Q36RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find the solution to the given initial value problem.

$\mathbf{9}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{12}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{3}$

Step-by-Step Solution

Answer

The solution to the given initial value problem is:

$\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{3}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{t}}\mathbf{+}{\mathbf{te}}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{t}}$

Write the auxiliary equation of the given differential equation

The given differential equation is,

 $\mathbf{9}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{12}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0} ......\left(\mathbf{1}\right)$

The auxiliary equation for the above equation,

$\begin{array}{rcl}\mathbf{9}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{12}\mathbf{m}\mathbf{+}\mathbf{4}& \mathbf{=}& \mathbf{0}\\ {\left(\mathbf{3}\mathbf{m}\mathbf{+}\mathbf{2}\right)}^{\mathbf{2}}& \mathbf{=}& \mathbf{0}\end{array}$

Now find the general solution of the given equation

The root of an auxiliary equation is, 

${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{,} {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}$

The general solution of the given equation is,

$\mathbf{y}\mathbf{=}{\mathbf{Ae}}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{t}}\mathbf{+}{\mathbf{Bte}}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{t}} ......\left(\mathbf{2}\right)$

Use the given initial condition

Given the initial condition,

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

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

 $\begin{array}{rcl}\mathbf{y}& \mathbf{=}& {\mathbf{Ae}}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{t}}\mathbf{+}{\mathbf{Bte}}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{t}}⩽\\ \mathbf{-}\mathbf{3}& \mathbf{=}& {\mathbf{Ae}}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\left(\mathbf{0}\right)}\mathbf{+}\mathbf{B}\left(\mathbf{0}\right){\mathbf{e}}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\left(\mathbf{0}\right)}\\ \mathbf{A}& \mathbf{=}& \mathbf{-}\mathbf{3}\end{array}$

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

$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}{\mathbf{Ae}}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{t}}\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}{\mathbf{Bte}}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{t}}$

Substitute the value of $y^{\text{'}}=3$ and $t=0$ in the above equation,

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

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

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

Final conclusion

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

 $\begin{array}{rcl}\mathbf{y}& \mathbf{=}& {\mathbf{Ae}}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{t}}\mathbf{+}{\mathbf{Bte}}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{t}}\\ \mathbf{y}& \mathbf{=}& \left(\mathbf{-}\mathbf{3}\right){\mathbf{e}}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{t}}\mathbf{+}\left(\mathbf{1}\right){\mathbf{te}}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{t}}\\ \mathbf{y}& \mathbf{=}& \mathbf{-}\mathbf{3}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{t}}\mathbf{+}{\mathbf{te}}^{\mathbf{-}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{t}}\end{array}$