e−2x +2xe−2x
y(0) = 1√
y′ = −2e−2x +2(
e−2x−2xe−2x)
y′(0) = −2+2 = 0√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Linear Homogeneous Constant Coefficient Differential Equations