Let $f$ be a twice differentiable function on $R$. If $f^{\text{'}}\left(0\right)=4$ and $f\left(x\right)+{\int }_0^x\left(x-t\right)f^{\text{'}}\left(t\right)dt$ $=\left(e^{2x}+e^{-2x}\right)\cos 2x+\frac{2}{a}x$, then ${\left(2a+1\right)}^5{a}^2$ is equal to _______.

Given,

$f\left(x\right)+{\int }_0^x\left(x-t\right)f^{\text{'}}\left(t\right)dt=\left(e^{2x}+e^{-2x}\right)\cos 2x+\frac{2x}{a} \cdots \left(\mathrm{i}\right)$

Putting $x=0$ both side we get, $f\left(0\right)=2 \cdots \left(\mathrm{ii}\right)$

On differentiating equation $\left(\mathrm{i}\right)$ w.r.t. $x$ we get:

$f^{\text{'}}\left(x\right)+{\int }_0^x{f}^{\text{'}}\left(t\right)dt+x{f}^{\text{'}}\left(x\right)-x{f}^{\text{'}}\left(x\right)=2\left(e^{2x}-e^{-2x}\right)\cos 2x-2\left(e^{2x}+e^{-2x}\right)\sin 2x+\frac{2}{a}$

$\Rightarrow f^{\text{'}}\left(x\right)+f\left(x\right)-f\left(0\right)=2\left(e^{2x}-e^{-2x}\right)\cos 2x-2\left(e^{2x}+e^{-2x}\right)\sin 2x+\frac{2}{a}$

Replace $x$ by $0$ we get:

$\Rightarrow 4=\frac{2}{a} \Rightarrow a=\frac{1}{2}$

Now putting the value of $a$ in${\left(2a+1\right)}^5·a^2$

We get, $2^5·\frac{1}{2^2}=2^3=8$