Find the solution to the initial value problem.

$\mathbf{z}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{+}\mathbf{z}\left(\mathbf{x}\right)\mathbf{=}\mathbf{2}{\mathbf{e}}^{-x}\mathbf{,}\mathbf{z}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{z}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{0}$

The differential equation is,

$\mathrm{z}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{z}\left(\mathrm{x}\right)=2{\mathrm{e}}^{-\mathrm{x}}\dots \left(1\right)$

Write the homogeneous differential equation of the equation (1),

$\mathrm{z}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{z}\left(\mathrm{x}\right)=0$

The auxiliary equation for the above equation,

${\mathrm{m}}^2+1=0$

The root of an auxiliary equation is,

The complementary solution of the given equation is,

${\mathrm{z}}_{\mathrm{c}}\left(\mathrm{x}\right)={\mathrm{c}}_1\mathrm{cosx}+{\mathrm{c}}_2\mathrm{sinx}$

Assume, the particular solution of equation (1),

${\mathrm{z}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{Ae}}^{-\mathrm{x}}......\left(2\right)$

Now find the first and second derivatives of the above equation,

$\begin{array}{rcl}{\mathrm{z}}_{\mathrm{p}}\text{'}\left(\mathrm{x}\right)& =& -{\mathrm{Ae}}^{-\mathrm{x}}\\ {\mathrm{z}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{x}\right)& =& {\mathrm{Ae}}^{-\mathrm{x}}\end{array}$

Substitute the value of and the equation (1),

$\begin{array}{c}\mathrm{z}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{z}\left(\mathrm{x}\right)=2{\mathrm{e}}^{-\mathrm{x}}\\ {\mathrm{Ae}}^{-\mathrm{x}}+{\mathrm{Ae}}^{-\mathrm{x}}=2{\mathrm{e}}^{-\mathrm{x}}\\ 2{\mathrm{Ae}}^{-\mathrm{x}}=2{\mathrm{e}}^{-\mathrm{x}}\\ \mathrm{A}=1\end{array}$

Substitute the value of A in the equation (2),

${\mathrm{z}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{e}}^{-\mathrm{x}}$

Therefore, the general solution is,

$\begin{array}{l}\mathrm{z}={\mathrm{z}}_{\mathrm{c}}\left(\mathrm{x}\right)+{\mathrm{z}}_{\mathrm{p}}\left(\mathrm{x}\right)\\ \mathrm{z}={\mathrm{c}}_1\mathrm{cosx}+{\mathrm{c}}_2\mathrm{sinx}+{\mathrm{e}}^{-\mathrm{x}}......\left(3\right)\end{array}$

Given the initial condition,

$\mathrm{z}\left(0\right)=0,\mathrm{z}\text{'}\left(0\right)=0$

Substitute the value of z = 0 and x = 0 in the equation (3),

$\begin{array}{c}\mathrm{z}={\mathrm{c}}_1\mathrm{cosx}+{\mathrm{c}}_2\mathrm{sinx}+{\mathrm{e}}^{-\mathrm{x}}\\ 0={\mathrm{c}}_1\cos \left(0\right)+{\mathrm{c}}_2\sin \left(0\right)+{\mathrm{e}}^{-0}\\ {\mathrm{c}}_1=-1\end{array}$

Now find the derivative of the equation (3),

$\mathrm{z}\text{'}=-{\mathrm{c}}_1\mathrm{sinx}+{\mathrm{c}}_2\mathrm{cosx}-{\mathrm{e}}^{-\mathrm{x}}$

Substitute the value of z’ = 0 and x = 0 in the above equation,

$\begin{array}{c}\mathrm{z}\text{'}=-{\mathrm{c}}_1\mathrm{sinx}+{\mathrm{c}}_2\mathrm{cosx}-{\mathrm{e}}^{-\mathrm{x}}\\ 0=-{\mathrm{c}}_1\sin \left(0\right)+{\mathrm{c}}_2\cos \left(0\right)-{\mathrm{e}}^{-0}\\ {\mathrm{c}}_2=1\end{array}$

Substitute the value of ${\mathrm{c}}_1=-1$and ${\mathrm{c}}_2=1$in the equation (3), we get:

$\begin{array}{c}\mathrm{z}={\mathrm{c}}_1\mathrm{cosx}+{\mathrm{c}}_2\mathrm{sinx}+{\mathrm{e}}^{-\mathrm{x}}\\ \mathrm{z}=(-1)\mathrm{cosx}+\left(1\right)\mathrm{sinx}+{\mathrm{e}}^{-\mathrm{x}}\\ \mathrm{z}=\mathrm{sinx}-\mathrm{cosx}+{\mathrm{e}}^{-\mathrm{x}}\end{array}$

Thus, the solution of the initial problem is:

$\mathrm{z}=\mathrm{sinx}-\mathrm{cosx}+{\mathrm{e}}^{-\mathrm{x}}$