# Solving Initial Boundary Value Problem (Second Order PDE)

My attempt at a solution for $$u(x,t)$$ is below, I don't think my $$G(t)$$ and $$\phi(x)$$ can satisfy the initial condition $$\dfrac{\mathrm du}{\mathrm dt}(x,0)=4 + 3\cos(2 \pi x)$$ when DSolve is giving me Cosh and Sinh but it's very likely that I'm using DSolve incorrectly even though after wrestling with it for hours.

 \$Version
(* 11.3.0 for Microsoft Windows (64-bit) (March 7, 2018) *)

pde = D[u[x, t], t, t] == D[u[x, t], x, x] - 4*u[x, t];
icb = {Derivative[1, 0][u][0, t] == 0, Derivative[1, 0][u][1, t] == 0,
u[x, 0] == 0, Derivative[0, 1][u][x, 0] == 5 + 3*Cos[2 Pi x]};

sol = DSolve[{pde, icb}, u, {x, t}]

(* {{u -> Function[{x, t}, 5/2 Sin[2 t] + (3 Cos[2 \[Pi] x] Sin[2 Sqrt[1 + \[Pi]^2] t])/
(2 Sqrt[1 + \[Pi]^2])]}}*)


$$u(x,t)=\frac{5}{2} \sin (2 t)+\frac{3 \cos (2 \pi x) \sin \left(2 \sqrt{1+\pi ^2} t\right)}{2 \sqrt{1+\pi ^2}}$$

Check pde equation:

 pde /. sol // FullSimplify
(* {True} *)(*OK*)


Check initial and boundary conditions: icb:

 icb /. sol
(* {{True, True, True, True}} *)(*OK*)