Another ODE, can't separate variables

  • #1

Main Question or Discussion Point

This is in a problem set for variables separate but I can't seem to separate them, and I do not know how to proceed.

(x^2)dy + 2xy dx = (x^2) dx

The solution given is: (3x^2)y = x^3 + c
 

Answers and Replies

  • #2
rock.freak667
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[tex]x^2dy + 2xy dx = x^2 dx[/tex]

[tex]\equiv x^2 \frac{dy}{dx}+2xy=x^2[/tex]

[tex]\equiv \frac{dy}{dx}+\frac{2}{x}y=1[/tex]


this is in a form that you should know how to solve and it isn't a separation of variables type.

If you don't know how to solve ODE's in this form check this link http://en.wikipedia.org/wiki/Integrating_factor" [Broken]
 
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  • #3
Thank you, yes I can solve from here. But I still don't understand why this was in the separation of variables chapter's problem set.
 
  • #4
arildno
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Well, introduce the new variable:
[tex]u=\frac{y}{x}[/tex]
Then, we have:
[tex]\frac{du}{dx}=\frac{1}{x}\frac{dy}{dx}-\frac{u}{x}\to\frac{dy}{dx}=x\frac{du}dx}+u[/tex]

We therefore get the diff.eq:
[tex]x\frac{du}{dx}+u+2u=1\to\frac{1}{1-3u}\frac{du}{dx}=\frac{1}{x}[/tex], which is separable.
 

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