# Need help on a nonlinear first order DE question

Hi Everyone

I tried all ways I can to solve (y^2+x^2+x)y'-y=0, but still cann't find a way to solve it.

Could anybody help on it?

Thank you very much!

## Answers and Replies

Try and get it exact. You can write it as:

$$(y^2+x^2)dy+xdy-ydx=0$$

Ok, divide by x^2 to get:

$$\left(1+\frac{y^2}{x^2}\right)dy+\frac{xdy-ydx}{x^2}=0$$

bingo-bango right?

you wouldn't go one further?

Hi !
Another method consists in converting to polar coordinates, which leads to a linear ODE very easy to solve.
The result can be expressed on a parametric form : x=(c-t)/tan(t) , y= c-t
or as x fonction on y :
x = y / tan(c-y)

you wouldn't go one further?

Well, one is a differential of just y and the other is a differential of the quantity y/x. So then if we have:

$$\left(1+\frac{y^2}{x^2}\right)dy+\frac{xdy-ydx}{x^2}=0$$

we could re-write that as:

$$dy=\frac{d(y/x)}{1+\left(\frac{y}{x}\right)^2}$$

Now, that right side is equivalent to the expression:

$$\frac{du}{f(u)}$$

right? So we just integrate it:

$$\int dy=\int\frac{d(y/x)}{1+\left(\frac{y}{x}\right)^2}$$

and get:

$$y=\arctan(y/x)+c$$