Post on 06-Feb-2022
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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
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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.
That’s it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
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.
That’s it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
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.That’s it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
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.That’s it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
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.
That’s it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
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.
That’s it.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Solve the Initial Value Problem
y′ = xsin(x)y+xsin(x)
y, y(0) = 1.
y′ = xsin(x)y+xsin(x)
y
y′ = 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
logo1
Solving Initial Value Problems An Example Double Check
Solve the Initial Value Problem
y′ = xsin(x)y+xsin(x)
y, y(0) = 1.
y′ = xsin(x)y+xsin(x)
y
y′ = 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
logo1
Solving Initial Value Problems An Example Double Check
Solve the Initial Value Problem
y′ = xsin(x)y+xsin(x)
y, y(0) = 1.
y′ = xsin(x)y+xsin(x)
y
y′ = 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
logo1
Solving Initial Value Problems An Example Double Check
Solve the Initial Value Problem
y′ = xsin(x)y+xsin(x)
y, y(0) = 1.
y′ = xsin(x)y+xsin(x)
y
y′ = 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
logo1
Solving Initial Value Problems An Example Double Check
Solve the Initial Value Problem
y′ = xsin(x)y+xsin(x)
y, y(0) = 1.
y′ = xsin(x)y+xsin(x)
y
y′ = 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
logo1
Solving Initial Value Problems An Example Double Check
Solve the Initial Value Problem
y′ = xsin(x)y+xsin(x)
y, y(0) = 1.
y′ = xsin(x)y+xsin(x)
y
y′ = 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
logo1
Solving Initial Value Problems An Example Double Check
The Integral∫ y
y2 +1dy.
Idea: Substitution u := y2 +1.dudy
= 2y, so dy =du2y
.∫ yy2 +1
dy =∫ y
udu2y
=12
∫ 1u
du
=12
ln |u|+ c
=12
ln∣∣∣y2 +1
∣∣∣+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
The Integral∫ y
y2 +1dy.
Idea: Substitution u := y2 +1.
dudy
= 2y, so dy =du2y
.∫ yy2 +1
dy =∫ y
udu2y
=12
∫ 1u
du
=12
ln |u|+ c
=12
ln∣∣∣y2 +1
∣∣∣+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
The Integral∫ y
y2 +1dy.
Idea: Substitution u := y2 +1.dudy
= 2y, so dy =du2y
.
∫ yy2 +1
dy =∫ y
udu2y
=12
∫ 1u
du
=12
ln |u|+ c
=12
ln∣∣∣y2 +1
∣∣∣+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
The Integral∫ y
y2 +1dy.
Idea: Substitution u := y2 +1.dudy
= 2y, so dy =du2y
.∫ yy2 +1
dy =
∫ yu
du2y
=12
∫ 1u
du
=12
ln |u|+ c
=12
ln∣∣∣y2 +1
∣∣∣+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
The Integral∫ y
y2 +1dy.
Idea: Substitution u := y2 +1.dudy
= 2y, so dy =du2y
.∫ yy2 +1
dy =∫ y
udu2y
=12
∫ 1u
du
=12
ln |u|+ c
=12
ln∣∣∣y2 +1
∣∣∣+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
The Integral∫ y
y2 +1dy.
Idea: Substitution u := y2 +1.dudy
= 2y, so dy =du2y
.∫ yy2 +1
dy =∫ y
udu2y
=12
∫ 1u
du
=12
ln |u|+ c
=12
ln∣∣∣y2 +1
∣∣∣+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
The Integral∫ y
y2 +1dy.
Idea: Substitution u := y2 +1.dudy
= 2y, so dy =du2y
.∫ yy2 +1
dy =∫ y
udu2y
=12
∫ 1u
du
=12
ln |u|+ c
=12
ln∣∣∣y2 +1
∣∣∣+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
The Integral∫ y
y2 +1dy.
Idea: Substitution u := y2 +1.dudy
= 2y, so dy =du2y
.∫ yy2 +1
dy =∫ y
udu2y
=12
∫ 1u
du
=12
ln |u|+ c
=12
ln∣∣∣y2 +1
∣∣∣+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
The Integral∫
xsin(x) dx.
Idea: Integration by parts.Integrate sin(x), differentiate x.∫
xsin(x) dx = x(− cos(x)
)−
∫1 ·
(− cos(x)
)dx
= −xcos(x)+∫
cos(x) dx
= −xcos(x)+ sin(x)+ c= sin(x)− xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
The Integral∫
xsin(x) dx.Idea: Integration by parts.
