Differential Equation : y/x substitution

Hey all,

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.

(y/x) sub difEq corrections made...please help me finish?

Thanks Danny,

I have seen the errors and redid it. If you don't mind I would like to stay with the y/x sub instead of the Bernoulli sub since we have not yet covered them ok?

Solve: (x^2 + 2y^2)dx + xydy = 0

xydy = -(x^2 + 2y^2)dx

dy/dx = -(x/y) - (2y/x)

let v = y/x, y = vx, dy/dx = v + xdv/dx

v + x(dv/dx) = -(1/v) - 2v

please check algebra here:

x(dv/dx) = (-1-3v^2)/v

-vdv/(1+3v^2) = dx/x

with a u substitution I get

LHS: -(1/6)ln|1 + 3y^2/x^2|

So:

ln|1 + 3y^2/x^2| = -6ln|x| + C

Is this correct so far? Also please could you show me some skills on how to wrap this up correctly. Thanks so much.