Question

Let $f$ be a twice differentiable function on $R$.
If $f ^{\prime}(0)=4$ and
$f(x)+\int\limits_0^x(x-t) f^{\prime}(t) d t=\left(e^{2 x}+e^{-2 x}\right) \cos 2 x+\frac{2}{a} x,$
then $(2 a+1)^5 a^2$ is equal to ______

Answer 1

$f ^{\prime}(0)=4$
$f(x)+\int\limits_0^x(x-t) f^{\prime}(t) d t=\left(e^{2 x}+e^{-2 x}\right) \cos 2 x+\frac{2}{a} x$
Put $x=0: f(0)=2$
$ f ^{\prime}(x)+x\left(f^{\prime}(x)\right)+\int\limits_0^x f^{\prime}(t) d t-x f^{\prime}(x) $
$ =\left(e^{2 x}+e^{-2 x}\right)(-2 \sin 2 x) $
$ +\cos 2 x\left(2 e^{2 x}-2 e^{-2 x}\right)+\frac{2}{a} $
$ \Rightarrow f^{\prime}(x)+f(x)-2=\left(e^{2 x}+e^{-2 x}\right)(-2 \sin 2 x) $
$ +\cos 2 x\left(2 e^{2 x}-2 e^{-2 x}\right)+\frac{2}{a}$
Put $x =0$
$4+2-2=0+(2-2)+2 / a$
$ \Rightarrow a =\frac{1}{2} $
$ (2 a +1)^5 a ^2=2^5 \cdot \frac{1}{2^2}=8$