Chapter 21 Exact Differential Equation Chapter 2 Exact Differential Equation.
An Initial Value Problem for a Separable Differential Equation
Transcript of An Initial Value Problem for a Separable Differential Equation
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Solving Initial Value Problems An Example Double Check
An Initial Value Problem for a SeparableDifferential Equation
Bernd Schroder
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 2: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/2.jpg)
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Solving Initial Value Problems An Example Double Check
Approach
This approach will work as long as the general solution of thedifferential equation can be computed.
1. Find the general solution of the differential equation.2. Use the initial conditions to determine the value(s) of the
constant(s) in the general solution.3. Double check if the solution works.
Thatβs it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 3: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/3.jpg)
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Solving Initial Value Problems An Example Double Check
ApproachThis approach will work as long as the general solution of thedifferential equation can be computed.
1. Find the general solution of the differential equation.2. Use the initial conditions to determine the value(s) of the
constant(s) in the general solution.3. Double check if the solution works.
Thatβs it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 4: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/4.jpg)
logo1
Solving Initial Value Problems An Example Double Check
ApproachThis approach will work as long as the general solution of thedifferential equation can be computed.
1. Find the general solution of the differential equation.
2. Use the initial conditions to determine the value(s) of theconstant(s) in the general solution.
3. Double check if the solution works.Thatβs it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 5: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/5.jpg)
logo1
Solving Initial Value Problems An Example Double Check
ApproachThis approach will work as long as the general solution of thedifferential equation can be computed.
1. Find the general solution of the differential equation.2. Use the initial conditions to determine the value(s) of the
constant(s) in the general solution.
3. Double check if the solution works.Thatβs it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 6: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/6.jpg)
logo1
Solving Initial Value Problems An Example Double Check
ApproachThis approach will work as long as the general solution of thedifferential equation can be computed.
1. Find the general solution of the differential equation.2. Use the initial conditions to determine the value(s) of the
constant(s) in the general solution.3. Double check if the solution works.
Thatβs it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 7: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/7.jpg)
logo1
Solving Initial Value Problems An Example Double Check
ApproachThis approach will work as long as the general solution of thedifferential equation can be computed.
1. Find the general solution of the differential equation.2. Use the initial conditions to determine the value(s) of the
constant(s) in the general solution.3. Double check if the solution works.
Thatβs it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 8: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/8.jpg)
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Solving Initial Value Problems An Example Double Check
Solve the Initial Value Problem
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1.
yβ² = xsin(x)y+xsin(x)
y
yβ² = xsin(x)(
y+1y
)dydx
= xsin(x)(
y2 +1y
)y
y2 +1dy = xsin(x) dx
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 9: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/9.jpg)
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Solving Initial Value Problems An Example Double Check
Solve the Initial Value Problem
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1.
yβ² = xsin(x)y+xsin(x)
y
yβ² = xsin(x)(
y+1y
)dydx
= xsin(x)(
y2 +1y
)y
y2 +1dy = xsin(x) dx
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 10: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/10.jpg)
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Solving Initial Value Problems An Example Double Check
Solve the Initial Value Problem
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1.
yβ² = xsin(x)y+xsin(x)
y
yβ² = xsin(x)(
y+1y
)
dydx
= xsin(x)(
y2 +1y
)y
y2 +1dy = xsin(x) dx
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 11: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/11.jpg)
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Solving Initial Value Problems An Example Double Check
Solve the Initial Value Problem
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1.
yβ² = xsin(x)y+xsin(x)
y
yβ² = xsin(x)(
y+1y
)dydx
= xsin(x)(
y2 +1y
)
yy2 +1
dy = xsin(x) dx
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 12: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/12.jpg)
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Solving Initial Value Problems An Example Double Check
Solve the Initial Value Problem
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1.
yβ² = xsin(x)y+xsin(x)
y
yβ² = xsin(x)(
y+1y
)dydx
= xsin(x)(
y2 +1y
)y
y2 +1dy = xsin(x) dx
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 13: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/13.jpg)
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Solving Initial Value Problems An Example Double Check
Solve the Initial Value Problem
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1.
yβ² = xsin(x)y+xsin(x)
y
yβ² = xsin(x)(
y+1y
)dydx
= xsin(x)(
y2 +1y
)β« y
y2 +1dy =
β«xsin(x) dx
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 14: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/14.jpg)
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Solving Initial Value Problems An Example Double Check
The Integralβ« y
y2 +1dy.
