Why multiplying the standard form by an integrating factor

Question

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?

Answer 1

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

As the equation is equaled to zero, you can always multiply or divide by almost everything you can desire, even by the previously calculated integrating factor! The technique is based exactly in this fact.

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.