Question

Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
$y=\sqrt{a^2-x^2}x\in (-a,a):x+y\frac{dy}{dx}=0(y\ne 0)$

Answer 1

Let $y=\sqrt{a^2-b^2}$
Differentiating both sides w.r.t. $x$ we get,
$\frac{dy}{dx}=\frac{d\left(\sqrt{a^2-b^2}\right)}{dx}$
$\frac{dy}{dx}=\frac{1}{2\sqrt{a^2-b^2}}\times (-2x)$
$\frac{dy}{dx}=\frac{-x}{\sqrt{(a^2-b^2)}}$
Taking LHS
Putting
$\frac{dy}{dx}=\frac{-x}{\sqrt{(a^2-b^2)}}$
$x+y\frac{dy}{dx}=x+y\left[\frac{-x}{\sqrt{a^2-b^2}}\right]$
Putting $y=\sqrt{a^2-b^2}$
$x+y\frac{dy}{dx}=x+\sqrt{a^2-x^2}\left[\frac{-x}{\sqrt{a^2-b^2}}\right]$
$x+y\frac{dy}{dx}=x-x$
$x+y\frac{dy}{dx}=0$
Thus $LHS=RHS$
Hence verified.

Final aswer:
Hence, the function $y=\sqrt{a^2-x^2}$ is a solution of the differernitial equation $x+y\frac{dy}{dx}=0$.