Let $f$ be a twice differentiable function on $\mathbb{R}$. If $f^{\prime}(0)=4$ and $$f(x) + \int\limits_0^x {(x - t)f'(t)dt = \left( {{e^{2x}} + {e^{ - 2x}}} \right)\cos 2x + {2 \over a}x} $$, then $(2 a+1)^{5}\, a^{2}$ is equal to _______________.

$$ \begin{aligned} \because f(x)+\int_0^x(x-t) f^{\prime}(t) d t & =\left(e^{2 x}+e^{-2 x}\right) \cos 2 x+\frac{2 x}{a} ~~...(i) \end{aligned} $$<br/><br/> Here $f(0)=2 \hspace{0.5cm} ...(ii)$<br/><br/> On differentiating equation (i) w.r.t. $x$ we get :<br/><br/> $$ \begin{aligned} & f^{\prime}(x)+\int_0^x f^{\prime}(t) d t+x f^{\prime}(x)-x f^{\prime}(x) \\\\ & = 2\left(e^{2 x}-e^{-2 x}\right) \cos 2 x-2\left(e^{2 x}+e^{-2 x}\right) \sin 2 x+\frac{2}{a} \\\\ & \Rightarrow \quad f(x)+f(x)-f(0)\\\\ &=2\left(e^{2 x}-e^{-2 x}\right) \cos 2 x-2\left(e^{2 x}+e^{-2 x}\right) \sin 2 x+\frac{2}{a} \\\\ & \quad \text { Replace } x \text { by } 0 \text { we get }: \\\\ & \Rightarrow \quad 4=\frac{2}{a} \Rightarrow a=\frac{1}{2} \cdot \\\\ & \therefore \quad(2 a+1)^5 \cdot a^2=2^5 \cdot \frac{1}{2^2}=2^3=8 \end{aligned} $$