**Another approach**

With equations like these, it is often good to check for **exactness**

An equation in this form:

#(x^2 + y^2) dx - 2xy dy = 0#

....can be related to level surface #\Phi(x,y) = const# such that #d\Phi = 0 = \Phi_x dx + \Phi_y dy#

So in this case

#\Phi_x =x^2 + y^2, \Phi_(xy) = 2y #

#\Phi_y =- 2xy, \Phi_(yx) = -2y #

But because #\Phi_(xy) ne \Phi_(yx) #, the equation is not exact

We can then re-arrange the equation to say:

#dy/dx = (x^2 +y^2)/(2xy) = x/(2y) + y/(2x)#

This is a **homogeneous** equation, ie

#dy/dx = f(x,y) = f(kx, ky)# for any constant #k#

in this case we make a substitution #v(x) = y/x, y = v x#, so that #y' = v' x + v# and the new equation becomes **separable**

So

#dy/dx = x/(2y) + y/(2x)#

#implies v'x + v = 1/(2v ) + v/2#

# v'x = (1-v^2)/(2v )#

# (2v)/(1-v^2) v' = 1/x#

We integrate wrt x:

#int( (2v)/(1-v^2) v' = 1/x ) dx#

#implies int (2v)/(1-v^2) dv =int 1/x dx#

Or

#-ln abs (1-v^2) = ln abs x + C \ \ \ [ = ln C abs x ]# **where, throughout, C is a generic constant**

#implies -1/ abs (1-v^2) = C abs x #

**For suitable v and x**

#1-v^2 = 1/ (C x) #

#1-(y/x)^2 = 1/ (C x) #

#y^2 =x^2 + C x #

#y = pm sqrt (x^2 + C x ) #