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

Originally, this is a question I asked on Quora.

https://www.quora.com/How-do-I-solve-dfrac-dy-dx-x-2+y-2?filter=all&nsrc=1&snid3=1500842823

Maybe this question has been solved but I was wondering if it can be done by integrating factor?

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

Using method of integrating factor

$$\text{I.F}= e^{\int-ydx}$$

$$\text{I.F}= e^{-yx}$$

Can it be done in this way?

Crazy
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  • It's NOT that easy

    https://www.wolframalpha.com/input/?i=solve+x%5E2%2By%5E2%3Dy%27

     – Raffaele Sep 19 '17 at 07:52
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    Questions on the same equation https://math.stackexchange.com/questions/2348022/riccati-d-e-vertical-asymptotes/2348258#2348258, https://math.stackexchange.com/questions/1353727/y-x2-y2-asymptote, https://math.stackexchange.com/questions/446926/riccati-differential-equation-y-x2y2 – Lutz Lehmann Sep 19 '17 at 12:54

1 Answers1

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The short answer is no it may not. It is not linear in $y$ in the sense that the thing multiplying $y(x)$ must be a function of $x$... i.e. $P(x)\cdot y(x)$ not $y(x)\cdot y(x)$.

If you could calculate $\displaystyle \int y(x)\,dx$ you would know $y(x)$.

It is not true that $\displaystyle \int y(x)\,dx=y(x)\cdot x$ by the way unless $y(x)=y$ is a constant... which is certainly not the case here.

JP McCarthy
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  • Does that mean that is must be false? Or just in some cases it can be valid? – Crazy Sep 19 '17 at 08:00
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    @Crazy The integrating factor only works if you have $y'(x)+P(x)\cdot y(x)=Q(x)$. You do not have an equation in this form. – JP McCarthy Sep 19 '17 at 10:09