logo1

Overview An Example Double Check

Linear First Order Differential Equations

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

What are Linear First Order DifferentialEquations?

1. A linear first order differential equation is of the formy′ +p(x)y = q(x).

2. Recognizing linear first order differential equationsrequires some pattern recognition.

3. To solve a linear first order differential equation, we turnthe left side into the derivative of a product.3.1 Compute the integrating factor µ(x) = e

∫p(x) dx,

3.2 Multiply the equation by the integrating factor,note that the left side e

∫p(x) dxy′ + e

∫p(x) dxp(x)y

is the derivative of the product e∫

p(x) dxy,3.3 Integrate both sides, solve for y.

That’s it.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

What are Linear First Order DifferentialEquations?

1. A linear first order differential equation is of the formy′ +p(x)y = q(x).

2. Recognizing linear first order differential equationsrequires some pattern recognition.

3. To solve a linear first order differential equation, we turnthe left side into the derivative of a product.3.1 Compute the integrating factor µ(x) = e

∫p(x) dx,

3.2 Multiply the equation by the integrating factor,note that the left side e

∫p(x) dxy′ + e

∫p(x) dxp(x)y

is the derivative of the product e∫

p(x) dxy,3.3 Integrate both sides, solve for y.

That’s it.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

What are Linear First Order DifferentialEquations?

1. A linear first order differential equation is of the formy′ +p(x)y = q(x).

2. Recognizing linear first order differential equationsrequires some pattern recognition.

3. To solve a linear first order differential equation, we turnthe left side into the derivative of a product.3.1 Compute the integrating factor µ(x) = e

∫p(x) dx,

3.2 Multiply the equation by the integrating factor,note that the left side e

∫p(x) dxy′ + e

∫p(x) dxp(x)y

is the derivative of the product e∫

p(x) dxy,3.3 Integrate both sides, solve for y.

That’s it.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

What are Linear First Order DifferentialEquations?

1. A linear first order differential equation is of the formy′ +p(x)y = q(x).

2. Recognizing linear first order differential equationsrequires some pattern recognition.

3. To solve a linear first order differential equation, we turnthe left side into the derivative of a product.

3.1 Compute the integrating factor µ(x) = e∫

p(x) dx,3.2 Multiply the equation by the integrating factor,

note that the left side e∫

p(x) dxy′ + e∫

p(x) dxp(x)yis the derivative of the product e

∫p(x) dxy,

3.3 Integrate both sides, solve for y.

That’s it.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

What are Linear First Order DifferentialEquations?

1. A linear first order differential equation is of the formy′ +p(x)y = q(x).

2. Recognizing linear first order differential equationsrequires some pattern recognition.

3. To solve a linear first order differential equation, we turnthe left side into the derivative of a product.3.1 Compute the integrating factor µ(x) = e

∫p(x) dx,

3.2 Multiply the equation by the integrating factor,note that the left side e

∫p(x) dxy′ + e

∫p(x) dxp(x)y

is the derivative of the product e∫

p(x) dxy,3.3 Integrate both sides, solve for y.

That’s it.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

What are Linear First Order DifferentialEquations?

1. A linear first order differential equation is of the formy′ +p(x)y = q(x).

2. Recognizing linear first order differential equationsrequires some pattern recognition.

3. To solve a linear first order differential equation, we turnthe left side into the derivative of a product.3.1 Compute the integrating factor µ(x) = e

∫p(x) dx,

3.2 Multiply the equation by the integrating factor,

note that the left side e∫

p(x) dxy′ + e∫

p(x) dxp(x)yis the derivative of the product e

∫p(x) dxy,

3.3 Integrate both sides, solve for y.

That’s it.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

What are Linear First Order DifferentialEquations?

1. A linear first order differential equation is of the formy′ +p(x)y = q(x).

2. Recognizing linear first order differential equationsrequires some pattern recognition.

3. To solve a linear first order differential equation, we turnthe left side into the derivative of a product.3.1 Compute the integrating factor µ(x) = e

∫p(x) dx,

3.2 Multiply the equation by the integrating factor,note that the left side e

∫p(x) dxy′ + e

∫p(x) dxp(x)y

is the derivative of the product e∫

p(x) dxy,

3.3 Integrate both sides, solve for y.

That’s it.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

What are Linear First Order DifferentialEquations?