Idea: Substitution u := y2 +1.dudy
= 2y, so dy =du2y
.β« yy2 +1
dy =β« y
udu2y
=12
β« 1u
du
=12
ln |u|+ c
=12
lnβ£β£β£y2 +1
β£β£β£+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 15: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/15.jpg)
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Solving Initial Value Problems An Example Double Check
The Integralβ« y
y2 +1dy.
Idea: Substitution u := y2 +1.
dudy
= 2y, so dy =du2y
.β« yy2 +1
dy =β« y
udu2y
=12
β« 1u
du
=12
ln |u|+ c
=12
lnβ£β£β£y2 +1
β£β£β£+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 16: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/16.jpg)
logo1
Solving Initial Value Problems An Example Double Check
The Integralβ« y
y2 +1dy.
Idea: Substitution u := y2 +1.dudy
= 2y, so dy =du2y
.
β« yy2 +1
dy =β« y
udu2y
=12
β« 1u
du
=12
ln |u|+ c
=12
lnβ£β£β£y2 +1
β£β£β£+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 17: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/17.jpg)
logo1
Solving Initial Value Problems An Example Double Check
The Integralβ« y
y2 +1dy.
Idea: Substitution u := y2 +1.dudy
= 2y, so dy =du2y
.β« yy2 +1
dy =
β« yu
du2y
=12
β« 1u
du
=12
ln |u|+ c
=12
lnβ£β£β£y2 +1
β£β£β£+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 18: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/18.jpg)
logo1
Solving Initial Value Problems An Example Double Check
The Integralβ« y
y2 +1dy.
Idea: Substitution u := y2 +1.dudy
= 2y, so dy =du2y
.β« yy2 +1
dy =β« y
udu2y
=12
β« 1u
du
=12
ln |u|+ c
=12
lnβ£β£β£y2 +1
β£β£β£+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 19: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/19.jpg)
logo1
Solving Initial Value Problems An Example Double Check
The Integralβ« y
y2 +1dy.
Idea: Substitution u := y2 +1.dudy
= 2y, so dy =du2y
.β« yy2 +1
dy =β« y
udu2y
=12
β« 1u
du
=12
ln |u|+ c
=12
lnβ£β£β£y2 +1
β£β£β£+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 20: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/20.jpg)
logo1
Solving Initial Value Problems An Example Double Check
The Integralβ« y
y2 +1dy.
Idea: Substitution u := y2 +1.dudy
= 2y, so dy =du2y
.β« yy2 +1
dy =β« y
udu2y
=12
β« 1u
du
=12
ln |u|+ c
=12
lnβ£β£β£y2 +1
β£β£β£+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 21: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/21.jpg)
logo1
Solving Initial Value Problems An Example Double Check
The Integralβ« y
y2 +1dy.
Idea: Substitution u := y2 +1.dudy
= 2y, so dy =du2y
.β« yy2 +1
dy =β« y
udu2y
=12
β« 1u
du
=12
ln |u|+ c
=12
lnβ£β£β£y2 +1
β£β£β£+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 22: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/22.jpg)
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Solving Initial Value Problems An Example Double Check
The Integralβ«
xsin(x) dx.
Idea: Integration by parts.Integrate sin(x), differentiate x.β«
xsin(x) dx = x(β cos(x)
)β
β«1 Β·
(β cos(x)
)dx
= βxcos(x)+β«
cos(x) dx
= βxcos(x)+ sin(x)+ c= sin(x)β xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 23: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/23.jpg)
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Solving Initial Value Problems An Example Double Check
The Integralβ«
xsin(x) dx.Idea: Integration by parts.
