A tricky Differential Equation

Question

How do you solve $$\frac{dy}{dx} = \frac{y^3}{e^{2x} + y^2}$$

I just need a hint. Its not an exact differential nor a linear D. E which I can solve...

Answer 1

$\dfrac{dy}{dx}=\dfrac{y^3}{e^{2x}+y^2}$

$\dfrac{dx}{dy}=\dfrac{e^{2x}+y^2}{y^3}$

$\dfrac{dx}{dy}=\dfrac{e^{2x}}{y^3}+\dfrac{1}{y}$

Which belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0142.pdf

Let $u=e^{-2x}$ ,

Then $\dfrac{du}{dy}=-2e^{-2x}\dfrac{dx}{dy}=-2u\dfrac{dx}{dy}$

$\therefore-\dfrac{1}{2u}\dfrac{du}{dy}=\dfrac{1}{y^3u}+\dfrac{1}{y}$

$\dfrac{du}{dy}+\dfrac{2u}{y}=-\dfrac{2}{y^3}$

I.F.$=e^{\int\frac{2~dy}{y}}=e^{2\ln y}=y^2$

$\therefore\dfrac{d(y^2u)}{dy}=-\dfrac{2}{y}$

$y^2u=-2\ln y+C$

$e^{-2x}=\dfrac{C-2\ln y}{y^2}$