dxdy=y+2x2y2x+2xy2
Let us solve this differential equation:
dxdy=y(1+2x2)2x(1+y2)
1+y2ydy=1+2x22xdx (1+y2>0)
∫1+y2ydy=∫1+2x22xdx
21∫1+y2d(1+y2)=21∫1+2x2d(1+2x2)
∫1+y2d(1+y2)=∫1+2x2d(1+2x2)
ln(1+y2)=ln(1+2x2)+ln(C)
ln(1+y2)=ln((1+2x2)C)
Therefore, the general solution is the following:
1+y2=C(1+2x2)
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