Solve xy^(3)y'=y^4+x^4 explicitly.

Question

anonymous · Answer

Another substitution like $y=ux$:
$$\begin{align*}
xy^3y'&=y^4+x^4\
y'&=\frac{y}{x}+\frac{x^3}{y^3}
\end{align*}$$
Letting $u=\dfrac{y}{x}$ and using $y'=u'x+u$, you have
$$\begin{align*}u'x+u&=u+\frac{1}{u^3}\\
u'x&=\frac{1}{u^3}\\
u^3~du&=\dfrac{dx}{x}\end{align*}$$

idealist10 · Answer

How did you came up with the substitution y=ux?

anonymous · Answer

$y=ux$ is a pretty standard sub (refer to the link I posted a while ago), and it comes from being able to write the equation in terms of the expression $\dfrac{y}{x}$. I saw that you could do that, so I used this sub.

idealist10 · Answer

Okay

idealist10 · Answer

After I integrate, I got u^4/4=ln abs(x)+C, and u^4=4ln abs(x)+C,

u=(4*ln abs(x)+C)^(1/4)

y/x=(4*ln abs(x)+C)^(1/4)

y=x(4*ln abs(x)+C)^(1/4)

idealist10 · Answer

Thanks!