Bernoulli Equations - University of Southern Mississippi

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Overview An Example Double Check

Bernoulli Equations

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Bernoulli Equations

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What are Bernoulli Equations?

1. A Bernoulli equation is of the form y′+p(x)y = q(x)yn,where n 6= 0,1.

2. Recognizing Bernoulli equations requires some patternrecognition.

3. To solve a Bernoulli equation, we translate the equationinto a linear equation.3.1 The substitution y = v

11−n turns the Bernoulli equation

y′ +p(x)y = q(x)yn into a linear first order equation for v,3.2 We can even write down the abstract form of the resulting

linear first order equation, but it is simpler to remember thesubstitution y = v

11−n ,

3.3 After we solve the equation for v, we obtain y as theappropriate power of v.

That’s it.


Bernoulli Equations

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What are Bernoulli Equations?1. A Bernoulli equation is of the form y′+p(x)y = q(x)yn,

where n 6= 0,1.

2. Recognizing Bernoulli equations requires some patternrecognition.





11−n ,


That’s it.


Bernoulli Equations

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where n 6= 0,1.2. Recognizing Bernoulli equations requires some pattern

recognition.





11−n ,


That’s it.


Bernoulli Equations

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recognition.3. To solve a Bernoulli equation, we translate the equation

into a linear equation.

3.1 The substitution y = v1

1−n turns the Bernoulli equationy′ +p(x)y = q(x)yn into a linear first order equation for v,

3.2 We can even write down the abstract form of the resultinglinear first order equation, but it is simpler to remember thesubstitution y = v

11−n ,


That’s it.


Bernoulli Equations

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into a linear equation.3.1 The substitution y = v


y′ +p(x)y = q(x)yn into a linear first order equation for v,

3.2 We can even write down the abstract form of the resultinglinear first order equation, but it is simpler to remember thesubstitution y = v

11−n ,


That’s it.


Bernoulli Equations

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11−n ,


That’s it.


Bernoulli Equations

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11−n ,


That’s it.


Bernoulli Equations

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11−n ,


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

Bernoulli Equations

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Solve the Initial Value Problem y′+ y = x2y5,y(0) = 1.

Reduction to a linear equation.

