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Differential equations help

  • Thread starter stefan10
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  • #1
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Homework Statement



Solve each of these differential equations by two different methods.

[tex] \frac{dy}{dx} = 4(y+1)x^3[/tex]


Homework Equations



Integrating factor

[tex] \rho = \exp (\int(p(x) dx)[/tex]

Linear Equation

[tex] \frac{dy}{dx} + p(x) y(x) = Q(x) [/tex]


The Attempt at a Solution



So I first solved it using separation. I get [tex] y=A \exp(x^4) -1 [/tex]

I then try to solve it using Integrating factors.

[tex] \frac{dy}{dx} + - 4x^3 y=4x^3 [/tex]
[tex] \rho = \exp (\int -4x^3 dx) = exp(-x^4) [/tex]

Multiplying through by the integrating factor I get.

[tex] \exp(-x^4) y= \int 4x^3 exp(-x^4) dx [/tex]

but when I solve it, I get

[tex] y = -1 [/tex]


What can I be doing wrong? I checked my steps multiple times.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Homework Statement



Solve each of these differential equations by two different methods.

[tex] \frac{dy}{dx} = 4(y+1)x^3[/tex]


Homework Equations



Integrating factor

[tex] \rho = \exp (\int(p(x) dx)[/tex]

Linear Equation

[tex] \frac{dy}{dx} + p(x) y(x) = Q(x) [/tex]


The Attempt at a Solution



So I first solved it using separation. I get [tex] y=A \exp(x^4) -1 [/tex]

I then try to solve it using Integrating factors.

[tex] \frac{dy}{dx} + - 4x^3 y=4x^3 [/tex]
[tex] \rho = \exp (\int -4x^3 dx) = exp(-x^4) [/tex]

Multiplying through by the integrating factor I get.

[tex] \exp(-x^4) y= \int 4x^3 exp(-x^4) dx [/tex]

but when I solve it, I get

[tex] y = -1 [/tex]


What can I be doing wrong? I checked my steps multiple times.
You are forgetting the constant of integration on ##\int 4x^3 exp(-x^4) dx##.
 
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  • #3
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You are forgetting the constant of integration on ##\int 4x^3 exp(-x^4) dx##.
Oh yeah!! I knew it was something so simple. Thank you very much! :)
 

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