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bryanosaurus
Jun14-08, 09:25 PM
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

rock.freak667
Jun14-08, 10:17 PM
x^2dy + 2xy dx = x^2 dx

\equiv x^2 \frac{dy}{dx}+2xy=x^2

\equiv \frac{dy}{dx}+\frac{2}{x}y=1


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 Integrating Factor. (http://en.wikipedia.org/wiki/Integrating_factor)

bryanosaurus
Jun15-08, 09:29 AM
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.

arildno
Jun15-08, 09:53 AM
Well, introduce the new variable:
u=\frac{y}{x}
Then, we have:
\frac{du}{dx}=\frac{1}{x}\frac{dy}{dx}-\frac{u}{x}\to\frac{dy}{dx}=x\frac{du}dx}+u

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