If ∫ e^a x cos b x d x=(e^2 x)/29 f(x)+C, then f''(x)=

We have,

$\int e^{a x} \cos b x d x=\frac{e^{a x}}{a^2+b^2}(a \cos b x+b \sin b x)+C$

$\Rightarrow \frac{e^{2 x}}{2^2+5^2} f(x)+C=\frac{e^{a x}}{a^2+b^2}(a \cos b x+b \sin b x)+C$

$\Rightarrow a=2, b=5$ and $f(x)=a \cos b x+b \sin b x$

$\Rightarrow f''(x)=-b^2 f(x) \Rightarrow f''(x)=-25 f(x)$