Find the solution to the initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{-}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{-}\mathbf{2}\mathbf{y}\left(\mathbf{x}\right)\mathbf{=}\mathbf{cosx}\mathbf{-}\mathbf{sin}\mathbf{2}\mathbf{x}\mathbf{;}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\frac{-7}{20}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\frac{1}{5}$

The differential equation is,

$\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)-\mathrm{y}\text{'}\left(\mathrm{x}\right)-2\mathrm{y}\left(\mathrm{x}\right)=\mathrm{cosx}-\sin 2\mathrm{x}......\left(1\right)$

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

$\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)-\mathrm{y}\text{'}\left(\mathrm{x}\right)-2\mathrm{y}\left(\mathrm{x}\right)=0$

The auxiliary equation for the above equation,

${\mathrm{m}}^2-\mathrm{m}-2=0$

Solve the above equation,

$\begin{array}{c}{\mathrm{m}}^2-\mathrm{m}-2=0\\ {\mathrm{m}}^2-2\mathrm{m}+\mathrm{m}-2=0\\ \mathrm{m}(\mathrm{m}-2)+1(\mathrm{m}-2)=0\\ (\mathrm{m}-2)(\mathrm{m}+1)=0\end{array}$

The root of an auxiliary equation is,

${\mathrm{m}}_1=2,{\mathrm{m}}_2=-1$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1{\mathrm{e}}^{2\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{x}}$

Assume, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{Acosx}+\mathrm{Bsin}\left(2\mathrm{x}\right)+\mathrm{Csinx}+\mathrm{Dcos}\left(2\mathrm{x}\right)......\left(2\right)$

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

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{x}\right)=-\mathrm{Asinx}+2\mathrm{Bcos}\left(2\mathrm{x}\right)+\mathrm{Ccosx}-2\mathrm{Dsin}\left(2\mathrm{x}\right)\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{x}\right)=-\mathrm{Acosx}-4\mathrm{Bsin}\left(2\mathrm{x}\right)-\mathrm{Csinx}-4\mathrm{Dcos}\left(2\mathrm{x}\right)\end{array}$

Substitute the value of $\mathrm{y}\left(\mathrm{x}\right),\mathrm{y}\text{'}\left(\mathrm{x}\right)$and ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$the equation (1),

$\begin{array}{l}\Rightarrow \mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)-\mathrm{y}\text{'}\left(\mathrm{x}\right)-2\mathrm{y}\left(\mathrm{x}\right)=\mathrm{cosx}-\sin 2\mathrm{x}\\ \Rightarrow -\mathrm{Acosx}-4\mathrm{Bsin}\left(2\mathrm{x}\right)-\mathrm{Csinx}-4\mathrm{Dcos}\left(2\mathrm{x}\right)-[-\mathrm{Asinx}+2\mathrm{Bcos}\left(2\mathrm{x}\right)+\mathrm{Ccosx}-2\mathrm{Dsin}\left(2\mathrm{x}\right)]\\ -2[\mathrm{Acosx}+\mathrm{Bsin}\left(2\mathrm{x}\right)+\mathrm{Csinx}+\mathrm{Dcos}\left(2\mathrm{x}\right)]=\mathrm{cosx}-\sin 2\mathrm{x}\\ \Rightarrow (-3\mathrm{A}-\mathrm{C})\mathrm{cosx}+(-3\mathrm{C}+\mathrm{A})\mathrm{sinx}+(-6\mathrm{D}-2\mathrm{B})\cos \left(2\mathrm{x}\right)+(-6\mathrm{B}+2\mathrm{D})\sin \left(2\mathrm{x}\right)=\mathrm{cosx}-\sin 2\mathrm{x}\end{array}$

Comparing all coefficients of the above equation,

$\begin{array}{c}-3\mathrm{A}-\mathrm{C}=1......\left(3\right)\\ -6\mathrm{B}+2\mathrm{D}=-1......\left(4\right)\\ -3\mathrm{C}+\mathrm{A}=0......\left(5\right)\\ -6\mathrm{D}-2\mathrm{B}=0......\left(6\right)\end{array}$

Solve the equation (3) and (5),

role="math" localid="1655096234005" $\begin{array}{c}3(-3\mathrm{A}-\mathrm{C})=1\times 3\\ -3\mathrm{C}-9\mathrm{A}=3\\ -3\mathrm{C}+\mathrm{A}=0\\ \mathrm{A}=\frac{-3}{10}\end{array}$

Substitute the value of A in the equation (3),

role="math" localid="1655096019737" $\begin{array}{c}-3\mathrm{A}-\mathrm{C}=1\\ -3\left(\frac{-3}{10}\right)-\mathrm{C}=1\\ \mathrm{c}=\frac{9}{10}-1\\ \mathrm{c}=\frac{-1}{10}\end{array}$

Solve the equation (4) and (6),

role="math" localid="1655096203374" $$\begin{gathered} 3(-6\mathrm{B}+2\mathrm{D})=-1\times 3 \\ \begin{array}{c}6\mathrm{D}-18\mathrm{B}=-3\\ -6\mathrm{D}-2\mathrm{B}=0\\ \mathrm{B}=\frac{3}{20}\end{array} \end{gathered}$$

