91Ó°ÊÓ

The given differential equation is,

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{8}\mathbf{y}\mathbf{=}\mathbf{0}                             ...\left(\mathbf{1}\right)$

The auxiliary equation for the above equation,

$$\begin{gathered} {\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{m}\mathbf{-}\mathbf{8}\mathbf{=}\mathbf{0} \\ {\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{m}\mathbf{-}\mathbf{2}\mathbf{m}\mathbf{-}\mathbf{8}\mathbf{=}\mathbf{0} \\ \mathbf{m}\left(\mathbf{m}\mathbf{+}\mathbf{4}\right)\mathbf{-}\mathbf{2}\left(\mathbf{m}\mathbf{+}\mathbf{4}\right)\mathbf{=}\mathbf{0} \\ \left(\mathbf{m}\mathbf{+}\mathbf{4}\right)\left(\mathbf{m}\mathbf{-}\mathbf{2}\right)\mathbf{=}\mathbf{0} \end{gathered}$$

The roots of an auxiliary equation are

${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{,} \& {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{2}.$

If the auxiliary equation has distinct real roots, then the general solution is given as;

$\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{{\mathbf{m}}_{\mathbf{1}}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{{\mathbf{m}}_{\mathbf{2}}\mathbf{t}}$

Therefore, the general solution of the given equation is,

$\mathbf{y}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{2}\mathbf{t}}                       ...\left(\mathbf{2}\right)$

Given the initial condition,

$\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{3}\mathbf{,}    \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{12}$

Substitute the value of y = 3 and t = 0 in the equation (2),

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{3}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\left(\mathbf{0}\right)}\mathbf{+}{\mathbf{Be}}^{\mathbf{2}\left(\mathbf{0}\right)} \\ \mathbf{A}\mathbf{+}\mathbf{B}\mathbf{=}\mathbf{3}                          ...\left(\mathbf{3}\right) \end{gathered}$$

Now find the derivative of the equation (2),

$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{4}{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{+}\mathbf{2}{\mathbf{Be}}^{\mathbf{2}\mathbf{t}}$

Substitute the value of y’ = -12 and t = 0 in the above equation,

$$\begin{gathered} \mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{4}{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{+}\mathbf{2}{\mathbf{Be}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{-}\mathbf{12}\mathbf{=}\mathbf{-}\mathbf{4}{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\left(\mathbf{0}\right)}\mathbf{+}\mathbf{2}{\mathbf{Be}}^{\mathbf{2}\left(\mathbf{0}\right)} \\ \mathbf{-}\mathbf{4}\mathbf{A}\mathbf{+}\mathbf{2}\mathbf{B}\mathbf{=}\mathbf{-}\mathbf{12}                           ...\left(\mathbf{4}\right) \end{gathered}$$

Solve the equation (3) and (4),

$$\begin{gathered}   \mathbf{4}\mathbf{A}\mathbf{+}\mathbf{4}\mathbf{B}\mathbf{=}\mathbf{12} \\ \mathbf{-}\mathbf{4}\mathbf{A}\mathbf{+}\mathbf{2}\mathbf{B}\mathbf{=}\mathbf{-}\mathbf{12} \\                 \mathbf{B}\mathbf{=}\mathbf{0} \end{gathered}$$

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

$$\begin{gathered} \mathbf{A}\mathbf{+}\mathbf{B}\mathbf{=}\mathbf{3} \\ \mathbf{A}\mathbf{+}\mathbf{0}\mathbf{=}\mathbf{3} \\ \mathbf{A}\mathbf{=}\mathbf{3} \end{gathered}$$

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

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{y}\mathbf{=}\mathbf{3}{\mathbf{e}}^{\mathbf{-}\mathbf{4}\mathbf{t}}\mathbf{+}\left(\mathbf{0}\right){\mathbf{e}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{y}\mathbf{=}\mathbf{3}{\mathbf{e}}^{\mathbf{-}\mathbf{4}\mathbf{t}} \end{gathered}$$

Thus, the solution to the given initial value problem is;

$y=3e^{-4t}$