Best Answer:
I think the ODE you meant to type was

y dx + 2x dy = 0.

Indeed, just plain "y" is an integrating factor, because if you write

y^2 dx + (2xy) dy = 0,

then the left side is the exact differential of xy^2.

Hence, a general solution to the ODE is xy^2 = constant.

Let's see about xy^3 as an integrating factor.

Once again I am assuming the original ODE was

y dx + 2x dy = 0.

Multiplying through by xy^3 gives

xy^4 dx + 2x^2 y^3 dy = 0.

Now the left side is the exact differential of (1/2)x^2 y^4,

so a general solution is (1/2) x^2 y^4 = constant.

You can see that this is in fact the same solution as was given before: it's just that the new set of constants the halves of the squares of the old set of constants.

In the third case, x^2 y^3 as an integ'g factor, the ODE becomes

x^2 y^4 dx + 2x^3 y^3 dy = 0,

but this time, the LHS isn't an exact differential, so this expression (x^2 y^3) didn't turn out to be an integrating factor.

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