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.

 A: the DE is separable so:
$\frac{dy}{dx} = - \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$ 
$\frac{dy}{\sqrt{1-y^2}} = - \frac{dx}{\sqrt{1-x^2}}$ 
$\arcsin(y) = - \arcsin(x) + c$
$y = \sin[-\arcsin(x) + c]$
A: $$\frac{dy}{dx}+ \sqrt\frac{1-y^2}{1-x^2}=0$$ 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.
A: First we note that $-1\leq x\leq 1$ and $-1\leq y\leq 1$ for the square roots to be real.
If you just want to be able to go from your solution
$$
c=\arcsin x+\arcsin y
$$
to the given solution you can apply sine to both sides and use the addition rule,
$$
\begin{aligned}
\sin c&=\sin(\arcsin x+\arcsin y)\\
&=\sin(\arcsin x)\cos(\arcsin y)+\sin(\arcsin y)\cos(\arcsin x)\\
&=x\sqrt{1-y^2}+y\sqrt{1-x^2}.
\end{aligned}
$$
Here we have used that $\arcsin x$ and $\arcsin y$ belongs to $[-\pi/2,\pi/2]$, so that, for example, $\cos(\arcsin x)=\sqrt{1-x^2}$.
As a final comment, if you solve the differential equation in both ways, as is done by @Aniket, then you could conclude the addition rule for sine, if you did not have that one before. Similar addition rules for elliptic functions can be derived this way.
A: You should try to solve it using variable separation.
Hint: take the under-root term to the right and then separate x and y containing terms. If you are still unable to solve it, consider the solution below (First TRY to do it yourself!)
Solution:
$\frac{dy}{dx}=-\sqrt\frac{1-y^2}{1-x^2}$
$\frac{dy}{\sqrt{1-y^2}}=-\frac{dx}{\sqrt{1-x^2}}$
On Integrating, 
$\sin^{-1} y = \cos^{-1}x\space $(or $-\sin^{-1}x) + C$
or $y=\sin(\cos^{-1}x + C)$
which is the required answer.
