Question:
$y=\sqrt{a^{2}-x^{2}} x \in(-a, a) \quad: x+y \frac{d y}{d x}=0(y \neq 0)$
Solution:
$y=\sqrt{a^{2}-x^{2}}$
Differentiating both sides of this equation with respect to x, we get:
$\frac{d y}{d x}=\frac{d}{d x}\left(\sqrt{a^{2}-x^{2}}\right)$
$\Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{a^{2}-x^{2}}} \cdot \frac{d}{d x}\left(a^{2}-x^{2}\right)$
$=\frac{1}{2 \sqrt{a^{2}-x^{2}}}(-2 x)$
$=\frac{-x}{\sqrt{a^{2}-x^{2}}}$
Substituting the value of $\frac{d y}{d x}$ in the given differential equation, we get:
L.H.S. $=x+y \frac{d y}{d x}=x+\sqrt{a^{2}-x^{2}} \times \frac{-x}{\sqrt{a^{2}-x^{2}}}$
$=x-x$
$=0$
$=$ R.H.S.
Hence, the given function is the solution of the corresponding differential equation.