Differential equation?

Question

(y^2 - x^2)y' + 2xy = 0

Can anyone pls help me solve with this differential equation? Thanks a lot

Answer 1

First solve for y':

y' = -2xy/(y^2-x^2)

Divide numerator and denominator of right side by x^2 to get

y' = -2(y/x) / [1 - (y/x)^2]

Make the substitution u = y/x. The equation becomes

y' = -2u/(1-u^2).

Differentiate the u = y/x wrt x:

u' = (1/x)y' - (1/x^2)y;

Multiply both sides by x:

xu' = y' - y/x

Note that y/x = u and y' = -2u/(1-u^2); then

xu' = -2u/(1-u^2) - u

This equation has separable variables and can be written

x*du/dx = -2u/(1-u^2) - u

du/[-2u/(1-u^2) - u] = dx/x;

Integrating both sides get

∫1/[-2u/(1-u^2) - u] du = ln|x| + C

I'll leave it to you to solve the integral.

EDIT: I've worked on this some more, and the result is either extremely messy or very simple, depending on initial conditions (which were not given). My attempt at solution is here
http://img254.imageshack.us/img254/5945/diffeqinxandyrv7.png
check it out.