Question

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

$y=\sqrt{1+x^2}and{y}^{\prime }=\frac{xy}{1+x^2}$

Answer 1

Given, $y=\sqrt{1+x^2}$
On differentiating both sides of this equation w.r.t. x, we have
$y^{\prime }=\frac{d}{dx}\left(\sqrt{1+x^2}\right)\Rightarrow y^{\prime }=\frac{1}{2}(1+x^2{)}^{\frac{1}{2}}(2x)\Rightarrow y^{\prime }=\frac{2x}{2\sqrt{1+x^2}}\Rightarrow y^{\prime }=\frac{x}{\sqrt{1+x^2}}$
But we have to verify, $y^{\prime }=\frac{xy}{1+x^2}....\left(i\right)$
On putting the values of y' and y Eq.(i), we get LHS=y'=$\frac{x}{\sqrt{1+x^2}})RHS=\frac{xy}{1+x^2}=\frac{x}{1+x^2}\times \sqrt{1+x^2}=\frac{x}{\sqrt{1+x^2}}\therefore LHS=RHS$
Hence, $y=\sqrt{1+x^2}$ is a solution of the given differential equation.