Answer:
lets take a differential equation
[tex]xy^3dx+(x^2y^2+1)dy=0\\let\\M= xy^3\\and\\N=(x^2y^2+1)[/tex]
taking derivative of M w.r.t y and N w.r.t x
[tex]dM/dy=3xy^2\\dN/dx=2xy^2\\so\\dM/dy\neq dN/dx[/tex]
so this is an in exact equation we will find integrating factor and make it exact equation
[tex]U=e^\int\limits^_{(-dN/dx-dM/dy)} \,/ M\\\\=e^\int\limits^ {2xy^2-3xy^2} \, /xy^3\\=e^\int\limits^ {-xy^2} /xy^3\\=e^\int\limits^{-1/y} \\=e^-^l^n^y\\=1/y[/tex]
so 1/y is our integrating factor multiply it by the equation we will get an exact equation:
[tex]1/y(xy^3)dx+1/y(x^2y^2+1)dy=0\\xy^2dx+(x^2y+1/y)dy=0\\let \\M=xy^2\\and \\N=(x^2y+1/y)[/tex]
take derivative
[tex]dM/dy=2xy\\dN/dx=2xy\\so\\dM/dy=dN/dx[/tex]
now solving this exact equation
[tex]f(x,y)=c=\int\limits{M} \, dx +g(y)\\=\int\limits {xy^2} \, dx +g(y)\\=\frac{1}{2}x^2y^2+g(y)\\ df/dy=d/dy(\frac{1}{2} x^2y^2+g(y))=N[/tex]
[tex]x^2y+g'(y)=x^2y+1/y\\g'(y)=1/y\\g(y)=lny\\f(x,y)=\frac{1}{2}x^2y^2+lny=C[/tex]