n = 5, y = v1

1−5 = v−14 , y′ =

ddx

v−14 = −1

4v−

54 v′

y′+ y = x2y5


Bernoulli Equations

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n = 5, y = v1

1−5 = v−14 , y′ =

ddx

v−14 = −1

4v−

54 v′

y′+ y = x2y5


Bernoulli Equations

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n = 5,

y = v1

1−5 = v−14 , y′ =

ddx

v−14 = −1

4v−

54 v′

y′+ y = x2y5


Bernoulli Equations

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n = 5, y = v1

1−5

= v−14 , y′ =

ddx

v−14 = −1

4v−

54 v′

y′+ y = x2y5


Bernoulli Equations

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n = 5, y = v1

1−5 = v−14 ,

y′ =ddx

v−14 = −1

4v−

54 v′

y′+ y = x2y5


Bernoulli Equations

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n = 5, y = v1

1−5 = v−14 , y′ =

ddx

v−14 = −1

4v−

54 v′

y′+ y = x2y5


Bernoulli Equations

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n = 5, y = v1

1−5 = v−14 , y′ =

ddx

v−14

= −14

v−54 v′

y′+ y = x2y5


Bernoulli Equations

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n = 5, y = v1

1−5 = v−14 , y′ =

ddx

v−14 = −1

4v−

54 v′

y′+ y = x2y5


Bernoulli Equations

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n = 5, y = v1

1−5 = v−14 , y′ =

ddx

v−14 = −1

4v−

54 v′

y′+ y = x2y5


Bernoulli Equations

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n = 5, y = v1

1−5 = v−14 , y′ =

ddx

v−14 = −1

4v−

54 v′

y′+ y = x2(

v−14

)5


Bernoulli Equations

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n = 5, y = v1

1−5 = v−14 , y′ =

ddx

v−14 = −1

4v−

54 v′

y′+(

v−14

)= x2

(v−

14

)5


Bernoulli Equations

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n = 5, y = v1

1−5 = v−14 , y′ =

ddx

v−14 = −1

4v−

54 v′

−14

v−54 v′+

(v−

14

)= x2

(v−

14

)5

−14

v−54 v′+ v−

14 = x2v−

54

−14

v′+ v = x2

v′−4v = −4x2


Bernoulli Equations

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n = 5, y = v1

1−5 = v−14 , y′ =

ddx

v−14 = −1

4v−

54 v′

−14

v−54 v′+

(v−

14

)= x2

(v−

14

)5

−14

v−54 v′+ v−

14 = x2v−

54

−14

v′+ v = x2

v′−4v = −4x2


Bernoulli Equations

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n = 5, y = v1

1−5 = v−14 , y′ =

ddx

v−14 = −1

4v−

54 v′

−14

v−54 v′+

(v−

14

)= x2

(v−

14

)5

−14

v−54 v′+ v−

14 = x2v−

54

−14

v′+ v = x2

v′−4v = −4x2


Bernoulli Equations

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n = 5, y = v1

1−5 = v−14 , y′ =

ddx

v−14 = −1

4v−

54 v′

−14

v−54 v′+

(v−

14

)= x2

(v−

14

)5

−14

v−54 v′+ v−

14 = x2v−

54

−14

v′+ v = x2

v′−4v = −4x2


Bernoulli Equations

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Solving the linear equation.

v′−4v = −4x2

µ(x) = e∫

p(x) dx = e∫−4 dx = e−4x

e−4x (v′−4v

)= −4x2e−4x

e−4xv′−4e−4xv = −4x2e−4x(e−4xv

)′= −4x2e−4x

e−4xv =∫−4x2e−4x dx

v = −4e4x∫

x2e−4x dx


Bernoulli Equations

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v′−4v = −4x2

µ(x) = e∫

p(x) dx = e∫−4 dx = e−4x

e−4x (v′−4v

)= −4x2e−4x


)′= −4x2e−4x


v = −4e4x∫

x2e−4x dx


Bernoulli Equations

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v′−4v = −4x2

µ(x)

= e∫

p(x) dx = e∫−4 dx = e−4x

e−4x (v′−4v

)= −4x2e−4x


)′= −4x2e−4x


v = −4e4x∫

x2e−4x dx


Bernoulli Equations

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v′−4v = −4x2

µ(x) = e∫

p(x) dx

= e∫−4 dx = e−4x

e−4x (v′−4v

)= −4x2e−4x


)′= −4x2e−4x


v = −4e4x∫

x2e−4x dx


Bernoulli Equations

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v′−4v = −4x2

µ(x) = e∫

p(x) dx = e∫−4 dx

= e−4x

e−4x (v′−4v

)= −4x2e−4x


)′= −4x2e−4x


v = −4e4x∫

x2e−4x dx


Bernoulli Equations

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v′−4v = −4x2

µ(x) = e∫

p(x) dx = e∫−4 dx = e−4x

e−4x (v′−4v

)= −4x2e−4x


)′= −4x2e−4x


v = −4e4x∫

x2e−4x dx


Bernoulli Equations

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v′−4v = −4x2

µ(x) = e∫

p(x) dx = e∫−4 dx = e−4x

e−4x (v′−4v

)= −4x2e−4x


)′= −4x2e−4x


v = −4e4x∫

x2e−4x dx


Bernoulli Equations

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v′−4v = −4x2

µ(x) = e∫

p(x) dx = e∫−4 dx = e−4x

e−4x (v′−4v

)= −4x2e−4x

e−4xv′−4e−4xv = −4x2e−4x

(e−4xv

)′= −4x2e−4x


v = −4e4x∫

x2e−4x dx


Bernoulli Equations

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v′−4v = −4x2

µ(x) = e∫

p(x) dx = e∫−4 dx = e−4x

e−4x (v′−4v

)= −4x2e−4x


)′= −4x2e−4x


v = −4e4x∫

x2e−4x dx


Bernoulli Equations

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v′−4v = −4x2

µ(x) = e∫

p(x) dx = e∫−4 dx = e−4x

e−4x (v′−4v

)= −4x2e−4x


)′= −4x2e−4x


v = −4e4x∫

x2e−4x dx


Bernoulli Equations

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v′−4v = −4x2

µ(x) = e∫

p(x) dx = e∫−4 dx = e−4x

e−4x (v′−4v

)= −4x2e−4x


)′= −4x2e−4x


v = −4e4x∫

x2e−4x dx


Bernoulli Equations

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Solving the linear equation (continued).∫x2e−4x dx =

