Q33RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find the solution to the given initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{12}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{27}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{40}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{6}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{12}$

Step-by-Step Solution

Answer

The general solution to the given initial value problem is $\mathrm{y}=-{\mathrm{e}}^{-\mathrm{t}}-3{\mathrm{e}}^{5\mathrm{t}}+{\mathrm{e}}^{8\mathrm{t}}$.

Write the auxiliary equation of the given differential equation

The given differential equation is,

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{12}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{27}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{40}\mathbf{y}\mathbf{=}\mathbf{0} ......\left(\mathbf{1}\right)$

The auxiliary equation for the above equation,

$\begin{array}{rcl}{\mathbf{m}}^{\mathbf{3}}\mathbf{-}\mathbf{12}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{27}\mathbf{m}\mathbf{+}\mathbf{40}& \mathbf{=}& \mathbf{0}\\ {\mathbf{m}}^{\mathbf{3}}\mathbf{-}\mathbf{13}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{40}\mathbf{m}\mathbf{-}\mathbf{13}\mathbf{m}\mathbf{+}\mathbf{40}& \mathbf{=}& \mathbf{0}\\ {\mathbf{m}}^{\mathbf{3}}\mathbf{+}{\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{13}{\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{13}\mathbf{m}\mathbf{+}\mathbf{40}\mathbf{m}\mathbf{+}\mathbf{40}& \mathbf{=}& \mathbf{0}\\ {\mathbf{m}}^{\mathbf{2}}\left(\mathbf{m}\mathbf{+}\mathbf{1}\right)\mathbf{-}\mathbf{13}\mathbf{m}\left(\mathbf{m}\mathbf{+}\mathbf{1}\right)\mathbf{+}\mathbf{40}\left(\mathbf{m}\mathbf{+}\mathbf{1}\right)& \mathbf{=}& \mathbf{0}\\ \left(\mathbf{m}\mathbf{+}\mathbf{1}\right)\left({\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{13}\mathbf{m}\mathbf{+}\mathbf{40}\right)& \mathbf{=}& \mathbf{0}\\ \left(\mathbf{m}\mathbf{+}\mathbf{1}\right)\left({\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{8}\mathbf{m}\mathbf{-}\mathbf{5}\mathbf{m}\mathbf{+}\mathbf{40}\right)& \mathbf{=}& \mathbf{0}\\ \left(\mathbf{m}\mathbf{+}\mathbf{1}\right)\left(\mathbf{m}\mathbf{-}\mathbf{5}\right)\left(\mathbf{m}\mathbf{-}\mathbf{8}\right)& \mathbf{=}& \mathbf{0}\end{array}$

Now find the general solution of the given equation

The root of an auxiliary equation is ${\mathrm{m}}_1=-1, {\mathrm{m}}_2=5, {\mathrm{m}}_3=8$

The general solution of the given equation is,

 $\mathrm{y}={\mathrm{Ae}}^{-\mathrm{t}}+{\mathrm{Be}}^{5\mathrm{t}}+{\mathrm{Ce}}^{8\mathrm{t}} ......\left(2\right)$

Use the given initial condition,

Given the initial condition,

$\mathrm{y}\left(0\right)=-3, \mathrm{y}\text{'}\left(0\right)=-6, \mathrm{y}\text{'}\text{'}\left(0\right)=-12$

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

$$\begin{gathered} -3={\mathrm{Ae}}^{-0}+{\mathrm{Be}}^{5\left(0\right)}+{\mathrm{Ce}}^{8\left(0\right)} \\ \mathrm{A}+\mathrm{B}+\mathrm{C}=-3 ......\left(3\right) \end{gathered}$$

Now find the first derivative of the equation (2),

$\mathrm{y}\text{'}=-{\mathrm{Ae}}^{-\mathrm{t}}+5{\mathrm{Be}}^{5\mathrm{t}}+8{\mathrm{Ce}}^{8\mathrm{t}}$

Substitute the value of $y^{\text{'}}=-6$ and $t=0$ in the above equation,

$$\begin{gathered} -6=-{\mathrm{Ae}}^{-\left(0\right)}+5{\mathrm{Be}}^{5\left(0\right)}+8{\mathrm{Ce}}^{8\left(0\right)} \\ -\mathrm{A}+5\mathrm{B}+8\mathrm{C}=-6 ......\left(4\right) \end{gathered}$$

Now find the second derivative of the equation (2),

$\mathrm{y}\text{'}\text{'}={\mathrm{Ae}}^{-\mathrm{t}}+25{\mathrm{Be}}^{5\mathrm{t}}+64{\mathrm{Ce}}^{8\mathrm{t}}$

Substitute the value of $y^{\text{'}\text{'}}=-12$and $t=0$ in the above equation,

$$\begin{gathered} -12={\mathrm{Ae}}^{-\left(0\right)}+25{\mathrm{Be}}^{5\left(0\right)}+64{\mathrm{Ce}}^{8\left(0\right)} \\ \mathrm{A}+25\mathrm{B}+64\mathrm{C}=-12 ......\left(5\right) \end{gathered}$$

Solve the equation (3) and (4),

$\begin{array}{rcl}\mathrm{A}+\mathrm{B}+\mathrm{C}& =& -3\\ -\mathrm{A}+5\mathrm{B}+8\mathrm{C}& =& -6\\ 6\mathrm{B}+9\mathrm{C}& =& -9 \dots \left(6\right)\\ & & \end{array}$

Solve the equation (4) and (5),

$$\begin{gathered} \mathrm{A}+25\mathrm{B}+64\mathrm{C}=-12 \\ -\mathrm{A}+5\mathrm{B}+8\mathrm{C}=-6 \\ \overline{ 30\mathrm{B}+72\mathrm{C}=-18} ......\left(7\right) \end{gathered}$$

Solve the equation (6) and (7),

$\begin{array}{rcl}30\mathrm{B}+45\mathrm{C}& =& -45\\ 30\mathrm{B}+72\mathrm{C}& =& -18\\ C& =& 1\\ & & \end{array}$

Substitute the value of C in the equation (6),

 $\begin{array}{rcl}6\mathrm{B}+9\mathrm{C}& =& -9\\ 6\mathrm{B}+9\left(1\right)& =& -9\\ 6\mathrm{B}& =& -18\\ \mathrm{B}& =& -3\\ & & \end{array}$

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

$\begin{array}{rcl}\mathrm{A}+\mathrm{B}+\mathrm{C}& =& -3\\ \mathrm{A}+\left(-3\right)+1& =& -3\\ \mathrm{A}& =& -1\\ & & \end{array}$

Final conclusion,

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