Integrate sin(x), differentiate x.β«xsin(x) dx = x
(β cos(x)
)β
β«1 Β·
(β cos(x)
)dx
= βxcos(x)+β«
cos(x) dx
= βxcos(x)+ sin(x)+ c= sin(x)β xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 24: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/24.jpg)
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Solving Initial Value Problems An Example Double Check
The Integralβ«
xsin(x) dx.Idea: Integration by parts.Integrate sin(x), differentiate x.
β«xsin(x) dx = x
(β cos(x)
)β
β«1 Β·
(β cos(x)
)dx
= βxcos(x)+β«
cos(x) dx
= βxcos(x)+ sin(x)+ c= sin(x)β xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 25: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/25.jpg)
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Solving Initial Value Problems An Example Double Check
The Integralβ«
xsin(x) dx.Idea: Integration by parts.Integrate sin(x), differentiate x.β«
xsin(x) dx =
x(β cos(x)
)β
β«1 Β·
(β cos(x)
)dx
= βxcos(x)+β«
cos(x) dx
= βxcos(x)+ sin(x)+ c= sin(x)β xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 26: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/26.jpg)
logo1
Solving Initial Value Problems An Example Double Check
The Integralβ«
xsin(x) dx.Idea: Integration by parts.Integrate sin(x), differentiate x.β«
xsin(x) dx = x(β cos(x)
)β
β«1 Β·
(β cos(x)
)dx
= βxcos(x)+β«
cos(x) dx
= βxcos(x)+ sin(x)+ c= sin(x)β xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 27: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/27.jpg)
logo1
Solving Initial Value Problems An Example Double Check
The Integralβ«
xsin(x) dx.Idea: Integration by parts.Integrate sin(x), differentiate x.β«
xsin(x) dx = x(β cos(x)
)β
β«1 Β·
(β cos(x)
)dx
= βxcos(x)+β«
cos(x) dx
= βxcos(x)+ sin(x)+ c= sin(x)β xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 28: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/28.jpg)
logo1
Solving Initial Value Problems An Example Double Check
The Integralβ«
xsin(x) dx.Idea: Integration by parts.Integrate sin(x), differentiate x.β«
xsin(x) dx = x(β cos(x)
)β
β«1 Β·
(β cos(x)
)dx
= βxcos(x)+β«
cos(x) dx
= βxcos(x)+ sin(x)+ c= sin(x)β xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 29: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/29.jpg)
logo1
Solving Initial Value Problems An Example Double Check
The Integralβ«
xsin(x) dx.Idea: Integration by parts.Integrate sin(x), differentiate x.β«
xsin(x) dx = x(β cos(x)
)β
β«1 Β·
(β cos(x)
)dx
= βxcos(x)+β«
cos(x) dx
= βxcos(x)+ sin(x)+ c
= sin(x)β xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 30: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/30.jpg)
logo1
Solving Initial Value Problems An Example Double Check
The Integralβ«
xsin(x) dx.Idea: Integration by parts.Integrate sin(x), differentiate x.β«
xsin(x) dx = x(β cos(x)
)β
β«1 Β·
(β cos(x)
)dx
= βxcos(x)+β«
cos(x) dx
= βxcos(x)+ sin(x)+ c= sin(x)β xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 31: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/31.jpg)
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Solving Initial Value Problems An Example Double Check
... Continuing Where We Left Off.β« y
y2 +1dy =
β«xsin(x) dx
12
lnβ£β£β£y2 +1
β£β£β£ = sin(x)β xcos(x)+ c
lnβ£β£β£y2 +1
β£β£β£ = 2sin(x)β2xcos(x)+ c
y2 +1 = e2sin(x)β2xcos(x)+c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 32: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/32.jpg)
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Solving Initial Value Problems An Example Double Check
... Continuing Where We Left Off.β« y
y2 +1dy =
β«xsin(x) dx
12
lnβ£β£β£y2 +1
β£β£β£ =
sin(x)β xcos(x)+ c
lnβ£β£β£y2 +1
β£β£β£ = 2sin(x)β2xcos(x)+ c
