Help with Exact Equation Problem

Question

I need help solving this exact equation problem:

x(xy+1)y' + y(xy+1) = 0

I solved the problem up to the function with respect to y and got this:

fy(x,y) = x^2/2 + x + h'(y)

I am not sure how to go from here or if I got this step right

Answer 1

x(xy+1)(dy/dx) + y(xy+1) = 0

(x²y + x)dy + (xy² + y)dx = 0

Let M(x,y) = xy² + y and N(x,y) = x²y + x

∂M/∂y = ∂N/∂x = 2xy + 1. So, the differential equation is exact.

So, there exists a function f(x,y) such that ∂f/∂x = xy²+y and ∂f/∂y = x²y + x

Integrating the first equation with respect to x, we get f(x,y) = (1/2)x²y² + xy + g(y)

So, ∂f/∂y = x²y + x + g'(y). But ∂f/∂y = N(x,y) = x²y + x

x²y + x + g'(y) = x²y + x. So, g'(y) = 0. Therefore, g(y) = C.

So, (1/2)x²y² + xy = C

Answer 2

I would approach this as the factored problem: (xy+1)(xy'+y) = 0 which leads to solutions -1/x and C/x

You could also define v =xy and dv=ydx+xdy

Equation becomes dv(v+1) = 0

This can be integrated as v²/2 + v + C = 0 (x≠0, which seems reasonable)

Plugging in x and y, I get y² + 2/x +2C/x²

Using quadratic formula to solve for y: 1/2 (-2/x ± sqrt(4/x²-8C/x²)) = -1/x ± (1/x)sqrt(1-2C)

or -1/x ± C'/x

Strange. Good luck!