− 14

e−4xx2−∫ (

−14

)e−4x2x dx

= −14

e−4xx2 +12

∫e−4xx dx

= −14

e−4xx2 +12

[−1

4e−4xx−

∫−1

4e−4x dx

]= −1

4e−4xx2− 1

8e−4xx+

18

∫e−4x dx

= −14

e−4xx2− 18

e−4xx− 132

e−4x + c


Bernoulli Equations

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Solving the linear equation (continued).∫x2e−4x dx = − 1

4e−4xx2−

∫ (−1

4

)e−4x2x dx

= −14

e−4xx2 +12

∫e−4xx dx

= −14

e−4xx2 +12

[−1

4e−4xx−

∫−1

4e−4x dx

]= −1

4e−4xx2− 1

8e−4xx+

18

∫e−4x dx

= −14

e−4xx2− 18

e−4xx− 132

e−4x + c


Bernoulli Equations

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4e−4xx2−

∫ (−1

4

)e−4x2x dx

= −14

e−4xx2 +12

∫e−4xx dx

= −14

e−4xx2 +12

[−1

4e−4xx−

∫−1

4e−4x dx

]= −1

4e−4xx2− 1

8e−4xx+

18

∫e−4x dx

= −14

e−4xx2− 18

e−4xx− 132

e−4x + c


Bernoulli Equations

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4e−4xx2−

∫ (−1

4

)e−4x2x dx

= −14

e−4xx2 +12

∫e−4xx dx

= −14

e−4xx2 +12

[−1

4e−4xx−

∫−1

4e−4x dx

]

= −14

e−4xx2− 18

e−4xx+18

∫e−4x dx

= −14

e−4xx2− 18

e−4xx− 132

e−4x + c


Bernoulli Equations

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4e−4xx2−

∫ (−1

4

)e−4x2x dx

= −14

e−4xx2 +12

∫e−4xx dx

= −14

e−4xx2 +12

[−1

4e−4xx−

∫−1

4e−4x dx

]= −1

4e−4xx2− 1

8e−4xx+

18

∫e−4x dx

= −14

e−4xx2− 18

e−4xx− 132

e−4x + c


Bernoulli Equations

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4e−4xx2−

∫ (−1

4

)e−4x2x dx

= −14

e−4xx2 +12

∫e−4xx dx

= −14

e−4xx2 +12

[−1

4e−4xx−

∫−1

4e−4x dx

]= −1

4e−4xx2− 1

8e−4xx+

18

∫e−4x dx

= −14

e−4xx2− 18

e−4xx− 132

e−4x + c


Bernoulli Equations

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Solving the linear equation (concluded), general solution of theoriginal equation.

v = −4e4x[−1

4e−4xx2− 1

8e−4xx− 1

32e−4x + c

]= x2 +

12

x+18

+ ce4x

y = v−14 =

(x2 +

12

x+18

+ ce4x)− 1

4


Bernoulli Equations

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v = −4e4x[−1

4e−4xx2− 1

8e−4xx− 1

32e−4x + c

]

= x2 +12

x+18

+ ce4x

y = v−14 =

(x2 +

12

x+18

+ ce4x)− 1

4


Bernoulli Equations

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v = −4e4x[−1

4e−4xx2− 1

8e−4xx− 1

32e−4x + c

]= x2 +

12

x+18

+ ce4x

y = v−14 =

(x2 +

12

x+18

+ ce4x)− 1

4


Bernoulli Equations

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v = −4e4x[−1

4e−4xx2− 1

8e−4xx− 1

32e−4x + c

]= x2 +

12

x+18

+ ce4x

y = v−14

=(

x2 +12

x+18

+ ce4x)− 1

4


Bernoulli Equations

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v = −4e4x[−1

4e−4xx2− 1

8e−4xx− 1

32e−4x + c

]= x2 +

12

x+18

+ ce4x

y = v−14 =

(x2 +

12

x+18

+ ce4x)− 1

4


Bernoulli Equations

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Finding c.

