Assume (2x²y-2xy²+y)dx + (x-2y)dy=0.
Let P(x,y)=2x²y-2xy²+y and Q(x,y)=x-2y.
If the DE is exact, ∂P/∂y=∂Q/∂x
The DE becomes:
P+Qdy/dx=0.
∂P/∂y=2x²-4xy+1, ∂Q/∂x=1, so the DE is not exact.
To convert it into an exact equation we want to find R₁(x) or R₂(y) where:
R₁=(1/Q)(∂P/∂y-∂Q/∂x) or R₂=(1/P)(∂Q/∂x-∂P/∂y).
So R₁=(1/(x-2y))(2x²-4xy+1-1)=2x(x-2y)/(x-2y)=2x.
R₂=(1/(2x²y-2xy²+y))(1-2x²+4xy-1)=-2x(x-2y)/(2x²y-2xy²+y).
Since only R₁ reduces to the requirement of a function of x alone, we can derive a multiplier e^∫R₁dx=e^∫2xdx=e^x².
e^x²P+e^x²Qdy/dx=0 is now an exact equation.
Let M=e^x²P=e^x²(2x²y-2xy²+y)
and N=e^x²(x-2y).
[Proof of exactness:
∂M/∂y=∂N/∂x=2x²e^x²-4xye^x²+e^x².]
For a function F(x,y), ∂F/∂x=M and ∂F/∂y=N.
If we use N to find F we have:
F=∫∂F/∂y.dy=∫(e^x²(x-2y))dy=xye^x²-y²e^x²+f(x). Here we are integrating as if x were a constant as we integrate with respect to y. Hence the constant of integration is a function of x, f(x).
But ∂F/∂x=M=e^x²(2x²y-2xy²+y)=2x²ye^x²-2xy²e^x²+ye^x².
Using F=xye^x²-y²e^x²+f(x), ∂F/∂x=ye^x²+2x²ye^x²-2xy²e^x²+df/dx.
These two expressions for ∂F/∂x are equal so df/dx=0, and f=k, a constant.
Therefore, F=xye^x²-y²e^x²+k.
By definition, ∂F/∂x+(∂F/dy)(dy/dx)=dF/dx, so dF/dx=0, because the original DE=0, and F=c, a constant.
Therefore, xye^x²-y²e^x²+k=c. We can combine the two constants into one, a.
So the DE solution is xye^x²-y²e^x²+a=0.
If we differentiate this we should get the original DE.
(ye^x²+2x²ye^x²-2xy²e^x²)dx+(xe^x²-2ye^x²)dy=0,
(y+2x²y-2xy²)dx+(x-2y)du=0.