Integrate sin(x), differentiate x.∫xsin(x) dx = x
(− cos(x)
)−
∫1 ·
(− cos(x)
)dx
= −xcos(x)+∫
cos(x) dx
= −xcos(x)+ sin(x)+ c= sin(x)− xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
The Integral∫
xsin(x) dx.Idea: Integration by parts.Integrate sin(x), differentiate x.
∫xsin(x) dx = x
(− cos(x)
)−
∫1 ·
(− cos(x)
)dx
= −xcos(x)+∫
cos(x) dx
= −xcos(x)+ sin(x)+ c= sin(x)− xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
The Integral∫
xsin(x) dx.Idea: Integration by parts.Integrate sin(x), differentiate x.∫
xsin(x) dx =
x(− cos(x)
)−
∫1 ·
(− cos(x)
)dx
= −xcos(x)+∫
cos(x) dx
= −xcos(x)+ sin(x)+ c= sin(x)− xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
The Integral∫
xsin(x) dx.Idea: Integration by parts.Integrate sin(x), differentiate x.∫
xsin(x) dx = x(− cos(x)
)−
∫1 ·
(− cos(x)
)dx
= −xcos(x)+∫
cos(x) dx
= −xcos(x)+ sin(x)+ c= sin(x)− xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
The Integral∫
xsin(x) dx.Idea: Integration by parts.Integrate sin(x), differentiate x.∫
xsin(x) dx = x(− cos(x)
)−
∫1 ·
(− cos(x)
)dx
= −xcos(x)+∫
cos(x) dx
= −xcos(x)+ sin(x)+ c= sin(x)− xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
The Integral∫
xsin(x) dx.Idea: Integration by parts.Integrate sin(x), differentiate x.∫
xsin(x) dx = x(− cos(x)
)−
∫1 ·
(− cos(x)
)dx
= −xcos(x)+∫
cos(x) dx
= −xcos(x)+ sin(x)+ c= sin(x)− xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
The Integral∫
xsin(x) dx.Idea: Integration by parts.Integrate sin(x), differentiate x.∫
xsin(x) dx = x(− cos(x)
)−
∫1 ·
(− cos(x)
)dx
= −xcos(x)+∫
cos(x) dx
= −xcos(x)+ sin(x)+ c
= sin(x)− xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
The Integral∫
xsin(x) dx.Idea: Integration by parts.Integrate sin(x), differentiate x.∫
xsin(x) dx = x(− cos(x)
)−
∫1 ·
(− cos(x)
)dx
= −xcos(x)+∫
cos(x) dx
= −xcos(x)+ sin(x)+ c= sin(x)− xcos(x)+ c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
... Continuing Where We Left Off.∫ y
y2 +1dy =
∫xsin(x) dx
12
ln∣∣∣y2 +1
∣∣∣ = sin(x)− xcos(x)+ c
ln∣∣∣y2 +1
∣∣∣ = 2sin(x)−2xcos(x)+ c
y2 +1 = e2sin(x)−2xcos(x)+c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
... Continuing Where We Left Off.∫ y
y2 +1dy =
∫xsin(x) dx
12
ln∣∣∣y2 +1
∣∣∣ =
sin(x)− xcos(x)+ c
ln∣∣∣y2 +1
∣∣∣ = 2sin(x)−2xcos(x)+ c
y2 +1 = e2sin(x)−2xcos(x)+c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
... Continuing Where We Left Off.∫ y
y2 +1dy =
∫xsin(x) dx
12
ln∣∣∣y2 +1
∣∣∣ = sin(x)− xcos(x)+ c
ln∣∣∣y2 +1
∣∣∣ = 2sin(x)−2xcos(x)+ c
y2 +1 = e2sin(x)−2xcos(x)+c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
... Continuing Where We Left Off.∫ y
y2 +1dy =
∫xsin(x) dx
12
ln∣∣∣y2 +1
∣∣∣ = sin(x)− xcos(x)+ c
ln∣∣∣y2 +1
∣∣∣ = 2sin(x)−2xcos(x)+ c
y2 +1 = e2sin(x)−2xcos(x)+c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
... Continuing Where We Left Off.∫ y
y2 +1dy =
∫xsin(x) dx
12
ln∣∣∣y2 +1
∣∣∣ = sin(x)− xcos(x)+ c
ln∣∣∣y2 +1
∣∣∣ = 2sin(x)−2xcos(x)+ c
y2 +1 = e2sin(x)−2xcos(x)+c
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
... Continuing Where We Left Off.∫ y
y2 +1dy =
