Find a particular solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{111}{\mathbf{e}}^{2t}\mathbf{cos}\mathbf{3}\mathbf{t}$

The differential equation is:

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{111}{\mathbf{e}}^{2t}\mathbf{cos}\mathbf{3}\mathbf{t}\dots \left(\mathbf{1}\right)$

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

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0}$

The auxiliary equation for the above equation,

${\mathbf{m}}^2\mathbf{+}\mathbf{2}\mathbf{m}\mathbf{+}\mathbf{4}\mathbf{=}\mathbf{0}$

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^2+2\mathrm{m}+4=0\\ \mathrm{m}=\frac{-2\pm \sqrt{2^2-4\left(1\right)\left(4\right)}}{2\left(1\right)}\\ \mathrm{m}=\frac{-2\pm \sqrt{-12}}{2}\\ \mathrm{m}=-1\pm \mathrm{i}\sqrt{3}\end{array}$

The roots of the auxiliary equation are,

${\mathrm{m}}_1=-1+\mathrm{i}\sqrt{3},\&{\mathrm{m}}_2=-1-\mathrm{i}\sqrt{3}$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{e}}^{-\mathrm{x}}({\mathrm{c}}_1\cos \sqrt{3}\mathrm{x}+{\mathrm{c}}_2\sin \sqrt{3}\mathrm{x})$

According to the method of undetermined coefficients, assume, the particular solution of equation (1),

${\mathbf{y}}_p\mathbf{=}{\mathbf{e}}^{2t}({\mathbf{A}}_0\mathbf{cos}\mathbf{3}\mathbf{t}\mathbf{+}{\mathbf{B}}_0\mathbf{sin}\mathbf{3}\mathbf{t})\mathrm{......}\left(\mathbf{2}\right)$

Now find the derivative of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}=2{\mathrm{e}}^{2\mathrm{t}}({\mathrm{A}}_0\cos 3\mathrm{t}+{\mathrm{B}}_0\sin 3\mathrm{t})+{\mathrm{e}}^{2\mathrm{t}}(-3{\mathrm{A}}_0\sin 3\mathrm{t}+3{\mathrm{B}}_0\cos 3\mathrm{t})\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}=4{\mathrm{e}}^{2\mathrm{t}}({\mathrm{A}}_0\cos 3\mathrm{t}+{\mathrm{B}}_0\sin 3\mathrm{t})+2{\mathrm{e}}^{2\mathrm{t}}(-3{\mathrm{A}}_0\sin 3\mathrm{t}+3{\mathrm{B}}_0\cos 3\mathrm{t})\\ +2{\mathrm{e}}^{2\mathrm{t}}(-3{\mathrm{A}}_0\sin 3\mathrm{t}+3{\mathrm{B}}_0\cos 3\mathrm{t})+{\mathrm{e}}^{2\mathrm{t}}(-9{\mathrm{A}}_0\cos 3\mathrm{t}-9{\mathrm{B}}_0\sin 3\mathrm{t})\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}={\mathrm{e}}^{2\mathrm{t}}(-5{\mathrm{A}}_0\cos 3\mathrm{t}-5{\mathrm{B}}_0\sin 3\mathrm{t})+{\mathrm{e}}^{2\mathrm{t}}(-12{\mathrm{A}}_0\sin 3\mathrm{t}+12{\mathrm{B}}_0\cos 3\mathrm{t})\end{array}$

From the equation (1), Substitute the value of ${\mathbf{y}}_p\mathbf{\text{'}}\mathbf{\text{'}},{\mathbf{y}}_p\mathbf{\text{'}}$and$\hspace{0.17em}{\mathbf{y}}_p$ in the equation (1),

$\begin{array}{l}\Rightarrow {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}+2{\mathrm{y}}_{\mathrm{p}}\text{'}+4{\mathrm{y}}_{\mathrm{p}}=111{\mathrm{e}}^{2\mathrm{t}}\cos 3\mathrm{t}\\ \Rightarrow {\mathrm{e}}^{2\mathrm{t}}(-5{\mathrm{A}}_0\cos 3\mathrm{t}-5{\mathrm{B}}_0\sin 3\mathrm{t})+{\mathrm{e}}^{2\mathrm{t}}(-12{\mathrm{A}}_0\sin 3\mathrm{t}+12{\mathrm{B}}_0\cos 3\mathrm{t})\\ +2[2{\mathrm{e}}^{2\mathrm{t}}({\mathrm{A}}_0\cos 3\mathrm{t}+{\mathrm{B}}_0\sin 3\mathrm{t})+{\mathrm{e}}^{2\mathrm{t}}(-3{\mathrm{A}}_0\sin 3\mathrm{t}+3{\mathrm{B}}_0\cos 3\mathrm{t})]\\ +4{\mathrm{e}}^{2\mathrm{t}}({\mathrm{A}}_0\cos 3\mathrm{t}+{\mathrm{B}}_0\sin 3\mathrm{t})=111{\mathrm{e}}^{2\mathrm{t}}\cos 3\mathrm{t}\\ \Rightarrow {\mathrm{e}}^{2\mathrm{t}}(3{\mathrm{A}}_0\cos 3\mathrm{t}+3{\mathrm{B}}_0\sin 3\mathrm{t})+{\mathrm{e}}^{2\mathrm{t}}(-18{\mathrm{A}}_0\sin 3\mathrm{t}+18{\mathrm{B}}_0\cos 3\mathrm{t})=111{\mathrm{e}}^{2\mathrm{t}}\cos 3\mathrm{t}\\ \Rightarrow {\mathrm{e}}^{2\mathrm{t}}\cos 3\mathrm{t}(3{\mathrm{A}}_0+18{\mathrm{B}}_0)+{\mathrm{e}}^{2\mathrm{t}}\sin 3\mathrm{t}(3{\mathrm{B}}_0-18{\mathrm{A}}_0)=111{\mathrm{e}}^{2\mathrm{t}}\cos 3\mathrm{t}\end{array}$

Comparing all coefficients of the above equation;

$\begin{array}{c}3{\mathrm{A}}_0+18{\mathrm{B}}_0=111......\left(3\right)\\ 3{\mathrm{B}}_0-18{\mathrm{A}}_0=0......\left(4\right)\end{array}$

Solve the above equations,

$\begin{array}{c}(3{\mathrm{A}}_0+18{\mathrm{B}}_0)=111\times 6\\ 18{\mathrm{A}}_0+108{\mathrm{B}}_0=666\\ -18{\mathrm{A}}_0+3{\mathrm{B}}_0=0\\ \overline{111{\mathrm{B}}_0=666}\\ {\mathrm{B}}_0=6\end{array}$

Substitute the value of${\mathrm{B}}_0$ in the equation (3),

$\begin{array}{c}3{\mathrm{A}}_0+18\left(6\right)=111\\ 3{\mathrm{A}}_0=3\\ {\mathrm{A}}_0=1\end{array}$

Therefore, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}={\mathrm{e}}^{2\mathrm{t}}({\mathrm{A}}_0\cos 3\mathrm{t}+{\mathrm{B}}_0\sin 3\mathrm{t})\\ {\mathrm{y}}_{\mathrm{p}}={\mathrm{e}}^{2\mathrm{t}}(\cos 3\mathrm{t}+6\sin 3\mathrm{t})\end{array}$