Differential Equation

[discoverer02]

I'm having trouble finding the general solution for the following problem:

(1-xy)y' + y^2 + 3xy^3

It's obviously not an exact equation.

I multiply through by an integrating factor (x^m)(y^n) and get

(x^m)(y^(n+2)) + [(3x^(m+1))(y^(n+3))] + [(x^m)(y^n) - (x^(m+1)y^(n+1)]y' (the forms aren't similar, I can't multiply one side to make them similar and have an equivalent equation)

The partial derivative with respect to y:
(n+2)[(x^m)(y^(n+1))] + 3(n+3)[x^(m+1)y^(n+2)]

The partial derivative with respect to x:
m(x^(m-1)y^n) - (m+1)[(x^m)y^(n+1)]

They are not of a similar form so it doesn't do any good to solve

n + 2 = m
3n + 9 = m + 1

What am I missing? Do I need to manipulate the equations somehow?

The General Solution in the book is:

y = [x+-(4x^2 + c)^(1/2)]^(-1) with an integrating factor of y^(-3)

Thanks in advance for the help.

[HallsofIvy]

The first thing I would see is that the very nice
(n+2)[(x^m)(y^(n+1))] and - (m+1)[(x^m)y^(n+1)] terms match up as long as n+2= -(m+1).

The second thing I would see is that in order to have
3(n+3)[x^(m+1)y^(n+2)] and m(x^(m-1)y^n) match up, we would need to have either m+1= m-1, which is impossible, or have those terms equal to 0: n+3= 0 and m= 0. That, of course gives m=0, n= -3 which, fortunately, also satisfy -3+2= -1= -(0+1).

Taking m=0, n= -3, the integrating factor is x0y-3= y-3 just as your book said.

[discoverer02]

Thanks for the help.

The mechanical approach the instructor took didn't make this clear, but I understand your explanation.