Find the solution to the given initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{7}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{2}$

The given differential equation is,

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{7}\mathbf{y}\mathbf{=}\mathbf{0} ......\left(\mathbf{1}\right)$

The auxiliary equation for the above equation,

$\begin{array}{rcl}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{m}\mathbf{+}\mathbf{7}& \mathbf{=}& \mathbf{0}\\ \mathbf{m}& \mathbf{=}& \frac{\mathbf{-}\mathbf{4}\mathbf{\pm }\sqrt{\mathbf{16}\mathbf{-}\mathbf{28}}}{\mathbf{2}}\\ \mathbf{m}& \mathbf{=}& \frac{\mathbf{-}\mathbf{4}\mathbf{\pm }\sqrt{\mathbf{-}\mathbf{12}}}{\mathbf{2}}\\ \mathbf{m}& \mathbf{=}& \mathbf{-}\mathbf{2}\mathbf{\pm }\mathbf{i}\sqrt{\mathbf{3}}\end{array}$

The root of an auxiliary equation is ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{+}\mathbf{i}\sqrt{\mathbf{3}}\mathbf{,} \& {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{-}\mathbf{i}\sqrt{\mathbf{3}}$

The general solution of the given equation is,

$\mathbf{y}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{cos}\left(\sqrt{\mathbf{3}}\mathbf{t}\right)\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{sin}\left(\sqrt{\mathbf{3}}\mathbf{t}\right) ......\left(\mathbf{2}\right)$

Given the initial condition,

$\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{2}$

Substitute the value of $y=1$and $t=0$in the equation (2),

role="math" $$\begin{gathered} \mathbf{1}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\mathbf{2}\left(\mathbf{0}\right)}\mathbf{cos}\left(\sqrt{\mathbf{3}}\left(\mathbf{0}\right)\right)\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{2}\left(\mathbf{0}\right)}\mathbf{sin}\left(\sqrt{\mathbf{3}}\left(\mathbf{0}\right)\right) \\ \mathbf{A}\mathbf{=}\mathbf{1} \end{gathered}$$

Now find the derivative of the equation (2),

$$\begin{gathered} \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{Ae}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{cos}\left(\sqrt{\mathbf{3}}\mathbf{t}\right)\mathbf{-}\sqrt{\mathbf{3}}{\mathbf{Ae}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{sin}\left(\sqrt{\mathbf{3}}\mathbf{t}\right)\mathbf{-}\mathbf{2}{\mathbf{Be}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{sin}\left(\sqrt{\mathbf{3}}\mathbf{t}\right)\mathbf{+}\sqrt{\mathbf{3}}{\mathbf{Be}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{cos}\left(\sqrt{\mathbf{3}}\mathbf{t}\right) \\ \mathbf{y}\mathbf{\text{'}}\mathbf{=}\left[\mathbf{-}\mathbf{2}{\mathbf{Ae}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{+}\sqrt{\mathbf{3}}{\mathbf{Be}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\right]\mathbf{cos}\left(\sqrt{\mathbf{3}}\mathbf{t}\right)\mathbf{+}\left[\mathbf{-}\sqrt{\mathbf{3}}{\mathbf{Ae}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{-}\mathbf{2}{\mathbf{Be}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\right]\mathbf{sin}\left(\sqrt{\mathbf{3}}\mathbf{t}\right) \end{gathered}$$

Substitute the value of $y^{\text{'}}=-2$and$t=0$in the above equation,

$$\begin{gathered} \mathbf{-}\mathbf{2}\mathbf{=}\left[\mathbf{-}\mathbf{2}{\mathbf{Ae}}^{\mathbf{-}\mathbf{2}\left(\mathbf{0}\right)}\mathbf{+}\sqrt{\mathbf{3}}{\mathbf{Be}}^{\mathbf{-}\mathbf{2}\left(\mathbf{0}\right)}\right]\mathbf{cos}\left(\sqrt{\mathbf{3}}\left(\mathbf{0}\right)\right)\mathbf{+}\left[\mathbf{-}\sqrt{\mathbf{3}}{\mathbf{Ae}}^{\mathbf{-}\mathbf{2}\left(\mathbf{0}\right)}\mathbf{-}\mathbf{2}{\mathbf{Be}}^{\mathbf{-}\mathbf{2}\left(\mathbf{0}\right)}\right]\mathbf{sin}\left(\sqrt{\mathbf{3}}\left(\mathbf{0}\right)\right) \\ \mathbf{-}\mathbf{2}\mathbf{A}\mathbf{+}\sqrt{\mathbf{3}}\mathbf{B}\mathbf{=}\mathbf{-}\mathbf{2} ......\left(\mathbf{3}\right) \end{gathered}$$

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

$\begin{array}{rcl}\mathbf{-}\mathbf{2}\mathbf{A}\mathbf{+}\sqrt{\mathbf{3}}\mathbf{B}& \mathbf{=}& \mathbf{-}\mathbf{2}\\ \mathbf{-}\mathbf{2}\left(\mathbf{1}\right)\mathbf{+}\sqrt{\mathbf{3}}\mathbf{B}& \mathbf{=}& \mathbf{-}\mathbf{2}\\ \mathbf{B}& \mathbf{=}& \mathbf{0}\\ & & \end{array}$

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

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{cos}\left(\sqrt{\mathbf{3}}\mathbf{t}\right)\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{sin}\left(\sqrt{\mathbf{3}}\mathbf{t}\right) \\ \mathbf{y}\mathbf{=}\left(\mathbf{1}\right){\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{cos}\left(\sqrt{\mathbf{3}}\mathbf{t}\right)\mathbf{+}\left(\mathbf{0}\right){\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{sin}\left(\sqrt{\mathbf{3}}\mathbf{t}\right) \\ \mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{cos}\left(\sqrt{\mathbf{3}}\mathbf{t}\right) \end{gathered}$$

Thus, the general solution is$\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{cos}\left(\sqrt{\mathbf{3}}\mathbf{t}\right)$.