Q34RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find the solution to the given initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{14}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\left(0\right)\mathbf{=}\mathbf{5}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{1}$

Step-by-Step Solution

Answer

The solution to the given initial value problem is; $\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{7}\mathbf{t}}\mathbf{+}\mathbf{4}{\mathbf{e}}^{\mathbf{2}\mathbf{t}}$

Write the auxiliary equation of the given differential equation

The given differential equation is,

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{14}\mathbf{y}\mathbf{=}\mathbf{0} ......\left(\mathbf{1}\right)$

The auxiliary equation for the above equation,

 $\begin{array}{rcl}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{5}\mathbf{m}\mathbf{-}\mathbf{14}& \mathbf{=}& \mathbf{0}\\ {\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{7}\mathbf{m}\mathbf{-}\mathbf{2}\mathbf{m}\mathbf{-}\mathbf{14}& \mathbf{=}& \mathbf{0}\\ \mathbf{m}\left(\mathbf{m}\mathbf{+}\mathbf{7}\right)\mathbf{-}\mathbf{2}\left(\mathbf{m}\mathbf{+}\mathbf{7}\right)& \mathbf{=}& \mathbf{0}\\ \left(\mathbf{m}\mathbf{+}\mathbf{7}\right)\left(\mathbf{m}\mathbf{-}\mathbf{2}\right)& \mathbf{=}& \mathbf{0}\end{array}$

Now find the general solution of the given equation

The root of an auxiliary equation is ${\mathrm{m}}_1=-7, {\mathrm{m}}_2=2$

The general solution of the given equation is,

$\mathrm{y}={\mathrm{Ae}}^{-7\mathrm{t}}+{\mathrm{Be}}^{2\mathrm{t}} ......\left(2\right)$

Use the given initial condition

Given the initial condition,

 $\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{5}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}$

Substitute the value of $y=5$ and $t=0$ in the equation (2),

$\begin{array}{rcl}\mathbf{y}& \mathbf{=}& {\mathbf{Ae}}^{\mathbf{-}\mathbf{7}\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{2}\mathbf{t}}\\ \mathbf{5}& \mathbf{=}& {\mathbf{Ae}}^{\mathbf{-}\mathbf{7}\left(\mathbf{0}\right)}\mathbf{+}{\mathbf{Be}}^{\mathbf{2}\left(\mathbf{0}\right)}\\ \mathbf{A}\mathbf{+}\mathbf{B}& \mathbf{=}& \mathbf{5} .......\left(\mathbf{3}\right)\\ & & \end{array}$

Now find the derivative of the equation (2),

$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{7}{\mathbf{Ae}}^{\mathbf{-}\mathbf{7}\mathbf{t}}\mathbf{+}\mathbf{2}{\mathbf{Be}}^{\mathbf{2}\mathbf{t}}$

Substitute the value of $y^{\text{'}}=1$ and $t=0$ in the above equation,

$\begin{array}{rcl}\mathbf{1}& \mathbf{=}& \mathbf{-}\mathbf{7}{\mathbf{Ae}}^{\mathbf{-}\mathbf{7}\left(\mathbf{0}\right)}\mathbf{+}\mathbf{2}{\mathbf{Be}}^{\mathbf{2}\left(\mathbf{0}\right)}\\ \mathbf{-}\mathbf{7}\mathbf{A}\mathbf{+}\mathbf{2}\mathbf{B}& \mathbf{=}& \mathbf{1} ......\left(\mathbf{4}\right)\\ & & \end{array}$

Solve the equation (3) and (4),

$$\begin{gathered} \left[\mathbf{A}\mathbf{+}\mathbf{B}\mathbf{=}\mathbf{5}\right]\mathbf{\times }\mathbf{7} \\ \mathbf{7}\mathbf{A}\mathbf{+}\mathbf{7}\mathbf{B}\mathbf{=}\mathbf{35} \\ \mathbf{-}\mathbf{7}\mathbf{A}\mathbf{+}\mathbf{2}\mathbf{B}\mathbf{=}\mathbf{1} \\ —————— \\ \mathbf{9}\mathbf{B}\mathbf{=}\mathbf{36} \\ \mathbf{B}\mathbf{=}\mathbf{4} \end{gathered}$$

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

$\begin{array}{rcl}\mathbf{A}\mathbf{+}\mathbf{B}& \mathbf{=}& \mathbf{5}\\ \mathbf{A}\mathbf{+}\mathbf{4}& \mathbf{=}& \mathbf{5}\\ \mathbf{A}& \mathbf{=}& \mathbf{1}\\ & & \end{array}$

Final conclusion

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

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\mathbf{7}\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{2}\mathbf{t}} \\ \mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{7}\mathbf{t}}\mathbf{+}\mathbf{4}{\mathbf{e}}^{\mathbf{2}\mathbf{t}} \end{gathered}$$