Please help with: $$\frac{dy}{dx} =\frac{x}{x^2y + y^3}$$
The hint says let $u = x^2$
I have tried all the possible substitutions and manipulations on this ODE and I just can't separate variables or use any methods on this! What am I missing? The closest is that I have tried to flip everything so
$$\frac{dx}{dy} = xy + \frac{y^3}{x}$$
and tried to do Bernoulli's method on it but failed..
Thank you..
Picking up from where you left, $$\frac{dx}{dy} = xy + \frac{y^3}{x}$$ or, $$x\frac{dx}{dy} = x^2y + y^3$$ Now put $u=x^2$ as the hint says. This gives $$\frac{du}{dy} = 2x\frac{dx}{dy}$$ Plugging that in, $$\frac{1}{2}\frac{du}{dy} = uy + y^3$$ Rearranging, $$\frac{du}{dy} - 2uy = 2y^3$$ which is a linear ODE which I'm sure you can solve.
(Here, $P(y) = -2y$ so the IF is $e^{\int P(y) dy} = e^{\int -2ydy} = e^{-y^2}$)
EDIT: (Reason why I multiplied by x)
I multiplied by $x$ because I would need a $du=2xdx$ for the substitution $x^2=u$. More over, it removes the term where both $x$ and $y$ appear together ($\frac{y^3}{x}$). The reasoning is quite similar to that for Bernoulli's method where you divide by $y^n$ and put $y^{-n+1} =t$ because it makes the last term (the one on the right hand side) "y-free" and the other parts match because $$\frac{d(y^{-n+1})}{dx} = (1-n)y^{-n}\frac{dy}{dx}$$ and you already have $y^{-n}\frac{dy}{dx}$