∫xsin(x) dx
12
ln∣∣∣y2 +1
∣∣∣ = sin(x)− xcos(x)+ c
ln∣∣∣y2 +1
∣∣∣ = 2sin(x)−2xcos(x)+ c
y2 +1 = ke2sin(x)−2xcos(x)
y2 = ke2sin(x)−2xcos(x)−1
y = ±√
ke2sin(x)−2xcos(x)−1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
... Continuing Where We Left Off.∫ y
y2 +1dy =
∫xsin(x) dx
12
ln∣∣∣y2 +1
∣∣∣ = sin(x)− xcos(x)+ c
ln∣∣∣y2 +1
∣∣∣ = 2sin(x)−2xcos(x)+ c
y2 +1 = ke2sin(x)−2xcos(x)
y2 = ke2sin(x)−2xcos(x)−1
y = ±√
ke2sin(x)−2xcos(x)−1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
... Continuing Where We Left Off.∫ y
y2 +1dy =
∫xsin(x) dx
12
ln∣∣∣y2 +1
∣∣∣ = sin(x)− xcos(x)+ c
ln∣∣∣y2 +1
∣∣∣ = 2sin(x)−2xcos(x)+ c
y2 +1 = ke2sin(x)−2xcos(x)
y2 = ke2sin(x)−2xcos(x)−1
y = ±√
ke2sin(x)−2xcos(x)−1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = ±√
ke2sin(x)−2xcos(x)−1
1 = ±√
ke2sin(0)−2·0·cos(0)−1
1 = ±√
ke0−1 = ±√
k−11 =
√k−1
1 = k−1k = 2
y(x) =√
2e2sin(x)−2xcos(x)−1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = ±√
ke2sin(x)−2xcos(x)−1
1 = ±√
ke2sin(0)−2·0·cos(0)−1
1 = ±√
ke0−1 = ±√
k−11 =
√k−1
1 = k−1k = 2
y(x) =√
2e2sin(x)−2xcos(x)−1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = ±√
ke2sin(x)−2xcos(x)−1
1 = ±√
ke2sin(0)−2·0·cos(0)−1
1 = ±√
ke0−1 = ±√
k−11 =
√k−1
1 = k−1k = 2
y(x) =√
2e2sin(x)−2xcos(x)−1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = ±√
ke2sin(x)−2xcos(x)−1
1 =
±√
ke2sin(0)−2·0·cos(0)−1
1 = ±√
ke0−1 = ±√
k−11 =
√k−1
1 = k−1k = 2
y(x) =√
2e2sin(x)−2xcos(x)−1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = ±√
ke2sin(x)−2xcos(x)−1
1 = ±√
ke2sin(0)−2·0·cos(0)−1
1 = ±√
ke0−1 = ±√
k−11 =
√k−1
1 = k−1k = 2
y(x) =√
2e2sin(x)−2xcos(x)−1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = ±√
ke2sin(x)−2xcos(x)−1
1 = ±√
ke2sin(0)−2·0·cos(0)−1
1 = ±√
ke0−1 =
±√
k−11 =
√k−1
1 = k−1k = 2
y(x) =√
2e2sin(x)−2xcos(x)−1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = ±√
ke2sin(x)−2xcos(x)−1
1 = ±√
ke2sin(0)−2·0·cos(0)−1
1 = ±√
ke0−1 = ±√
k−1
1 =√
k−11 = k−1k = 2
y(x) =√
2e2sin(x)−2xcos(x)−1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = ±√
ke2sin(x)−2xcos(x)−1
1 = ±√
ke2sin(0)−2·0·cos(0)−1
1 = ±√
ke0−1 = ±√
k−11 =
√k−1
1 = k−1k = 2
y(x) =√
2e2sin(x)−2xcos(x)−1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = ±√
ke2sin(x)−2xcos(x)−1
1 = ±√
ke2sin(0)−2·0·cos(0)−1
1 = ±√
ke0−1 = ±√
k−11 =
√k−1
1 = k−1
k = 2
y(x) =√
2e2sin(x)−2xcos(x)−1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = ±√
ke2sin(x)−2xcos(x)−1
1 = ±√
ke2sin(0)−2·0·cos(0)−1
1 = ±√
ke0−1 = ±√
k−11 =
√k−1
1 = k−1k = 2
y(x) =√
2e2sin(x)−2xcos(x)−1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
1 = y(0)
y(x) = ±√
ke2sin(x)−2xcos(x)−1
1 = ±√
ke2sin(0)−2·0·cos(0)−1
1 = ±√
ke0−1 = ±√
k−11 =
√k−1
1 = k−1k = 2
y(x) =√
2e2sin(x)−2xcos(x)−1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
y(x) =√
2e2sin(x)−2xcos(x)−1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Solving the Initial Value Problem
y(x) =√
2e2sin(x)−2xcos(x)−1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =√
2e2sin(x)−2xcos(x)−1 Solve the IVP
y′ = xsin(x)y+xsin(x)
y, y(0) = 1?