y2 +1 = e2sin(x)β2xcos(x)+c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 33: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/33.jpg)
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Solving Initial Value Problems An Example Double Check
... Continuing Where We Left Off.β« y
y2 +1dy =
β«xsin(x) dx
12
lnβ£β£β£y2 +1
β£β£β£ = sin(x)β xcos(x)+ c
lnβ£β£β£y2 +1
β£β£β£ = 2sin(x)β2xcos(x)+ c
y2 +1 = e2sin(x)β2xcos(x)+c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 34: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/34.jpg)
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Solving Initial Value Problems An Example Double Check
... Continuing Where We Left Off.β« y
y2 +1dy =
β«xsin(x) dx
12
lnβ£β£β£y2 +1
β£β£β£ = sin(x)β xcos(x)+ c
lnβ£β£β£y2 +1
β£β£β£ = 2sin(x)β2xcos(x)+ c
y2 +1 = e2sin(x)β2xcos(x)+c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 35: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/35.jpg)
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Solving Initial Value Problems An Example Double Check
... Continuing Where We Left Off.β« y
y2 +1dy =
β«xsin(x) dx
12
lnβ£β£β£y2 +1
β£β£β£ = sin(x)β xcos(x)+ c
lnβ£β£β£y2 +1
β£β£β£ = 2sin(x)β2xcos(x)+ c
y2 +1 = e2sin(x)β2xcos(x)+c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 36: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/36.jpg)
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Solving Initial Value Problems An Example Double Check
... Continuing Where We Left Off.β« y
y2 +1dy =
β«xsin(x) dx
12
lnβ£β£β£y2 +1
β£β£β£ = sin(x)β xcos(x)+ c
lnβ£β£β£y2 +1
β£β£β£ = 2sin(x)β2xcos(x)+ c
y2 +1 = ke2sin(x)β2xcos(x)
y2 = ke2sin(x)β2xcos(x)β1
y = Β±β
ke2sin(x)β2xcos(x)β1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 37: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/37.jpg)
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Solving Initial Value Problems An Example Double Check
... Continuing Where We Left Off.β« y
y2 +1dy =
β«xsin(x) dx
12
lnβ£β£β£y2 +1
β£β£β£ = sin(x)β xcos(x)+ c
lnβ£β£β£y2 +1
β£β£β£ = 2sin(x)β2xcos(x)+ c
y2 +1 = ke2sin(x)β2xcos(x)
y2 = ke2sin(x)β2xcos(x)β1
y = Β±β
ke2sin(x)β2xcos(x)β1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 38: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/38.jpg)
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Solving Initial Value Problems An Example Double Check
... Continuing Where We Left Off.β« y
y2 +1dy =
β«xsin(x) dx
12
lnβ£β£β£y2 +1
β£β£β£ = sin(x)β xcos(x)+ c
lnβ£β£β£y2 +1
β£β£β£ = 2sin(x)β2xcos(x)+ c
y2 +1 = ke2sin(x)β2xcos(x)
y2 = ke2sin(x)β2xcos(x)β1
y = Β±β
ke2sin(x)β2xcos(x)β1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 39: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/39.jpg)
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Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = Β±β
ke2sin(x)β2xcos(x)β1
1 = Β±β
ke2sin(0)β2Β·0Β·cos(0)β1
1 = Β±β
ke0β1 = Β±β
kβ11 =
βkβ1
1 = kβ1k = 2
y(x) =β
2e2sin(x)β2xcos(x)β1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 40: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/40.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = Β±β
ke2sin(x)β2xcos(x)β1
1 = Β±β
ke2sin(0)β2Β·0Β·cos(0)β1
1 = Β±β
ke0β1 = Β±β
kβ11 =
βkβ1
1 = kβ1k = 2
y(x) =β
2e2sin(x)β2xcos(x)β1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 41: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/41.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = Β±β
ke2sin(x)β2xcos(x)β1
1 = Β±β
ke2sin(0)β2Β·0Β·cos(0)β1
1 = Β±β
ke0β1 = Β±β
kβ11 =
βkβ1
1 = kβ1k = 2
y(x) =β
2e2sin(x)β2xcos(x)β1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 42: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/42.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = Β±β
