I need help with this problem:
Setting y/x=u, solve:
xy' = (y-x)^3 + y; y(1) = 3/2
Thanks!
Using the hint given and assuming u is a function of x.
$\displaystyle \frac{y}{x}=u \iff y=xu$
Taking the derivative we get
$\displaystyle y'=u+x\frac{du}{dx}$
Now subbing this into the equation we get
$\displaystyle x\left( u+x\frac{du}{dx}\right)=(ux-x)^3+ux$
Simplifing and factoring we get
$\displaystyle x^2\frac{du}{dx}=x^3(u-1)^3 \iff \frac{du}{dx}=x(u-1)^3$
$\displaystyle \frac{du}{(u-1)^3}=xdx$
Integrating both sides we get
$\displaystyle \frac{1}{(u-1)^2}=\frac{1}{2}x^2+c$
From here it is just back subbing good luck.