1. A linear first order differential equation is of the formy′ +p(x)y = q(x).

2. Recognizing linear first order differential equationsrequires some pattern recognition.

3. To solve a linear first order differential equation, we turnthe left side into the derivative of a product.3.1 Compute the integrating factor µ(x) = e

∫p(x) dx,

3.2 Multiply the equation by the integrating factor,note that the left side e

∫p(x) dxy′ + e

∫p(x) dxp(x)y

is the derivative of the product e∫

p(x) dxy,3.3 Integrate both sides, solve for y.

That’s it.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

What are Linear First Order DifferentialEquations?

1. A linear first order differential equation is of the formy′ +p(x)y = q(x).

2. Recognizing linear first order differential equationsrequires some pattern recognition.

3. To solve a linear first order differential equation, we turnthe left side into the derivative of a product.3.1 Compute the integrating factor µ(x) = e

∫p(x) dx,

3.2 Multiply the equation by the integrating factor,note that the left side e

∫p(x) dxy′ + e

∫p(x) dxp(x)y

is the derivative of the product e∫

p(x) dxy,3.3 Integrate both sides, solve for y.

That’s it.Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.

Integrating factor:∫p(x) dx =

∫ cos(x)sin(x)

dx u := sin(x),dudx

= cos(x),

dx =du

cos(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.Integrating factor:∫

p(x) dx =

∫ cos(x)sin(x)

dx u := sin(x),dudx

= cos(x),

dx =du

cos(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.Integrating factor:∫

p(x) dx =∫ cos(x)

sin(x)dx

u := sin(x),dudx

= cos(x),

dx =du

cos(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.Integrating factor:∫

p(x) dx =∫ cos(x)

sin(x)dx u := sin(x),

dudx

= cos(x),

dx =du

cos(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.Integrating factor:∫

p(x) dx =∫ cos(x)

sin(x)dx u := sin(x),

dudx

= cos(x),

dx =du

cos(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.Integrating factor:∫

p(x) dx =∫ cos(x)

sin(x)dx u := sin(x),

dudx

= cos(x),

dx =du

cos(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.Integrating factor:∫

p(x) dx =∫ cos(x)

sin(x)dx u := sin(x),

dudx

= cos(x),

=∫ cos(x)

udu

cos(x)dx =

ducos(x)

=∫ 1

udu

= ln |u|+ c

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.Integrating factor:∫

p(x) dx =∫ cos(x)

sin(x)dx u := sin(x),

dudx

= cos(x),

=∫ cos(x)

udu

cos(x)dx =

ducos(x)

=∫ 1

udu

= ln |u|+ c

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.Integrating factor:∫

p(x) dx =∫ cos(x)

sin(x)dx u := sin(x),

dudx

= cos(x),

=∫ cos(x)

udu

cos(x)dx =

ducos(x)

=∫ 1

udu

= ln |u|+ c

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.Integrating factor:∫

p(x) dx =∫ cos(x)

sin(x)dx u := sin(x),

dudx

= cos(x),

=∫ cos(x)

udu

cos(x)dx =

ducos(x)

=∫ 1

udu

= ln |u|

= ln∣∣sin(x)

∣∣µ(x) = e

∫p(x) dx = eln

∣∣sin(x)∣∣= sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.Integrating factor:∫

p(x) dx =∫ cos(x)

sin(x)dx u := sin(x),

dudx

= cos(x),

=∫ cos(x)

udu

cos(x)dx =

ducos(x)

=∫ 1

udu

= ln |u|= ln

∣∣sin(x)∣∣

µ(x) = e∫

p(x) dx = eln∣∣sin(x)

∣∣= sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.Integrating factor:∫

p(x) dx =∫ cos(x)

sin(x)dx u := sin(x),

dudx

= cos(x),

=∫ cos(x)

udu

cos(x)dx =

ducos(x)

=∫ 1

udu

= ln |u|= ln

∣∣sin(x)∣∣

µ(x) =

e∫

p(x) dx = eln∣∣sin(x)

∣∣= sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.Integrating factor:∫

p(x) dx =∫ cos(x)

sin(x)dx u := sin(x),

dudx

= cos(x),

=∫ cos(x)

udu

cos(x)dx =

ducos(x)