ke2sin(x)β2xcos(x)β1
1 =
Β±β
ke2sin(0)β2Β·0Β·cos(0)β1
1 = Β±β
ke0β1 = Β±β
kβ11 =
βkβ1
1 = kβ1k = 2
y(x) =β
2e2sin(x)β2xcos(x)β1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 43: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/43.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = Β±β
ke2sin(x)β2xcos(x)β1
1 = Β±β
ke2sin(0)β2Β·0Β·cos(0)β1
1 = Β±β
ke0β1 = Β±β
kβ11 =
βkβ1
1 = kβ1k = 2
y(x) =β
2e2sin(x)β2xcos(x)β1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 44: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/44.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = Β±β
ke2sin(x)β2xcos(x)β1
1 = Β±β
ke2sin(0)β2Β·0Β·cos(0)β1
1 = Β±β
ke0β1 =
Β±β
kβ11 =
βkβ1
1 = kβ1k = 2
y(x) =β
2e2sin(x)β2xcos(x)β1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 45: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/45.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = Β±β
ke2sin(x)β2xcos(x)β1
1 = Β±β
ke2sin(0)β2Β·0Β·cos(0)β1
1 = Β±β
ke0β1 = Β±β
kβ1
1 =β
kβ11 = kβ1k = 2
y(x) =β
2e2sin(x)β2xcos(x)β1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 46: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/46.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = Β±β
ke2sin(x)β2xcos(x)β1
1 = Β±β
ke2sin(0)β2Β·0Β·cos(0)β1
1 = Β±β
ke0β1 = Β±β
kβ11 =
βkβ1
1 = kβ1k = 2
y(x) =β
2e2sin(x)β2xcos(x)β1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 47: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/47.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = Β±β
ke2sin(x)β2xcos(x)β1
1 = Β±β
ke2sin(0)β2Β·0Β·cos(0)β1
1 = Β±β
ke0β1 = Β±β
kβ11 =
βkβ1
1 = kβ1
k = 2
y(x) =β
2e2sin(x)β2xcos(x)β1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 48: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/48.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = Β±β
ke2sin(x)β2xcos(x)β1
1 = Β±β
ke2sin(0)β2Β·0Β·cos(0)β1
1 = Β±β
ke0β1 = Β±β
kβ11 =
βkβ1
1 = kβ1k = 2
y(x) =β
2e2sin(x)β2xcos(x)β1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 49: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/49.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = Β±β
ke2sin(x)β2xcos(x)β1
1 = Β±β
ke2sin(0)β2Β·0Β·cos(0)β1
1 = Β±β
ke0β1 = Β±β
kβ11 =
βkβ1
1 = kβ1k = 2
y(x) =β
2e2sin(x)β2xcos(x)β1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 50: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/50.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
y(x) =β
2e2sin(x)β2xcos(x)β1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 51: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/51.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
y(x) =β
2e2sin(x)β2xcos(x)β1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 52: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/52.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =β
2e2sin(x)β2xcos(x)β1 Solve the IVP
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1?
y(x) =β
2e2sin(x)β2xcos(x)β1
y(0) =β
2e2sin(0)β2Β·0Β·cos(0)β1 =β
2e0β1 = 1β
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 53: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/53.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =β
2e2sin(x)β2xcos(x)β1 Solve the IVP
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1?
y(x) =β
2e2sin(x)β2xcos(x)β1
y(0) =β
2e2sin(0)β2Β·0Β·cos(0)β1
=β
2e0β1 = 1β
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 54: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/54.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =β
2e2sin(x)β2xcos(x)β1 Solve the IVP
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1?
