Let $f(x)$ be a polynomial satisfying $f(0)=2$, $f'(0)=3$ and $f''(x)=f(x)$. Then, $f(4)=$

$\frac{5\left(e^8-1\right)}{2 e^4}$

We have,

$f''(x)=f(x)$

$\Rightarrow 2 f'(x) f''(x)=2 f(x) f'(x)$

$\Rightarrow \frac{d}{d x}\left\{f'(x)\right\}^2=\frac{d}{d x}\{f(x)\}^2 $

$\Rightarrow \left\{f'(x)\right\}^2=\{f(x)\}^2+C$            .....(i)

Now, $f(0)=2$ and $f'(0)=3$. Therefore, from (i), we get

$\left\{f'(0)\right\}^2=\{f(0)\}^2+C$

$\Rightarrow 9=4+C \Rightarrow C=5$

∴  $\left\{f'(x)\right\}^2=\{f(x)\}^2+5 $

$\Rightarrow f'(x)=\sqrt{5+\{f(x)\}^2}$

$\Rightarrow \int \frac{1}{\sqrt{(\sqrt{5})^2+\{f(x)\}^2}} d\{f(x)\}=\int d x$

$\Rightarrow \log \left\{f(x)+\sqrt{5+\{f(x)\}^2}\right\}=x+C_1$

∴  $f(0)=2 \Rightarrow \log \{2+3\}=C_1 \Rightarrow C_1=\log 5$

∴  $\log \left\{f(x)+\sqrt{5+\{f(x)\}^2}\right\}=x+\log 5$

$\Rightarrow \log \left\{\frac{f(x)+\sqrt{5+\{f(x)\}^2}}{5}\right\}=x$

$\Rightarrow f(x)+\sqrt{5+\{f(x)\}^2}=5 e^x$

$\Rightarrow \sqrt{5+\{f(x)\}^2}+f(x)=5 e^x$ and $\sqrt{5+\{f(x)\}^2}-f(x)=5 e^{-x}$

$\Rightarrow 2 f(x)=5\left(e^x-e^{-x}\right)$

$\Rightarrow f(x)=\frac{5}{2}\left(e^x-e^{-x}\right)$

$\Rightarrow f(4)=\frac{5}{2}\left(e^4-e^{-4}\right)=\frac{5\left(e^8-1\right)}{2 e^4}$