y =(

x2 +12

x+18

+ ce4x)− 1

4

1 = y(0) =(

02 +12·0+

18

+ ce4·0)− 1

4

1 =(

18

+ c)− 1

4

1 =18

+ c

c =78, y =

(x2 +

12

x+18

+78

e4x)− 1

4


Bernoulli Equations

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Finding c.y =

(x2 +

12

x+18

+ ce4x)− 1

4

1 = y(0) =(

02 +12·0+

18

+ ce4·0)− 1

4

1 =(

18

+ c)− 1

4

1 =18

+ c

c =78, y =

(x2 +

12

x+18

+78

e4x)− 1

4


Bernoulli Equations

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Finding c.y =

(x2 +

12

x+18

+ ce4x)− 1

4

1 = y(0)

=(

02 +12·0+

18

+ ce4·0)− 1

4

1 =(

18

+ c)− 1

4

1 =18

+ c

c =78, y =

(x2 +

12

x+18

+78

e4x)− 1

4


Bernoulli Equations

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Finding c.y =

(x2 +

12

x+18

+ ce4x)− 1

4

1 = y(0) =(

02 +12·0+

18

+ ce4·0)− 1

4

1 =(

18

+ c)− 1

4

1 =18

+ c

c =78, y =

(x2 +

12

x+18

+78

e4x)− 1

4


Bernoulli Equations

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Finding c.y =

(x2 +

12

x+18

+ ce4x)− 1

4

1 = y(0) =(

02 +12·0+

18

+ ce4·0)− 1

4

1 =(

18

+ c)− 1

4

1 =18

+ c

c =78, y =

(x2 +

12

x+18

+78

e4x)− 1

4


Bernoulli Equations

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Finding c.y =

(x2 +

12

x+18

+ ce4x)− 1

4

1 = y(0) =(

02 +12·0+

18

+ ce4·0)− 1

4

1 =(

18

+ c)− 1

4

1 =18

+ c

c =78, y =

(x2 +

12

x+18

+78

e4x)− 1

4


Bernoulli Equations

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Finding c.y =

(x2 +

12

x+18

+ ce4x)− 1

4

1 = y(0) =(

02 +12·0+

18

+ ce4·0)− 1

4

1 =(

18

+ c)− 1

4

1 =18

+ c

c =78,

y =(

x2 +12

x+18

+78

e4x)− 1

4


Bernoulli Equations

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Finding c.y =

(x2 +

12

x+18

+ ce4x)− 1

4

1 = y(0) =(

02 +12·0+

18

+ ce4·0)− 1

4

1 =(

18

+ c)− 1

4

1 =18

+ c

c =78, y =

(x2 +

12

x+18

+78

e4x)− 1

4


Bernoulli Equations

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y =(

x2 +12

x+18

+78

e4x)− 1

4


Bernoulli Equations

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Does y =(

x2 +12

x+18

+78

e4x)−1

4

Really Solve

the Initial Value Problem y′+ y = x2y5, y(0) = 1?

ddx

[(x2 +

12

x+18

+78

e4x)− 1

4]

+(

x2 +12

x+18

+78

e4x)− 1

4

= −14

(x2 +

12

x+18

+78

e4x)− 5

4(

2x+12

+72

e4x)

+(

x2 +12

x+18

+78

e4x)− 1

4

=(

x2 +12

x+18

+78

e4x)− 5

4(− x

2− 1

8− 7

8e4x + x2 +

12

x+18

+78

e4x)

=(

x2 +12

x+18

+78

e4x)− 5

4x2 = y5x2 √


Bernoulli Equations

logo1


Does y =(

x2 +12

x+18

+78

e4x)−1

4

Really Solve


ddx

[(x2 +

12

x+18

+78

e4x)− 1

4]

+(

x2 +12

x+18

+78

e4x)− 1

4

= −14

(x2 +

12

x+18

+78

e4x)− 5

4(

2x+12

+72

e4x)

+(

x2 +12

x+18

+78

e4x)− 1

4

=(

x2 +12

x+18

+78

e4x)− 5

4(− x

2− 1

8− 7

8e4x + x2 +

12

x+18

+78

e4x)