y(x) =β
2e2sin(x)β2xcos(x)β1
y(0) =β
2e2sin(0)β2Β·0Β·cos(0)β1 =β
2e0β1
= 1β
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 55: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/55.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =β
2e2sin(x)β2xcos(x)β1 Solve the IVP
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1?
y(x) =β
2e2sin(x)β2xcos(x)β1
y(0) =β
2e2sin(0)β2Β·0Β·cos(0)β1 =β
2e0β1 = 1
β
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 56: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/56.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =β
2e2sin(x)β2xcos(x)β1 Solve the IVP
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1?
y(x) =β
2e2sin(x)β2xcos(x)β1
y(0) =β
2e2sin(0)β2Β·0Β·cos(0)β1 =β
2e0β1 = 1β
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 57: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/57.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =β
2e2sin(x)β2xcos(x)β1 Solve the IVP
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1?
ddx
(β2e2sin(x)β2xcos(x)β1
)
=1
2β
2e2sin(x)β2xcos(x)β12e2sin(x)β2xcos(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 58: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/58.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =β
2e2sin(x)β2xcos(x)β1 Solve the IVP
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1?
ddx
(β2e2sin(x)β2xcos(x)β1
)=
1
2β
2e2sin(x)β2xcos(x)β1
2e2sin(x)β2xcos(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 59: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/59.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =β
2e2sin(x)β2xcos(x)β1 Solve the IVP
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1?
ddx
(β2e2sin(x)β2xcos(x)β1
)=
1
2β
2e2sin(x)β2xcos(x)β12e2sin(x)β2xcos(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 60: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/60.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =β
2e2sin(x)β2xcos(x)β1 Solve the IVP
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1?
ddx
(β2e2sin(x)β2xcos(x)β1
)=
2e2sin(x)β2xcos(x)
2β
2e2sin(x)β2xcos(x)β1
(2cos(x)β2cos(x)+2xsin(x)
)=
2e2sin(x)β2xcos(x)
2β
2e2sin(x)β2xcos(x)β12xsin(x)
=2e2sin(x)β2xcos(x)β
2e2sin(x)β2xcos(x)β1xsin(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 61: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/61.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =β
2e2sin(x)β2xcos(x)β1 Solve the IVP
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1?
ddx
(β2e2sin(x)β2xcos(x)β1
)=
2e2sin(x)β2xcos(x)
2β
2e2sin(x)β2xcos(x)β1
(2cos(x)β2cos(x)+2xsin(x)
)
=2e2sin(x)β2xcos(x)
2β
2e2sin(x)β2xcos(x)β12xsin(x)
=2e2sin(x)β2xcos(x)β
2e2sin(x)β2xcos(x)β1xsin(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 62: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/62.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =β
2e2sin(x)β2xcos(x)β1 Solve the IVP
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1?
ddx
(β2e2sin(x)β2xcos(x)β1
)=
2e2sin(x)β2xcos(x)
2β
2e2sin(x)β2xcos(x)β1
(2cos(x)β2cos(x)+2xsin(x)
)=
2e2sin(x)β2xcos(x)
2β
2e2sin(x)β2xcos(x)β12xsin(x)
=2e2sin(x)β2xcos(x)β
2e2sin(x)β2xcos(x)β1xsin(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 63: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/63.jpg)
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =β
2e2sin(x)β2xcos(x)β1 Solve the IVP
yβ² = xsin(x)y+xsin(x)
y, y(0) = 1?
