How do I solve this D.E. using integrating factor: y^2 dx+(3xy+y^2 -1)dy=0 ans:xy^3 +(y^2 -1)^2 /4=c but sol'n missing. tnx.?

Let M = y^2 , N = 3xy+y^2 -1
∂M/∂y = 2y , ∂N/∂x = 3y
∂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.
If ( ∂M/∂y - ∂N/∂x ) / M = g(y)
Then I = I(y) = e^( - ∫ g(y)dy )

( ∂M/∂y - ∂N/∂x ) / M
= ( 2y - 3y ) / y^2
= - y / y^2
= - 1/y
≡ g(y)

So, the integrating factor is
I
= e^[ - ∫ (-1/y) dy ]
= e^[ ∫ (1/y) dy ]
= e^( ln y )
= y

Multiply the eq. by y , we obtain
y^3 dx + ( 3xy^2 + y^3 - y ) dy = 0
Now, M = y^3 , N = 3xy^2 + y^3 - y
∂M/∂y = 3y^2 , ∂N/∂x = 3y^2
∂M/∂y = ∂N/∂x, so the new eq. is exact.

Hence, a function f exists for which
∂f/∂x = y^3 , ∂f/∂y = 3xy^2 + y^3 - y
we start with ∂f/∂x = y^3 , so
f = xy^3 + h(y)
∂f/∂y = 3xy^2 + h'(y) = 3xy^2 + y^3 - y
h'(y) = y^3 - y
h(y) = (1/4)y^4 - (1/2)y^2
f = xy^3 + (1/4)y^4 - (1/2)y^2

Ans: xy^3 + (1/4)y^4 - (1/2)y^2 = c

Note that xy^3 + (1/4)y^4 - (1/2)y^2 = c
is equivalent to xy^3 +(y^2 -1)^2 /4 = c