How do I solve this D.E. using integrating factor: 2y(x^2 - y+x)dx + (x^2 -2y)dy = 0 the answer is actually (x^2 y-y)e^(2x) sol'n missing.?

Let M = 2y(x^2-y+x) = 2x^2 y - 2y^2 + 2xy , N = x^2 - 2y
∂M/∂y = 2x^2 - 4y + 2x , ∂N/∂x = 2x
∂M/∂y ≠ ∂N/∂x, so the eq. is not an exact eq.

To find the integrating factor, we use the following rule:
M(x,y) dx + N(x,y) dy = 0 is not an exact eq.
Let U = ∂M/∂y - ∂N/∂x
(1) If U/N = f(x) , then I = I(x) = e^( ∫ f(x)dx )
(2) If U/M = g(y) , then I = I(y) = e^( - ∫ g(y)dy )

Now,
U / N
= ( ∂M/∂y - ∂N/∂x ) / ( x^2 - 2y )
= ( 2x^2 - 4y ) / ( x^2 - 2y )
= 2
≡ f(x)

by part (1) of the rule, the integrating factor is
I
= I(x)
= e^( ∫ f(x)dx )
= e^( ∫ 2 dx )
= e^(2x)

Multiply the eq. by e^(2x) , we have
e^(2x)*[ 2x^2 y - 2y^2 + 2xy ] dx + e^(2x)*[ x^2 - 2y ] dy = 0

Checking if the new eq. is exact :
Now, M = e^(2x)*[ 2x^2 y - 2y^2 + 2xy ] , N = e^(2x)*[ x^2 - 2y ]
∂M/∂y = e^(2x)*[ 2x^2 - 4y + 2x ]
∂N/∂x = 2*e^(2x)*[ x^2 - 2y ] + e^(2x)*[ 2x ] = e^(2x)*[ 2x^2 - 4y + 2x ]
∂M/∂y = ∂N/∂x, so the new eq. is exact.

Suppose the solution is Φ(x,y) = c1 , then
∂Φ/∂x = e^(2x)*[ 2x^2 y - 2y^2 + 2xy ] , ∂Φ/∂y = e^(2x)*[ x^2 - 2y ]
we start with ∂Φ/∂y = e^(2x)*[ x^2 - 2y ] , so
Φ = e^(2x)*[ x^2 y - y^2 ] + h(x)

∂Φ/∂x
= 2*e^(2x)*[ x^2 y - y^2 ] + e^(2x)*[ 2xy ] + h'(x)
= e^(2x)*[ 2x^2 y - 2y^2 + 2xy ] + h'(x)
= e^(2x)*[ 2x^2 y - 2y^2 + 2xy ]

h'(x) = 0
h(x) = c2 , so
Φ = e^(2x)*( x^2 y - y^2 ) + h(x) = e^(2x)*( x^2 y - y^2 ) + c2
and Φ(x,y) = c1 , thus
Φ = e^(2x)*( x^2 y - y^2 ) + c2 = c1
e^(2x)*( x^2 y - y^2 ) = c1 - c2 = c

Ans: e^(2x)*( x^2 y - y^2 ) = c