Originally Posted by

**cYn** OK so here is my problem:

$\displaystyle \frac{dy}{dx} = \frac{(x+y)^2}{2x}$

My steps:

$\displaystyle y = vx$

$\displaystyle y' = v + v'x$

$\displaystyle y' = \frac{(x+y)^2}{2x}$

$\displaystyle v + v'x = \frac{(x+y)^2}{2x}$

$\displaystyle v + \frac{dv}{dx}x = \frac{(x+y)^2}{2x}$

$\displaystyle v + \frac{dv}{dx}x = \frac{(x+vx)^2}{2x}$

$\displaystyle v + \frac{dv}{dx}x = \frac{x^2 + 2vx^2 +v^2x^2}{2x}$

$\displaystyle v + \frac{dv}{dx}x = \frac{x^2}{2x} + \frac{2vx^2}{2x} + \frac{v^2x^2}{2x}$

$\displaystyle v + \frac{dv}{dx}x = \frac{x}{2} + vx + \frac{v^2x}{2}$

If I divide both sides by x I get:

$\displaystyle \frac{v}{x} + \frac{dv}{dx} = \frac{1}{2} + v + \frac{v^2}{2}$

$\displaystyle \frac{v}{x} + dv = (\frac{1}{2} + v + \frac{v^2}{2})dx$

This is the part that I'm having trouble with. I haven't done homogeneous/seperable in a long time. Can someone help me with the next step? And possibly a solution?