Q: x²y³+x(1+y²)y′=0, µ(x,y)=1/(xy³)
A:
Let M=x²y³, N=x(1+y²)
My=2x²y², Nx=1+y²
Since My≠Nx, this equation is not exact.
Then, multiple both sides by µ(x,y)=1/(xy³)
We have, x+(1/y³+1/y)y′=0
Let A=x, B=1/y³+1/y
Ay=Bx=0, thus, this equation now is exact.
therefore, ∃Ф(x,y), s.t. Фx=A, Фy=B
Фx=A=x
Ф(x,y)=∫xdx=1/2x²+h(y)
Фy=h′(y)=B=1/y³+1/y
h(y)=∫(1/y³+1/y)dy=-1/(2y²)+ln|y|+C
Ф(x,y)=1/2x²-1/(2y²)+ln|y|+C
General solution: 1/2x²-1/(2y²)+ln|y|=C
