Solve: dy/dx = 2xy/(x^2 - y^2)
[solution]
(x^2-y^2) dy = 2xy dx = y d(x^2)
Let X = x^2, then after dividing y on both sides, one can get
(X/y - y) dy = dX
Let Y = y^2, then
y = Y^(1/2), and dY = 2y dy = 2Y^(1/2) dy.
So dy =(1/2) *Y(-1/2) dY and then
[ X/Y^(1/2) - Y^(1/2) ] dY = 2*Y^(1/2) dX, or
(X-Y) dY = 2Y dX.
Add 2YdY on both sides, one can get
(X+Y) dY = 2Y(dX+dY) = 2Y d(X+Y).
Let Z = X+Y, then one can get separable differential equation as
Z dY = 2Y dZ, or
d(lnY) = 2 d(lnZ)
ln Y = 2 lnZ + C1
Y = BZ^2.
Rewrite the solution using the (x,y) variables:
y^2 = B(x^2 + y^2)^2
Take square root on both sides,
y = c (x^2 + y^2)
[QED]