The differential equation whose general solution is given by $y=c_{1}cos\left(x+c_2\right)-c_3{e}^{\left(-x+c_4\right)}+c_{5}sinx$  where $c_1,c_2,c_3,c_4,c_5$ are arbitrary constants, is

$$\begin{gathered} y=c_1\cos \left(x+c_2\right)-\left(c_3{e}^{-x+c_4}\right)+\left(c_5\sin x\right) \\ =c_1\left(\cos x\cos c_2-\sin x\sin c_2\right)-\left(c_3{e}^{c_4}{e}^{-x}\right)+\left(c_5\sin x\right) \end{gathered}$$

$=\left(c_1\cos c_2\right)\cos x-\left(c_1\sin c_2-c_5\right)\sin x-\left(c_3{e}^{c_4}\right)e^{-x}$

Or $y=l\cos x+m\sin x-n{e}^{-x} \to \left(1\right)$

Where $l,m,n$ are arbitrary constants

$\therefore \frac{dy}{dx}=-l\sin x+m\cos x+n{e}^{-x} \to \left(2\right)$

Or $\frac{d^{2}y}{d{x}^2}=-l\cos x-m\sin x-n{e}^{-x} \to \left(3\right)$

Or $\frac{d^{3}y}{d{x}^3}=l\sin x-m\cos x+n{e}^{-x} \to \left(4\right)$

From equation  $\left(1\right) + \left(3\right) , \frac{d^{2}y}{d{x}^2}+y=-2n{e}^{-x} \left(5\right)$

From equation $\left(2\right)+\left(4\right),\frac{d^{3}y}{d{x}^3}+\frac{dy}{dx}=2n{e}^{-x} \left(6\right)$

From equation $\left(5\right)+\left(6\right),\frac{d^{3}y}{d{x}^3}+\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}+y=0$