Math Help - diff eq's

1. diff eq's

The original problem: x(dy/dx)=y+xe^(5y/x)

so far i got...
(dy/dx)=y/x+e^(5y/x)

u=y/x, xu=y therefore x(du/dx)=dy/dx

x(du/dx)=u + e^(5u)

I'm stuck here.

By the way, I'm new and I'm assuming you guys don't just give out solutions so a push in the right direction will do. Thanks

2. Originally Posted by aaronb
The original problem: x(dy/dx)=y+xe^(5y/x)

so far i got...
(dy/dx)=y/x+e^(5y/x)

u=y/x, xu=y therefore x(du/dx)=dy/dx

x(du/dx)=u + e^(5u)

I'm stuck here.

By the way, I'm new and I'm assuming you guys don't just give out solutions so a push in the right direction will do. Thanks
$u = \frac{y}{x} \Rightarrow y = x u$. To get $\frac{dy}{dx}$ you have to use the product rule: $\frac{dy}{dx} = u + x \frac{du}{dx}$.

So your DE becomes $u + x \frac{du}{dx} = u + e^{5u} \, ....$