# Another ODE, can't separate variables

1. Jun 14, 2008

### bryanosaurus

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

2. Jun 14, 2008

### rock.freak667

$$x^2dy + 2xy dx = x^2 dx$$

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

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

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 http://en.wikipedia.org/wiki/Integrating_factor" [Broken]

Last edited by a moderator: May 3, 2017
3. Jun 15, 2008

### bryanosaurus

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. Jun 15, 2008

### arildno

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.