=(

x2 +12

x+18

+78

e4x)− 5

4x2 = y5x2 √


Bernoulli Equations

logo1


Does y =(

x2 +12

x+18

+78

e4x)−1

4

Really Solve


ddx

[(x2 +

12

x+18

+78

e4x)− 1

4]

+(

x2 +12

x+18

+78

e4x)− 1

4

= −14

(x2 +

12

x+18

+78

e4x)− 5

4(

2x+12

+72

e4x)

+(

x2 +12

x+18

+78

e4x)− 1

4

=(

x2 +12

x+18

+78

e4x)− 5

4(− x

2− 1

8− 7

8e4x + x2 +

12

x+18

+78

e4x)

=(

x2 +12

x+18

+78

e4x)− 5

4x2 = y5x2 √


Bernoulli Equations

logo1


Does y =(

x2 +12

x+18

+78

e4x)−1

4

Really Solve


ddx

[(x2 +

12

x+18

+78

e4x)− 1

4]

+(

x2 +12

x+18

+78

e4x)− 1

4

= −14

(x2 +

12

x+18

+78

e4x)− 5

4(

2x+12

+72

e4x)

+(

x2 +12

x+18

+78

e4x)− 1

4

=(

x2 +12

x+18

+78

e4x)− 5

4(− x

2− 1

8− 7

8e4x + x2 +

12

x+18

+78

e4x)

=(

x2 +12

x+18

+78

e4x)− 5

4x2

= y5x2 √


Bernoulli Equations

logo1


Does y =(

x2 +12

x+18

+78

e4x)−1

4

Really Solve


ddx

[(x2 +

12

x+18

+78

e4x)− 1

4]

+(

x2 +12

x+18

+78

e4x)− 1

4

= −14

(x2 +

12

x+18

+78

e4x)− 5

4(

2x+12

+72

e4x)

+(

x2 +12

x+18

+78

e4x)− 1

4

=(

x2 +12

x+18

+78

e4x)− 5

4(− x

2− 1

8− 7

8e4x + x2 +

12

x+18

+78

e4x)

=(

x2 +12

x+18

+78

e4x)− 5

4x2 = y5x2

√


Bernoulli Equations

logo1


Does y =(

x2 +12

x+18

+78

e4x)−1

4

Really Solve


ddx

[(x2 +

12

x+18

+78

e4x)− 1

4]

+(

x2 +12

x+18

+78

e4x)− 1

4

= −14

(x2 +

12

x+18

+78

e4x)− 5

4(

2x+12

+72

e4x)

+(

x2 +12

x+18

+78

e4x)− 1

4

=(

x2 +12

x+18

+78

e4x)− 5

4(− x

2− 1

8− 7

8e4x + x2 +

12

x+18

+78

e4x)

=(

x2 +12

x+18

+78

e4x)− 5

4x2 = y5x2 √


Bernoulli Equations

logo1


Does y =(

x2 +12

x+18

+78

e4x)−1

4

Really Solve


y(0) =(

02 +12·0+

18

+78

e4·0)− 1

4

=(

18

+78

)− 14

= 1√

Yes, it does.


Bernoulli Equations

logo1


Does y =(

x2 +12

x+18

+78

e4x)−1

4

Really Solve


y(0) =(

02 +12·0+

18

+78

e4·0)− 1

4

=(

18

+78

)− 14

= 1√

Yes, it does.


Bernoulli Equations

logo1


Does y =(

x2 +12

x+18

+78

e4x)−1

4

Really Solve


y(0) =(

02 +12·0+

18

+78

e4·0)− 1

4

=(

18

+78

)− 14

= 1

√

Yes, it does.


Bernoulli Equations

logo1


Does y =(

x2 +12

x+18

+78

e4x)−1

4

Really Solve


y(0) =(

02 +12·0+

18

+78

e4·0)− 1

4

=(

18

+78

)− 14

= 1√

Yes, it does.


Bernoulli Equations

logo1


Does y =(

x2 +12

x+18

+78

e4x)−1

4

Really Solve


y(0) =(

02 +12·0+

18

+78

e4·0)− 1

4

=(

18

+78

)− 14

= 1√

Yes, it does.


Bernoulli Equations

Bernoulli Equations - University of Southern Mississippi

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