solve-
(x2 + xy)dy = ( x2 + y2)dx
Look,
dy/dx=x2+y2/x2+xy -(1)
Thus,
dy/dx is a homogeneous function.
Put y=vx in the eq.
dy/dx=v+xdv/dx
Put in eqn.(1),
v+xdv/dx=x2+y2/x2+xy
on solving, we get,
1+v/1-vdv=1/xdx
Integrating both sides we get,
-2log|v-1|-v=log|x|+c
Put v=y/x,
-2log|(y/x)-1|-y/x=log|x|+c,
I hope it was helpful buddy!