Rewriting equation for solving differential equation

Question

When going through a solution concerning a differential equation we have the following expression:

$(1-\frac{x^2}{y^2})\frac{dy}{dx} + \frac{2x}{y} = 0$

and then its said to note that:

$ \frac{d}{dx}(\frac{x^2}{y}+y)=\frac{2x}{y}-\frac{x^2}{y^2}\frac{dy}{dx} + \frac{dy}{dx} = (1-\frac{x^2}{y^2})\frac{dy}{dx} + \frac{2x}{y}$

so that we later can use the expression to the left, but I don't really understand how the $\frac{dy}{dx}$ is used here, could someone explain how ? Specifically what is going on with the $-\frac{x^2}{y^2}\frac{dy}{dx}+ \frac{dy}{dx}$ terms in the middle part.

Thanks

Answer 1

I'm not exactly sure which part you are struggling with, but my guess is that he is taking some additional steps that you aren't seeing. Also, be sure to note that $\frac{d}{dx}$ is the derivative operation.

Basically, he is using the quotient rule and simplifying without showing you all the steps. Here are more steps expanded:

$$ \frac{d}{dx}\left(\frac{x^2}{y} + y\right) = \frac{d}{dx}\left(\frac{x^2}{y}\right) + \frac{d}{dx}\biggl(y\biggr) = \frac{d}{dx}\left(\frac{x^2}{y}\right) + \frac{dy}{dx} $$

So that is where the isolated $\frac{dy}{dx}$ comes from - using the addition rule to split the derivative at the + sign. Now we need to perform the quotient rule:

$$ \frac{d}{dx}\left(\frac{x^2}{y}\right) + \frac{dy}{dx} = \frac{2y x - x^2 \frac{dy}{dx} }{y^2} + \frac{dy}{dx}$$

Splitting the fraction yields:

$$ \frac{2y x}{y^2} - \frac{x^2 \frac{dy}{dx}}{y^2} + \frac{dy}{dx}$$

This can be further simplified to the final form of the middle part of your question above:

$$ \frac{2x}{y} - \frac{x^2}{y^2}\frac{dy}{dx} + \frac{dy}{dx}$$