<s>[INST] How do I solve the following differential equation? 
$\frac{dy}{dx}+ \sqrt\frac{1-y^2}{1-x^2}=0$. How do I substitute $y$? Any help would be appreciated thanks.

 [/INST] $$\frac{dy}{dx}+ \sqrt\frac{1-y^2}{1-x^2}=0$$ or,
$$\frac{dy}{dx}=- \sqrt\frac{1-y^2}{1-x^2}$$ or,
$$\frac{dy}{\sqrt{1-y^2}}=- \frac{dx}{\sqrt{1-x^2}}$$ or,
$$\sin^{-1}y=-\sin^{-1}x+c$$ or,
$$\sin^{-1}y+\sin^{-1}x=c$$
EDIT:
$$\frac{dy}{dx}+ \sqrt\frac{1-y^2}{1-x^2}=0$$ or,
$$\frac{dy}{dx}=- \sqrt\frac{1-y^2}{1-x^2}$$ or,
$$\frac{dy}{dx}=- \sqrt\frac{1-y^2}{1-x^2} \left(\frac{xy-\sqrt{(1-x^2)(1-y^2)}}{xy-\sqrt{(1-x^2)(1-y^2)}}\right)$$ or,
$$\frac{dy}{dx}\left(\frac{xy-\sqrt{(1-x^2)(1-y^2)}}{\sqrt{1-y^2}}\right)=- \left(\frac{xy-\sqrt{(1-x^2)(1-y^2)}}{\sqrt{1-x^2}}\right)$$ or,
$$\frac{dy}{dx}\left(\frac{xy}{\sqrt{1-y^2}}-\sqrt{1-x^2}\right)=- \left(\frac{xy}{\sqrt{1-x^2}}-\sqrt{1-y^2}\right)$$ or,
$$-\sqrt{1-y^2}+\frac{dy}{dx}\cdot \frac{xy}{\sqrt{1-y^2}}=- \frac{xy}{\sqrt{1-x^2}}+\frac{dy}{dx}\cdot \sqrt{1-x^2}$$ or,
$$d\left(-x\sqrt{1-y^2}\right)=d\left(y\sqrt{1-x^2}\right)$$ or,
$$c-x\sqrt{1-y^2}=y\sqrt{1-x^2}$$ or,
$$x\sqrt{1-y^2}+y\sqrt{1-x^2}=c$$
The method is quite laborious and boring. At present, I cannot remember the shortcut. I'll post it when it comes to my mind.
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