Question

If $x={\int }_0^y\frac{\mathrm{dt}}{\sqrt{1+9t^2}}\mathrm{and}\frac{d^{2}y}{{\mathrm{dx}}^2}=\mathrm{ay}$, then the value of a is equal to

Answer 1

$x={\int }_0^y\frac{\mathrm{dt}}{\sqrt{1+9t^2}}\Rightarrow \frac{\mathrm{dx}}{\mathrm{dy}}=\frac{1}{\sqrt{1+9y^2}}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{1+9y^2}\Rightarrow \frac{d^{2}y}{{\mathrm{dx}}^2}=\frac{18y}{2\sqrt{1+9y^2}}\cdot \frac{\mathrm{dy}}{\mathrm{dx}}=9y\therefore a=9$