=∫ 1

udu

= ln |u|= ln

∣∣sin(x)∣∣

µ(x) = e∫

p(x) dx

= eln∣∣sin(x)

∣∣= sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.Integrating factor:∫

p(x) dx =∫ cos(x)

sin(x)dx u := sin(x),

dudx

= cos(x),

=∫ cos(x)

udu

cos(x)dx =

ducos(x)

=∫ 1

udu

= ln |u|= ln

∣∣sin(x)∣∣

µ(x) = e∫

p(x) dx = eln∣∣sin(x)

∣∣

= sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.Integrating factor:∫

p(x) dx =∫ cos(x)

sin(x)dx u := sin(x),

dudx

= cos(x),

=∫ cos(x)

udu

cos(x)dx =

ducos(x)

=∫ 1

udu

= ln |u|= ln

∣∣sin(x)∣∣

µ(x) = e∫

p(x) dx = eln∣∣sin(x)

∣∣= sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.

y′ +cos(x)sin(x)

y = 1

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.y′ +

cos(x)sin(x)

y = 1

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.µ(x)

(y′ +

cos(x)sin(x)

y)

= 1 ·µ(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.sin(x)

(y′ +

cos(x)sin(x)

y)

= 1 · sin(x)

sin(x)y′ + cos(x)y = sin(x)ddx

(sin(x)y

)= sin(x)

sin(x)y = −cos(x)+ c

y = −cos(x)sin(x)

+c

sin(x)

= −cot(x)+c

sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.sin(x)

(y′ +

cos(x)sin(x)

y)

= 1 · sin(x)

sin(x)y′ + cos(x)y = sin(x)

ddx

(sin(x)y

)= sin(x)

sin(x)y = −cos(x)+ c

y = −cos(x)sin(x)

+c

sin(x)

= −cot(x)+c

sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.sin(x)

(y′ +

cos(x)sin(x)

y)

= 1 · sin(x)

sin(x)y′ + cos(x)y = sin(x)ddx

(sin(x)y

)= sin(x)

sin(x)y = −cos(x)+ c

y = −cos(x)sin(x)

+c

sin(x)

= −cot(x)+c

sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.sin(x)

(y′ +

cos(x)sin(x)

y)

= 1 · sin(x)

sin(x)y′ + cos(x)y = sin(x)ddx

(sin(x)y

)= sin(x)

sin(x)y = −cos(x)+ c

y = −cos(x)sin(x)

+c

sin(x)

= −cot(x)+c

sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.sin(x)

(y′ +

cos(x)sin(x)

y)

= 1 · sin(x)

sin(x)y′ + cos(x)y = sin(x)ddx

(sin(x)y

)= sin(x)

sin(x)y = −cos(x)+ c

y = −cos(x)sin(x)

+c

sin(x)

= −cot(x)+c

sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.sin(x)

(y′ +

cos(x)sin(x)

y)

= 1 · sin(x)

sin(x)y′ + cos(x)y = sin(x)ddx

(sin(x)y

)= sin(x)

sin(x)y = −cos(x)+ c

y = −cos(x)sin(x)

+c

sin(x)

= −cot(x)+c

sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.

y(x) = −cot(x)+c

sin(x)

0 = y(1) = −cot(1)+c

sin(1)c

sin(1)= cot(1)

c = cot(1)sin(1) = cos(1)

y(x) = −cot(x)+cos(1)sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.

y(x) = −cot(x)+c

sin(x)

0 = y(1) = −cot(1)+c

sin(1)c

sin(1)= cot(1)

c = cot(1)sin(1) = cos(1)

y(x) = −cot(x)+cos(1)sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.

y(x) = −cot(x)+c

sin(x)

0 = y(1)

= −cot(1)+c

sin(1)c

sin(1)= cot(1)

c = cot(1)sin(1) = cos(1)

y(x) = −cot(x)+cos(1)sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.

y(x) = −cot(x)+c

sin(x)

0 = y(1) = −cot(1)+c

sin(1)

csin(1)

= cot(1)

c = cot(1)sin(1) = cos(1)

y(x) = −cot(x)+cos(1)sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.

y(x) = −cot(x)+c

sin(x)

