<s>[INST] Differential equation $xy^3y'=2y^4+x^4$ Solve the differential equation
$$xy^3y'=2y^4+x^4$$
 [/INST] Another method is to realise that $y^3y'$ is, up to a constant factor, the derivative of $y^4$. So we obtain:
\begin{align*}
xy^3y' &= 2y^4 + x^4\\
x\frac{d}{dx}\left(\frac{y^4}{4}\right) - 2y^4 &= x^4\\
\frac{d}{dx}(y^4) - \frac{8}{x}y^4 &= 4x^3\\
\end{align*}
where, in the last step, we need $x \neq 0$. Now you can solve this by using an integrating factor.
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