# First order differential equation $y'y^2=y+xy'$, switching dependent and independent variable

I want to solve the differential equation $y'y^2=y+xy'$. I notice that the differential is not in the general form of first order linear equations which is $y'+P(x)y=Q(x)$. In fact the above equation is not linear.

I read that I have to switch the dependent and independent variable of the problem and then the problem will turn into linear first order differential equation. I don't quite understand the concept of switching dependent and independent variable, can someone explain?

• Direct integration yields $y^3-3xy=3k$ for some constant $k$, hence $x=\frac{y^2}{3}-\frac{k}{y}$. – Omran Kouba Sep 16 '15 at 20:29

I'm not sure, what do you mean by switching variables, but if you simply need to solve it, here's what I'd do $$y'y^2 = y + xy' \implies \left(\frac {y^3}3 \right)' = (xy)' \implies \frac {y^3}3 = xy + C$$ You can leave it as it is in the form of implicit function, or solve for $x$ $$x = \frac {y^3 - 3C}{3y}$$
To follow on from the comment above. Two things to know $$\dfrac{d}{dx} g(y) = \dfrac{dg}{dy}y'$$ And $$f'(x)y + f(x)y' = (f(x)y)'$$ Always look for tricks like the above when solving (or trying to solve) odes.