0 = y(1) = −cot(1)+c

sin(1)c

sin(1)= cot(1)

c = cot(1)sin(1) = cos(1)

y(x) = −cot(x)+cos(1)sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.

y(x) = −cot(x)+c

sin(x)

0 = y(1) = −cot(1)+c

sin(1)c

sin(1)= cot(1)

c = cot(1)sin(1)

= cos(1)

y(x) = −cot(x)+cos(1)sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.

y(x) = −cot(x)+c

sin(x)

0 = y(1) = −cot(1)+c

sin(1)c

sin(1)= cot(1)

c = cot(1)sin(1) = cos(1)

y(x) = −cot(x)+cos(1)sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.

y(x) = −cot(x)+c

sin(x)

0 = y(1) = −cot(1)+c

sin(1)c

sin(1)= cot(1)

c = cot(1)sin(1) = cos(1)

y(x) = −cot(x)+cos(1)sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Solve the Initial Value Problem y′+cos(x)sin(x)

y = 1,

y(1) = 0.

y(x) = −cot(x)+cos(1)sin(x)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Does y(x) = −cot(x)+cos(1)sin(x)

Really Solve the

Initial Value Problem y′+cos(x)sin(x)

y = 1, y(1) = 0?

y′ +cos(x)sin(x)

y ?= 1

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Does y(x) = −cot(x)+cos(1)sin(x)

Really Solve the

Initial Value Problem y′+cos(x)sin(x)

y = 1, y(1) = 0?

y′ +cos(x)sin(x)

y ?= 1

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Does y(x) = −cot(x)+cos(1)sin(x)

Really Solve the

Initial Value Problem y′+cos(x)sin(x)

y = 1, y(1) = 0?

ddx

(−cot(x)+

cos(1)sin(x)

)+

cos(x)sin(x)

(−cot(x)+

cos(1)sin(x)

)?= 1

1sin2(x)

− cos(1)cos(x)sin2(x)

− cos2(x)sin2(x)

+cos(1)cos(x)

sin2(x)?= 1

1− cos2(x)sin2(x)

?= 1

sin2(x)sin2(x)

?= 1

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Does y(x) = −cot(x)+cos(1)sin(x)

Really Solve the

Initial Value Problem y′+cos(x)sin(x)

y = 1, y(1) = 0?

ddx

(−cot(x)+

cos(1)sin(x)

)+

cos(x)sin(x)

(−cot(x)+

cos(1)sin(x)

)?= 1

1sin2(x)

− cos(1)cos(x)sin2(x)

− cos2(x)sin2(x)

+cos(1)cos(x)

sin2(x)?= 1

1− cos2(x)sin2(x)

?= 1

sin2(x)sin2(x)

?= 1

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Does y(x) = −cot(x)+cos(1)sin(x)

Really Solve the

Initial Value Problem y′+cos(x)sin(x)

y = 1, y(1) = 0?

ddx

(−cot(x)+

cos(1)sin(x)

)+

cos(x)sin(x)

(−cot(x)+

cos(1)sin(x)

)?= 1

1sin2(x)

− cos(1)cos(x)sin2(x)

− cos2(x)sin2(x)

+cos(1)cos(x)

sin2(x)?= 1

1− cos2(x)sin2(x)

?= 1

sin2(x)sin2(x)

?= 1

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Does y(x) = −cot(x)+cos(1)sin(x)

Really Solve the

Initial Value Problem y′+cos(x)sin(x)

y = 1, y(1) = 0?

ddx

(−cot(x)+

cos(1)sin(x)

)+

cos(x)sin(x)

(−cot(x)+

cos(1)sin(x)

)?= 1

1sin2(x)

− cos(1)cos(x)sin2(x)

− cos2(x)sin2(x)

+cos(1)cos(x)

sin2(x)?= 1

1− cos2(x)sin2(x)

?= 1

sin2(x)sin2(x)

?= 1

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Does y(x) = −cot(x)+cos(1)sin(x)

Really Solve the

Initial Value Problem y′+cos(x)sin(x)

y = 1, y(1) = 0?

ddx

(−cot(x)+

cos(1)sin(x)

)+

cos(x)sin(x)

(−cot(x)+

cos(1)sin(x)

)?= 1

1sin2(x)

