# Math Help - "simple" initial value problem

1. ## "simple" initial value problem

I need help with this problem:

Setting y/x=u, solve:

xy' = (y-x)^3 + y; y(1) = 3/2

Thanks!

2. Originally Posted by ntfabolous
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.

$\frac{y}{x}=u \iff y=xu$

Taking the derivative we get

$y'=u+x\frac{du}{dx}$

Now subbing this into the equation we get

$x\left( u+x\frac{du}{dx}\right)=(ux-x)^3+ux$

Simplifing and factoring we get

$x^2\frac{du}{dx}=x^3(u-1)^3 \iff \frac{du}{dx}=x(u-1)^3$

$\frac{du}{(u-1)^3}=xdx$

Integrating both sides we get

$\frac{1}{(u-1)^2}=\frac{1}{2}x^2+c$

From here it is just back subbing good luck.