# Why multiplying the standard form by an integrating factor

I use the algorithm to solve a First Order Linear Differential equation for the following problem

$$2(y-4x^2)dx + xdy=0$$

It is a trivial case and we can easily transform it into the standard form:

$$\frac{dy}{dx} + \frac{2}{x}y=8x \rightarrow standard$$ From here we obtain the integrating factor $x^2$ by aplying the general procedure. We know that if we multiply the standard form by the integrating factor, we will obtain an exact equation. And it is really the case: $$x^2 dy+(2xy-8x^3)dx = 0$$ is exact. But if we try to multiply the initial equation by $x^2$, we won't get an exact one: $$(2x^2y-8x^4)dx + x^3dy=0$$ Obviously yields different second order partial derivatives of a possible solution. My question is: Why does it happen?

$2(y-4x^2)dx + xdy=0$ has as integrating factor $x$
Consider $2(xy-4x^3)dx + x^2dy=0$, put it in standard form and you get $\dfrac{dy}{dx} + \dfrac{2}{x}y=8x$, that you know it has as integrating factor $x^2$, that is exactly the factor you have simplified.
Anyway, get the solution for $\dfrac{dy}{dx} + \dfrac{2}{x}y=8x$ and check it in $2(y-4x^2)dx + xdy=0$ and even in $(2x^2y-8x^4)dx + x^3dy=0$ , you'll see that satisfies both.