Question

If $x=\underset{0}{\overset{y}{\int }}\frac{dt}{\sqrt{1+9t^2}}$ and $\frac{d^{2}y}{d{x}^2}=ay,$ then the value of $a$ is:

Answer 1

Given, $x=\underset{0}{\overset{y}{\int }}\frac{dt}{\sqrt{1+9t^2}}$
Differentiating with respect to $y$ using Leibnitz theorem, we get
$\Rightarrow \frac{dx}{dy}=\frac{1}{\sqrt{1+9y^2}}$
$\Rightarrow \frac{dy}{dx}=\sqrt{1+9y^2}$
$\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dy}\left(\sqrt{1+9y^2}\right)\frac{dy}{dx}\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{18y}{2\sqrt{1+9y^2}}\sqrt{1+9y^2}=9y$
$\therefore a=9$