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

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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

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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

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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

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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