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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

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  • $\begingroup$ Hello welcome to MSE! Have you used product rule? Or the linearity of the derivative operator to split derivatives at $+$ signs? $\endgroup$ Commented Aug 12, 2022 at 10:03
  • $\begingroup$ Familiar with the product rule, but not the second part $\endgroup$
    – Emelie
    Commented Aug 12, 2022 at 10:28
  • $\begingroup$ If I have one function $f$ which is the sum of two other functions $f=g+h$, then what we have is $$\frac{d}{dx}f = \frac{d}{dx}(g+h) = \frac{d}{dx}g + \frac{d}{dx}h$$ $\endgroup$ Commented Aug 12, 2022 at 10:58

1 Answer 1

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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:

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

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