# Differential equation $xy^3y'=2y^4+x^4$

Solve the differential equation $$xy^3y'=2y^4+x^4$$

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Hint: Introduce the new function $z(x)$ with $y(x)= x z(x)$ to make the ODE separable.

Ater some calculation, you should obtain $$\frac{ z^3z'}{1+z^4} = \frac1x.$$

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Ah, got it! Very helpful! Thanks! (+1) – user 1618033 Jan 2 '13 at 10:49

Dividing $xy^3$ on both sides, we get $$\tag{1}y'=2\frac{y}{x}+\frac{x^3}{y^3}$$ which is homogenous differential equation. We can use the substitution $u=\frac{y}{x}$. Then $y=ux$ and $\frac{dy}{dx}=x\frac{du}{dx}+u$. Substituting it into $(1)$, we obtain $$x\frac{du}{dx}+u=2u+\frac{1}{u^3}$$ or $$x\frac{du}{dx}=u+\frac{1}{u^3}$$ which can be solved by separation of variables.

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nice way to choose (+1) – user 1618033 Jan 2 '13 at 10:50

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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