Substitute the value of B in the equation (4),

role="math" localid="1655096279255" $\begin{array}{c}-6\mathrm{B}+2\mathrm{D}=-1\\ -6\left(\frac{3}{20}\right)+2\mathrm{D}=-1\\ 2\mathrm{D}=-1+\frac{9}{10}\\ \mathrm{D}=\frac{-1}{20}\end{array}$

Substitute the value of A, B, C, and D in the equation (2),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{Acosx}+\mathrm{Bsin}\left(2\mathrm{x}\right)+\mathrm{Csinx}+\mathrm{Dcos}\left(2\mathrm{x}\right)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=-\frac{3}{10}\mathrm{cosx}+\frac{3}{20}\sin \left(2\mathrm{x}\right)-\frac{1}{10}\mathrm{sinx}-\frac{1}{20}\cos \left(2\mathrm{x}\right)\end{array}$

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{x}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)\\ \mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{2\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{x}}-\frac{3}{10}\mathrm{cosx}+\frac{3}{20}\sin \left(2\mathrm{x}\right)-\frac{1}{10}\mathrm{sinx}-\frac{1}{20}\cos \left(2\mathrm{x}\right).....\left(7\right)\end{array}$

Given the initial condition,

$\mathrm{y}\left(0\right)=\frac{-7}{20},\mathrm{y}\text{'}\left(0\right)=\frac{1}{5}$

Substitute the value of $\mathrm{y}=\frac{-7}{20}$and x = 0 in the equation (7),

$\begin{array}{c}\frac{-7}{20}={\mathrm{c}}_1{\mathrm{e}}^{2\left(0\right)}+{\mathrm{c}}_2{\mathrm{e}}^{-0}-\frac{3}{10}\cos \left(0\right)+\frac{3}{20}\sin \left(0\right)-\frac{1}{10}\sin \left(0\right)-\frac{1}{20}\cos \left(0\right)\\ \frac{-7}{2}={\mathrm{c}}_1+{\mathrm{c}}_2-\frac{3}{10}-\frac{1}{20}\\ {\mathrm{c}}_1+{\mathrm{c}}_2=-\frac{7}{20}+\frac{3}{10}+\frac{1}{20}\\ {\mathrm{c}}_1+{\mathrm{c}}_2=0......\left(8\right)\end{array}$

Now find the derivative of the equation (7),

$\mathrm{y}\text{'}=2{\mathrm{c}}_1{\mathrm{e}}^{2\mathrm{x}}-{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{x}}+\frac{3}{10}\mathrm{sinx}+\frac{3}{10}\cos \left(2\mathrm{x}\right)-\frac{1}{10}\mathrm{cosx}+\frac{1}{10}\sin \left(2\mathrm{x}\right)$

Substitute the value of $\mathrm{y}\text{'}=\frac{1}{5}$and x = 0 in the above equation,

$\begin{array}{c}\frac{1}{5}=2{\mathrm{c}}_1{\mathrm{e}}^{2\left(0\right)}-{\mathrm{c}}_2{\mathrm{e}}^{-0}+\frac{3}{10}\sin \left(0\right)+\frac{3}{10}\cos \left(0\right)-\frac{1}{10}\cos \left(0\right)+\frac{1}{10}\sin \left(0\right)\\ \frac{1}{5}=2{\mathrm{c}}_1-{\mathrm{c}}_2+\frac{3}{10}-\frac{1}{10}\\ 2{\mathrm{c}}_1-{\mathrm{c}}_2=\frac{1}{5}-\frac{3}{10}+\frac{1}{10}\\ 2{\mathrm{c}}_1-{\mathrm{c}}_2=0......\left(9\right)\end{array}$

Solve the equation (8) and (9),

$\begin{array}{c}{\mathrm{c}}_1+{\mathrm{c}}_2=0\\ 2{\mathrm{c}}_1-{\mathrm{c}}_2=0\\ c_1=0\end{array}$

Substitute the value of ${\mathrm{C}}_1=0$ in the equation (8),

$\begin{array}{c}{\mathrm{c}}_1+{\mathrm{c}}_2=0\\ {\mathrm{c}}_2=0\end{array}$

Substitute the value of ${\mathrm{c}}_1=0$and ${\mathrm{c}}_2=0$in the equation (7),

role="math" localid="1655097899726" $\begin{array}{c}\mathrm{y}=\left(0\right){\mathrm{e}}^{2\mathrm{x}}+\left(0\right){\mathrm{e}}^{-\mathrm{x}}-\frac{3}{10}\mathrm{cosx}+\frac{3}{20}\sin \left(2\mathrm{x}\right)-\frac{1}{10}\mathrm{sinx}-\frac{1}{20}\cos \left(2\mathrm{x}\right)\\ \mathrm{y}=-\frac{3}{10}\mathrm{cosx}+\frac{3}{20}\sin \left(2\mathrm{x}\right)-\frac{1}{10}\mathrm{sinx}-\frac{1}{20}\cos \left(2\mathrm{x}\right)\end{array}$

Thus, the initial solution to the differential equation is:

$\mathrm{y}=-\frac{3}{10}\mathrm{cosx}+\frac{3}{20}\sin \left(2\mathrm{x}\right)-\frac{1}{10}\mathrm{sinx}-\frac{1}{20}\cos \left(2\mathrm{x}\right)$