Question

solve the differential equation:
(1-2x^2-2y)dy/dx=4x^3+4xy

Answer 1

\[(1-2x^2-2y)y'=4x^3+4xy\]\[(1-2x^2-2y)dy=(4x^3+4xy)dx\]\[(4x^3+4xy)dx+(-1+2x^2+2y)dy=0\]\[f(x,y)=C\quad;\quad df(x,y)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=0\]\[\frac{\partial f}{\partial x}=4x^3+4xy\quad\Rightarrow\quad f(x,y)=x^4+2x^2y+g(y)\]\[\frac{\partial f}{\partial y}=-1+2x^2+2y=2x^2+\frac{\partial g}{\partial y}\quad\Rightarrow\quad\frac{\partial g}{\partial y}=2y-1\]\[g(y)=y^2-y\]\[\Rightarrow\quad f(x,y)=x^4+2x^2y+y^2-y=(x^2+y)^2-y=C\]