Is it possible to solve?

  • Thread starter busyocean
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  • #1
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Main Question or Discussion Point

I've tried almost everything but i still get stuck when finding the integral of :

(2x^2y-e^(-x^2))dx+(x+1)dy =0

To attempt to solve it i did the following:----> since the DE is not exact :

1) [M(x,y)/dy - N(x,y)/dx]/N(x,y) = [2x^2-1/x+1]
2) ∫[2x^2-1/x+1] =
3) Integrating Factor: e^(x^2 - 2x +ln|x|) to make it smaller I'm gonna call

{e^(x^2 - 2x +ln|x|)} = P

4) P*[(2x^2-e^-(x^2))dx + (x+1)dy] =0

After simplifying I get: [(2x^2)ye^(x^2)-1]dx = [e^(x^2)](x+1)dy

After that I don't know what to do.
 

Answers and Replies

  • #2
1,796
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I get the integrating factor:

[tex]\mu=e^{x^2-2x+\ln(1+x)}[/tex]

Once you get the integrating factor, multiply both sides of equation by it, then use the techniques of solving an exact equation (taking those partial integrals and all).
 
  • #3
798
34
Hi !
(2x²y-e^(-x²))dx+(x+1)dy = 0
is a linear ODE :
(x+1)y' +(2x²)y = e^(-x²)
You may solve it in using the classical method to solve linear ODEs.
 
  • #4
22,097
3,277
I've tried almost everything but i still get stuck when finding the integral of :

(2x^2y-e^(-x^2))dx+(x+1)dy =0

To attempt to solve it i did the following:----> since the DE is not exact :

1) [M(x,y)/dy - N(x,y)/dx]/N(x,y) = [2x^2-1/x+1]
2) ∫[2x^2-1/x+1] =
3) Integrating Factor: e^(x^2 - 2x +ln|x|) to make it smaller I'm gonna call

{e^(x^2 - 2x +ln|x|)} = P

4) P*[(2x^2-e^-(x^2))dx + (x+1)dy] =0

After simplifying I get: [(2x^2)ye^(x^2)-1]dx = [e^(x^2)](x+1)dy

After that I don't know what to do.
Here is a cool LaTeX guide: https://www.physicsforums.com/showpost.php?p=3977517&postcount=3
 

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