Solve y^2 + xy dy/dx = 0 by integrating factor?

**a) Given (dN/dx - dM/dy) = Q, which is a function of only y. Prove that**

e^$\displaystyle \int $Q dy is an integrating factor of the equation

M dx + N dy = 0, where M and N are functions of x and y.

I have already answered this.

**b) Solve**

y^2 + x y dy/dx = 0

by finding an integrating factor to make it exact.

This is how I did:

M = y^2

N = x y

dN/dx = y

dM/dy = 2y

dN/dx - dM/dy = y - 2y = -y

(dN/dx - dM/dy)/M = -y/y^2 = -1/y, which is a function of y

Therefore, from part a,

an integrating factor is

e^[$\displaystyle \int $-1/y dy] = e^[-ln |y|]

Multiplying both sides of the given equation by the integrating factor,

y^2 e^[-ln |y|] + x y e^[-ln |y|] dy/dx = 0

d/dx (x y^2 e^[-ln|y|]) = 0

**x y^2 e^[-ln|y|] = C ( a constant)**

**y^2 e^[-ln|y|] = C/x**Is this correct?

c) Solve the above differential equation by finding am integrating factor to make it separable.

I think the integrating factor in part b also makes it separable. So can I use the same answer?

But I think the equation can be solved much easier as follows:

y^2 + x y dy/dx = 0

y (y + x dy/dx) = 0

y = 0 Or

y + x dy/dx = 0

Or y = -x dy/dx

Or, dy/y = -1/x dx

Or,dy/dy + dx/x = 0

ln y + ln x = C (a constant)

ln(xy) = C

xy = e^C = K, where K is a constant

y = K/x, if X is not zero

If x is zero, then from the given equation

y^2 + 0 = 0

Or, y = 0

Is this correct? Can I use this in the answer?