Let f be a twice differentiable function on R. If f^′(0)=4 and f(x)+∫_0^x(x-t) f^′(t) d t=(e^2 x+e^-2 x) cos 2 x+2/a x, then (2 a+1)^5 a^2 is equal…

Let $\displaystyle f$ be a twice differentiable function on $\displaystyle \mathbb{R}$. If $\displaystyle f^{\prime}(0)=4$ and $\displaystyle f(x)+\int_0^x(x-t) f^{\prime}(t) d t=\left(e^{2 x}+e^{-2 x}\right) \cos 2 x+\frac{2}{a} x$, then $\displaystyle (2 a+1)^5 a^2$ is equal to \_\_\_\_.