Question

For the following question verify that the given function (explicit or implicit) is a solution of the corresponding differential equation.

$y=\sqrt{a^2-x^2}andx+y\frac{dy}{dx}=0(y\ne 0).$

Answer 1

Given, $x+y\frac{dy}{dx}=0...\left(i\right)$
$y=\sqrt{a^2-x^2}$
On differentiating both sides w.r.t. x, we get
$\frac{dy}{dx}=y^{\prime }=\frac{d}{dx}\sqrt{a^2-x^2}\Rightarrow y^{\prime }=\frac{1}{2}(a^2-x^2{)}^{\frac{1}{2}-1}\frac{d}{dx}(a^2-x^2)\Rightarrow y^{\prime }=\frac{1}{2\sqrt{a^2-x^2}}(-2x)=\frac{-x}{\sqrt{a^2-x^2}}$
On substituting the value of y' Eq. (i), we get
$LHS=x+y{y}^{\prime }=x+\sqrt{a^2-x^2}\times \frac{-x}{\sqrt{a^2-x^2}}=x-x=0[\because y=\sqrt{a^2-x^2}]\Rightarrow x+y{y}^{\prime }=0\therefore y=\sqrt{a^2-x^2}$ is a solution of the given differential equation.