Question

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

Answer 1

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
\[\begin{align*}u'x+u&=u+\frac{1}{u^3}\\\\
u'x&=\frac{1}{u^3}\\\\
u^3~du&=\dfrac{dx}{x}\end{align*}\]

Answer 2

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

Answer 3

\(y=ux\) 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.

Answer 4

Okay

Answer 5

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)

Answer 6

Thanks!