y(x) =√
2e2sin(x)−2xcos(x)−1
y(0) =√
2e2sin(0)−2·0·cos(0)−1 =√
2e0−1 = 1√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =√
2e2sin(x)−2xcos(x)−1 Solve the IVP
y′ = xsin(x)y+xsin(x)
y, y(0) = 1?
y(x) =√
2e2sin(x)−2xcos(x)−1
y(0) =√
2e2sin(0)−2·0·cos(0)−1
=√
2e0−1 = 1√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
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Solving Initial Value Problems An Example Double Check
Does y(x) =√
2e2sin(x)−2xcos(x)−1 Solve the IVP
y′ = xsin(x)y+xsin(x)
y, y(0) = 1?
y(x) =√
2e2sin(x)−2xcos(x)−1
y(0) =√
2e2sin(0)−2·0·cos(0)−1 =√
2e0−1
= 1√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =√
2e2sin(x)−2xcos(x)−1 Solve the IVP
y′ = xsin(x)y+xsin(x)
y, y(0) = 1?
y(x) =√
2e2sin(x)−2xcos(x)−1
y(0) =√
2e2sin(0)−2·0·cos(0)−1 =√
2e0−1 = 1
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =√
2e2sin(x)−2xcos(x)−1 Solve the IVP
y′ = xsin(x)y+xsin(x)
y, y(0) = 1?
y(x) =√
2e2sin(x)−2xcos(x)−1
y(0) =√
2e2sin(0)−2·0·cos(0)−1 =√
2e0−1 = 1√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =√
2e2sin(x)−2xcos(x)−1 Solve the IVP
y′ = xsin(x)y+xsin(x)
y, y(0) = 1?
ddx
(√2e2sin(x)−2xcos(x)−1
)
=1
2√
2e2sin(x)−2xcos(x)−12e2sin(x)−2xcos(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =√
2e2sin(x)−2xcos(x)−1 Solve the IVP
y′ = xsin(x)y+xsin(x)
y, y(0) = 1?
ddx
(√2e2sin(x)−2xcos(x)−1
)=
1
2√
2e2sin(x)−2xcos(x)−1
2e2sin(x)−2xcos(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =√
2e2sin(x)−2xcos(x)−1 Solve the IVP
y′ = xsin(x)y+xsin(x)
y, y(0) = 1?
ddx
(√2e2sin(x)−2xcos(x)−1
)=
1
2√
2e2sin(x)−2xcos(x)−12e2sin(x)−2xcos(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =√
2e2sin(x)−2xcos(x)−1 Solve the IVP
y′ = xsin(x)y+xsin(x)
y, y(0) = 1?
ddx
(√2e2sin(x)−2xcos(x)−1
)=
2e2sin(x)−2xcos(x)
2√
2e2sin(x)−2xcos(x)−1
(2cos(x)−2cos(x)+2xsin(x)
)=
2e2sin(x)−2xcos(x)
2√
2e2sin(x)−2xcos(x)−12xsin(x)
=2e2sin(x)−2xcos(x)√
2e2sin(x)−2xcos(x)−1xsin(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =√
2e2sin(x)−2xcos(x)−1 Solve the IVP
y′ = xsin(x)y+xsin(x)
y, y(0) = 1?
ddx
(√2e2sin(x)−2xcos(x)−1
)=
2e2sin(x)−2xcos(x)
2√
2e2sin(x)−2xcos(x)−1
(2cos(x)−2cos(x)+2xsin(x)
)
=2e2sin(x)−2xcos(x)
2√
2e2sin(x)−2xcos(x)−12xsin(x)
=2e2sin(x)−2xcos(x)√
2e2sin(x)−2xcos(x)−1xsin(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =√
2e2sin(x)−2xcos(x)−1 Solve the IVP
y′ = xsin(x)y+xsin(x)
y, y(0) = 1?
