Differential equation $\frac{dx}{\sqrt{1-x^4}}+\frac{dy}{\sqrt{1-y^4}}=0$

Asked Dec 14 '23 at 17:31

Active Dec 14 '23 at 17:31

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Prove that the general integral of the equation $\frac{dx}{\sqrt{1-x^4}}+\frac{dy}{\sqrt{1-y^4}}=0$ can be represented in the form $y\sqrt{1-x^4}+x\sqrt{1-y^4}=C\left ( 1+x^2y^2 \right )$

I tried to take the integral $\int \frac{dx}{\sqrt{1-x^4}}$, but it turns out to be an elliptic function and it is not yet clear how to express the required equality from it. Can you show how you can prove it?

asked Dec 14 '23 at 17:31

Dmitry

1,238

That's indeed an elliptic integral (of the first kind, I believe). The equation is exact, but the resulting integral is nothing nice. – DonAntonio Dec 14 '23 at 18:00
Could you please include some personal work please? – Soham Saha Dec 14 '23 at 18:19
2

This is connected to the lemnicate curve : see here. – Jean Marie Dec 14 '23 at 18:38

Differential equation $\frac{dx}{\sqrt{1-x^4}}+\frac{dy}{\sqrt{1-y^4}}=0$

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