− cos(1)cos(x)sin2(x)

− cos2(x)sin2(x)

+cos(1)cos(x)

sin2(x)

?= 1

1− cos2(x)sin2(x)

?= 1

sin2(x)sin2(x)

?= 1

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Does y(x) = −cot(x)+cos(1)sin(x)

Really Solve the

Initial Value Problem y′+cos(x)sin(x)

y = 1, y(1) = 0?

ddx

(−cot(x)+

cos(1)sin(x)

)+

cos(x)sin(x)

(−cot(x)+

cos(1)sin(x)

)?= 1

1sin2(x)

− cos(1)cos(x)sin2(x)

− cos2(x)sin2(x)

+cos(1)cos(x)

sin2(x)?= 1

1− cos2(x)sin2(x)

?= 1

sin2(x)sin2(x)

?= 1

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Does y(x) = −cot(x)+cos(1)sin(x)

Really Solve the

Initial Value Problem y′+cos(x)sin(x)

y = 1, y(1) = 0?

ddx

(−cot(x)+

cos(1)sin(x)

)+

cos(x)sin(x)

(−cot(x)+

cos(1)sin(x)

)?= 1

1sin2(x)

− cos(1)cos(x)sin2(x)

− cos2(x)sin2(x)

+cos(1)cos(x)

sin2(x)?= 1

1− cos2(x)sin2(x)

?= 1

sin2(x)sin2(x)

?= 1

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Does y(x) = −cot(x)+cos(1)sin(x)

Really Solve the

Initial Value Problem y′+cos(x)sin(x)

y = 1, y(1) = 0?

ddx

(−cot(x)+

cos(1)sin(x)

)+

cos(x)sin(x)

(−cot(x)+

cos(1)sin(x)

)?= 1

1sin2(x)

− cos(1)cos(x)sin2(x)

− cos2(x)sin2(x)

+cos(1)cos(x)

sin2(x)?= 1

1− cos2(x)sin2(x)

?= 1

sin2(x)sin2(x)

?= 1

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Does y(x) = −cot(x)+cos(1)sin(x)

Really Solve the

Initial Value Problem y′+cos(x)sin(x)

y = 1, y(1) = 0?

ddx

(−cot(x)+

cos(1)sin(x)

)+

cos(x)sin(x)

(−cot(x)+

cos(1)sin(x)

)?= 1

1sin2(x)

− cos(1)cos(x)sin2(x)

− cos2(x)sin2(x)

+cos(1)cos(x)

sin2(x)?= 1

1− cos2(x)sin2(x)

?= 1

sin2(x)sin2(x)

= 1√

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Does y(x) = −cot(x)+cos(1)sin(x)

Really Solve the

Initial Value Problem y′+cos(x)sin(x)

y = 1, y(1) = 0?

y(1) = − cot(1)+cos(1)sin(1)

= 0√

Yes, it does

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Does y(x) = −cot(x)+cos(1)sin(x)

Really Solve the

Initial Value Problem y′+cos(x)sin(x)

y = 1, y(1) = 0?

y(1) =

− cot(1)+cos(1)sin(1)

= 0√

Yes, it does

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Does y(x) = −cot(x)+cos(1)sin(x)

Really Solve the

Initial Value Problem y′+cos(x)sin(x)

y = 1, y(1) = 0?

y(1) = − cot(1)+cos(1)sin(1)

= 0√

Yes, it does

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Does y(x) = −cot(x)+cos(1)sin(x)

Really Solve the

Initial Value Problem y′+cos(x)sin(x)

y = 1, y(1) = 0?

y(1) = − cot(1)+cos(1)sin(1)

= 0

√

Yes, it does

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Does y(x) = −cot(x)+cos(1)sin(x)

Really Solve the

Initial Value Problem y′+cos(x)sin(x)

y = 1, y(1) = 0?

y(1) = − cot(1)+cos(1)sin(1)

= 0√

Yes, it does

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations

logo1

Overview An Example Double Check

Does y(x) = −cot(x)+cos(1)sin(x)

Really Solve the

Initial Value Problem y′+cos(x)sin(x)

y = 1, y(1) = 0?

y(1) = − cot(1)+cos(1)sin(1)

= 0√

Yes, it does

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Linear First Order Differential Equations