ddx
(β2e2sin(x)β2xcos(x)β1
)=
2e2sin(x)β2xcos(x)
2β
2e2sin(x)β2xcos(x)β1
(2cos(x)β2cos(x)+2xsin(x)
)=
2e2sin(x)β2xcos(x)
2β
2e2sin(x)β2xcos(x)β12xsin(x)
=2e2sin(x)β2xcos(x)β
2e2sin(x)β2xcos(x)β1xsin(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 64: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/64.jpg)
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Solving Initial Value Problems An Example Double Check
xsin(x)y+xsin(x)
y
= xsin(x)β
2e2sin(x)β2xcos(x)β1+xsin(x)β
2e2sin(x)β2xcos(x)β1
=xsin(x)
(β2e2sin(x)β2xcos(x)β1
)2+ xsin(x)β
2e2sin(x)β2xcos(x)β1
=xsin(x)
(2e2sin(x)β2xcos(x)β1+1
)β
2e2sin(x)β2xcos(x)β1
= xsin(x)2e2sin(x)β2xcos(x)β
2e2sin(x)β2xcos(x)β1
β
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 65: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/65.jpg)
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Solving Initial Value Problems An Example Double Check
xsin(x)y+xsin(x)
y
= xsin(x)β
2e2sin(x)β2xcos(x)β1+xsin(x)β
2e2sin(x)β2xcos(x)β1
=xsin(x)
(β2e2sin(x)β2xcos(x)β1
)2+ xsin(x)β
2e2sin(x)β2xcos(x)β1
=xsin(x)
(2e2sin(x)β2xcos(x)β1+1
)β
2e2sin(x)β2xcos(x)β1
= xsin(x)2e2sin(x)β2xcos(x)β
2e2sin(x)β2xcos(x)β1
β
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 66: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/66.jpg)
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Solving Initial Value Problems An Example Double Check
xsin(x)y+xsin(x)
y
= xsin(x)β
2e2sin(x)β2xcos(x)β1+xsin(x)β
2e2sin(x)β2xcos(x)β1
=xsin(x)
(β2e2sin(x)β2xcos(x)β1
)2+ xsin(x)β
2e2sin(x)β2xcos(x)β1
=xsin(x)
(2e2sin(x)β2xcos(x)β1+1
)β
2e2sin(x)β2xcos(x)β1
= xsin(x)2e2sin(x)β2xcos(x)β
2e2sin(x)β2xcos(x)β1
β
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 67: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/67.jpg)
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Solving Initial Value Problems An Example Double Check
xsin(x)y+xsin(x)
y
= xsin(x)β
2e2sin(x)β2xcos(x)β1+xsin(x)β
2e2sin(x)β2xcos(x)β1
=xsin(x)
(β2e2sin(x)β2xcos(x)β1
)2+ xsin(x)β
2e2sin(x)β2xcos(x)β1
=xsin(x)
(2e2sin(x)β2xcos(x)β1+1
)β
2e2sin(x)β2xcos(x)β1
= xsin(x)2e2sin(x)β2xcos(x)β
2e2sin(x)β2xcos(x)β1
β
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 68: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/68.jpg)
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Solving Initial Value Problems An Example Double Check
xsin(x)y+xsin(x)
y
= xsin(x)β
2e2sin(x)β2xcos(x)β1+xsin(x)β
2e2sin(x)β2xcos(x)β1
=xsin(x)
(β2e2sin(x)β2xcos(x)β1
)2+ xsin(x)β
2e2sin(x)β2xcos(x)β1
=xsin(x)
(2e2sin(x)β2xcos(x)β1+1
)β
2e2sin(x)β2xcos(x)β1
= xsin(x)2e2sin(x)β2xcos(x)β
2e2sin(x)β2xcos(x)β1
β
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
![Page 69: An Initial Value Problem for a Separable Differential Equation](https://reader031.fdocuments.us/reader031/viewer/2022020707/61ffc77422b5b0371c5c3f7a/html5/thumbnails/69.jpg)
logo1
Solving Initial Value Problems An Example Double Check
xsin(x)y+xsin(x)
y
= xsin(x)β
2e2sin(x)β2xcos(x)β1+xsin(x)β
2e2sin(x)β2xcos(x)β1
=xsin(x)
(β2e2sin(x)β2xcos(x)β1
)2+ xsin(x)β
2e2sin(x)β2xcos(x)β1
=xsin(x)
(2e2sin(x)β2xcos(x)β1+1
)β
2e2sin(x)β2xcos(x)β1
= xsin(x)2e2sin(x)β2xcos(x)β
2e2sin(x)β2xcos(x)β1
β
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation