\(x=\int_{0}^{y}\frac{1}{\sqrt{1+9t^2}}dt\) and \(\frac{d^2y}{dx^2}=ay\). Then \(a=\)

3 6 9 1

9

By Leibnitz rule \(\frac{dx}{dy}=\frac{1}{\sqrt{1+9y^2}}\) \(\frac{d^2y}{dx^2}=9y\)