ddx
(√2e2sin(x)−2xcos(x)−1
)=
2e2sin(x)−2xcos(x)
2√
2e2sin(x)−2xcos(x)−1
(2cos(x)−2cos(x)+2xsin(x)
)=
2e2sin(x)−2xcos(x)
2√
2e2sin(x)−2xcos(x)−12xsin(x)
=2e2sin(x)−2xcos(x)√
2e2sin(x)−2xcos(x)−1xsin(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
Does y(x) =√
2e2sin(x)−2xcos(x)−1 Solve the IVP
y′ = xsin(x)y+xsin(x)
y, y(0) = 1?
ddx
(√2e2sin(x)−2xcos(x)−1
)=
2e2sin(x)−2xcos(x)
2√
2e2sin(x)−2xcos(x)−1
(2cos(x)−2cos(x)+2xsin(x)
)=
2e2sin(x)−2xcos(x)
2√
2e2sin(x)−2xcos(x)−12xsin(x)
=2e2sin(x)−2xcos(x)√
2e2sin(x)−2xcos(x)−1xsin(x)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
xsin(x)y+xsin(x)
y
= xsin(x)√
2e2sin(x)−2xcos(x)−1+xsin(x)√
2e2sin(x)−2xcos(x)−1
=xsin(x)
(√2e2sin(x)−2xcos(x)−1
)2+ xsin(x)√
2e2sin(x)−2xcos(x)−1
=xsin(x)
(2e2sin(x)−2xcos(x)−1+1
)√
2e2sin(x)−2xcos(x)−1
= xsin(x)2e2sin(x)−2xcos(x)√
2e2sin(x)−2xcos(x)−1
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
xsin(x)y+xsin(x)
y
= xsin(x)√
2e2sin(x)−2xcos(x)−1+xsin(x)√
2e2sin(x)−2xcos(x)−1
=xsin(x)
(√2e2sin(x)−2xcos(x)−1
)2+ xsin(x)√
2e2sin(x)−2xcos(x)−1
=xsin(x)
(2e2sin(x)−2xcos(x)−1+1
)√
2e2sin(x)−2xcos(x)−1
= xsin(x)2e2sin(x)−2xcos(x)√
2e2sin(x)−2xcos(x)−1
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
xsin(x)y+xsin(x)
y
= xsin(x)√
2e2sin(x)−2xcos(x)−1+xsin(x)√
2e2sin(x)−2xcos(x)−1
=xsin(x)
(√2e2sin(x)−2xcos(x)−1
)2+ xsin(x)√
2e2sin(x)−2xcos(x)−1
=xsin(x)
(2e2sin(x)−2xcos(x)−1+1
)√
2e2sin(x)−2xcos(x)−1
= xsin(x)2e2sin(x)−2xcos(x)√
2e2sin(x)−2xcos(x)−1
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
xsin(x)y+xsin(x)
y
= xsin(x)√
2e2sin(x)−2xcos(x)−1+xsin(x)√
2e2sin(x)−2xcos(x)−1
=xsin(x)
(√2e2sin(x)−2xcos(x)−1
)2+ xsin(x)√
2e2sin(x)−2xcos(x)−1
=xsin(x)
(2e2sin(x)−2xcos(x)−1+1
)√
2e2sin(x)−2xcos(x)−1
= xsin(x)2e2sin(x)−2xcos(x)√
2e2sin(x)−2xcos(x)−1
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
xsin(x)y+xsin(x)
y
= xsin(x)√
2e2sin(x)−2xcos(x)−1+xsin(x)√
2e2sin(x)−2xcos(x)−1
=xsin(x)
(√2e2sin(x)−2xcos(x)−1
)2+ xsin(x)√
2e2sin(x)−2xcos(x)−1
=xsin(x)
(2e2sin(x)−2xcos(x)−1+1
)√
2e2sin(x)−2xcos(x)−1
= xsin(x)2e2sin(x)−2xcos(x)√
2e2sin(x)−2xcos(x)−1
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation
logo1
Solving Initial Value Problems An Example Double Check
xsin(x)y+xsin(x)
y
= xsin(x)√
2e2sin(x)−2xcos(x)−1+xsin(x)√
2e2sin(x)−2xcos(x)−1
=xsin(x)
(√2e2sin(x)−2xcos(x)−1
)2+ xsin(x)√
2e2sin(x)−2xcos(x)−1
=xsin(x)
(2e2sin(x)−2xcos(x)−1+1
)√
2e2sin(x)−2xcos(x)−1
= xsin(x)2e2sin(x)−2xcos(x)√
2e2sin(x)−2xcos(x)−1
√
Bernd Schroder Louisiana Tech University, College of Engineering and Science
An Initial Value Problem for a Separable Differential Equation