Question

heLlo, can someone help me out on the following equation. change it to 1st order homogeneous then use v= y/x substitution method
xy^2y'=X^3 + Y^3

Answer 1

i divided both side with xy^2 then the ODE becomes 1st order homogeneous
y'= x^3/(xy^2) + y^3/(xy^2)
after i worked it out my answer is y = x/(ln |x| +C) but the answer is wrong when compare to the solution manual

Answer 2

\[xy^2y'=x^3+y^3\]
Letting \(y=xv\) (the same substitution, pretty much), you have \(y'=v+xv'\).
\[\begin{align*}x^3v^2(v+xv')&=x^3+x^3v^3\\
v^2(v+xv')&=1+v^3\\
v+xv'&=\frac{1+v^3}{v^2}\\
v+xv'&=\frac{1}{v^2}+v\\
xv'&=\frac{1}{v^2}\\
v^2~dv&=\frac{dx}{x}
\end{align*}\]
Is this the answer you're given (or something similar)?
\[y=x\left(\ln (Cx^3)\right)^{1/3}\]