Find the solution to the initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{27}\mathbf{;}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{4}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{6}$

The differential equation is,

$\mathrm{y}\text{'}\text{'}+9\mathrm{y}=27......\left(1\right)$

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

$\mathrm{y}\text{'}\text{'}+9\mathrm{y}=0$

The auxiliary equation for the above equation,

${\mathrm{m}}^2+9=0$

The root of an auxiliary equation is,

${\mathrm{m}}_1=3\mathrm{i},{\mathrm{m}}_2=-3\mathrm{i}$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1\cos \left(3\mathrm{x}\right)+{\mathrm{c}}_2\sin \left(3\mathrm{x}\right)$

Assume, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{A}.....\left(2\right)$

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

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{x}\right)=0\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{x}\right)=0\end{array}$

Substitute the value of ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{x}\right)$and ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)$the equation (1),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}+9\mathrm{y}=27\\ 0+9\mathrm{A}=27\\ \mathrm{A}=3\end{array}$

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

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=3$

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{x}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)\\ \mathrm{y}={\mathrm{c}}_1\cos \left(3\mathrm{x}\right)+{\mathrm{c}}_2\sin \left(3\mathrm{x}\right)+3......\left(3\right)\end{array}$

Given the initial condition,

$\mathrm{y}\left(0\right)=4,\mathrm{y}\text{'}\left(0\right)=6$

Substitute the value of y = 4 and x = 0 in the equation (3),

$\begin{array}{c}4={\mathrm{c}}_1\cos \left(0\right)+{\mathrm{c}}_2\sin \left(0\right)+3\\ 4={\mathrm{c}}_1+3\\ {\mathrm{c}}_1=1\end{array}$

Now find the derivative of the equation (3),

$\mathrm{y}\text{'}=-3{\mathrm{c}}_1\sin \left(3\mathrm{x}\right)+3{\mathrm{c}}_2\cos \left(3\mathrm{x}\right)$

Substitute the value of y’ = 6 and x = 0 in the above equation,

$\begin{array}{c}6=-3{\mathrm{c}}_1\sin \left(0\right)+3{\mathrm{c}}_2\cos \left(0\right)\\ 3{\mathrm{c}}_2=6\\ {\mathrm{c}}_2=2\end{array}$

Substitute the value of ${\mathrm{c}}_1=1$and ${\mathrm{c}}_2=2$in the equation (3),

$\begin{array}{c}\mathrm{y}={\mathrm{c}}_1\cos \left(3\mathrm{x}\right)+{\mathrm{c}}_2\sin \left(3\mathrm{x}\right)+3\\ \mathrm{y}=\left(1\right)\cos \left(3\mathrm{x}\right)+\left(2\right)\sin \left(3\mathrm{x}\right)+3\\ \mathrm{y}=\cos \left(3\mathrm{x}\right)+2\sin \left(3\mathrm{x}\right)+3\end{array}$

Thus, the solution to the initial problem is$\mathrm{y}=\cos \left(3\mathrm{x}\right)+2\sin \left(3\mathrm{x}\right)+3.$