Solve the differential equation y√(1-x^2) dy+x√(1-y^2) dx=0

Solve the differential equation $y\sqrt{1-x^2}\,dy+x\sqrt{1-y^2}\,dx=0$.

Step 1: Separate variables

\[\frac{y\,dy}{\sqrt{1-y^2}}=-\frac{x\,dx}{\sqrt{1-x^2}},\qquad |x|<1,\ |y|<1.\]

Step 2: Integrate

With $u=1-y^2$, $\displaystyle\int\frac{y\,dy}{\sqrt{1-y^2}}=-\sqrt{1-y^2}$; similarly $\displaystyle-\int\frac{x\,dx}{\sqrt{1-x^2}}=\sqrt{1-x^2}$. Hence

\[-\sqrt{1-y^2}=\sqrt{1-x^2}+C \quad\Longrightarrow\quad \boxed{\,\sqrt{1-x^2}+\sqrt{1-y^2}=k.}\]

Step 3: Range of $k$

Since each square root lies in $[0,1]$, the constant must satisfy $0\le k\le2$: $k=0$ corresponds to the corners of the unit square, and $k=2$ to the origin.