Functional Equation

[jacks]

If \(f(0) = 2\) and \(f^{'}(0) = 3\) and \(f^{''}(x) = f(x)\). Then value of \(f(4) =\)

[Papapetros Vaggelis]

For each \(\displaystyle{x\in\mathbb{R}}\) holds :

\(\displaystyle{\begin{aligned} f^{\prime \prime}(x)=f(x)&\implies f^{\prime \prime}(x)+f^\prime(x)=f^\prime(x)+f(x)\\&\implies \left(f^\prime(x)+f(x)\right)'-\left(f^\prime(x)+f(x)\right)=0\\&\implies e^{-x}\,\left[\left(f^\prime(x)+f(x)\right)'-\left(f^\prime(x)+f(x)\right)\right]=0\\&\implies \left[e^{-x}\,\left(f^\prime(x)+f(x)\right)\right]'=0\end{aligned}}\)

so

there is \(\displaystyle{c_1\in\mathbb{R}}\) such that \(\displaystyle{e^{-x}\,\left(f^\prime(x)+f(x)\right)=c_1\,,\forall\,x\in\mathbb{R}}\) .

Now, \(\displaystyle{e^{0}\,\left(f^\prime(0)+f(0)\right)=c_1\implies c_1=5\implies f^\prime(x)+f(x)=5\,e^{x}\,,\forall\,x\in\mathbb{R}}\) .

\(\displaystyle{\begin{aligned} x\in\mathbb{R}&\implies f^\prime(x)+f(x)=5\,e^{x}\\&\implies e^{x}\,\left(f^\prime(x)+f(x)\right)=5\,e^{2\,x}\\&\implies \left(e^{x}\,f(x)\right)'=\left(\frac{5}{2}\,e^{2\,x}\right)'\end{aligned}}\)

There is \(\displaystyle{c_2\in\mathbb{R}}\) such that \(\displaystyle{e^{x}\,f(x)=\frac{5}{2}\,e^{2\,x}+c_2\,,\forall\,x\in\mathbb{R}}\) ,

thus : \(\displaystyle{e^{0}\,f(0)=\dfrac{5}{2}\,e^{0}+c_2\iff c_2+\dfrac{5}{2}=2\iff c_2=-\dfrac{1}{2}}\) .

So, \(\displaystyle{f(x)=\dfrac{5\,e^{x}-e^{-x}}{2}\,,x\in\mathbb{R}}\) and

\(\displaystyle{f(4)=\dfrac{5\,e^{4}-e^{-4}}{2}=\dfrac{5\,e^{8}-1}{2\,e^{4}}}\) .