I used a v = y/x substitution on this and I got a really weird looking answer so I just want to see how right or wrong I am.
Solve: (x^2 + 2y^2)dx + xydx = 0
v = y/x and dy/dx = v +x(dv/dx)
So: x(dv/dx) = (-1/v) - v
after integrating I get:
(-1/2)ln|1 + v^2| = ln(x) + C
subbing back gives
(1 + y^2/x^2)^2 = x + e^c
Can someone please check this work and then come up with an equation for y? I have an answer but it looks so weird I don't think its right. Thanks.