Q22RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the given differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{8}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{33}\mathbf{y}\mathbf{=}\mathbf{546}\mathbf{sint}$

Step-by-Step Solution

Answer

$\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{11}\left(\mathbf{t}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{3}\left(\mathbf{t}\right)}\mathbf{+}\frac{\mathbf{1092}}{\mathbf{305}}\mathbf{cost}\mathbf{-}\frac{\mathbf{4641}}{\mathbf{305}}\mathbf{sint}$

Write the auxiliary equation of the given differential equation

The differential equation is,

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{8}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{33}\mathbf{y}\mathbf{=}\mathbf{546}\mathbf{sint} ......\left(\mathbf{1}\right)$

Write the homogeneous differential equation of the equation (1),

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{8}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{33}\mathbf{y}\mathbf{=}\mathbf{0}$

The auxiliary equation for the above equation, ${\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{8}\mathbf{m}\mathbf{-}\mathbf{33}\mathbf{=}\mathbf{0}$

Find the roots of the auxiliary equation.

Solve the auxiliary equation,

$\begin{array}{rcl}{\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{8}\mathbf{m}\mathbf{-}\mathbf{33}& \mathbf{=}& \mathbf{0}\\ {\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{11}\mathbf{m}\mathbf{+}\mathbf{3}\mathbf{m}\mathbf{-}\mathbf{33}& \mathbf{=}& \mathbf{0}\\ \mathbf{m}\left(\mathbf{m}\mathbf{-}\mathbf{11}\right)\mathbf{+}\mathbf{3}\left(\mathbf{m}\mathbf{-}\mathbf{11}\right)& \mathbf{=}& \mathbf{0}\\ \left(\mathbf{m}\mathbf{-}\mathbf{11}\right)\left(\mathbf{m}\mathbf{+}\mathbf{3}\right)& \mathbf{=}& \mathbf{0}\\ & & \end{array}$

The roots of the auxiliary equation are ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{11}\mathbf{,} \& {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{3}$.

The complementary solution of the given equation is ${\mathbf{y}}_{\mathbf{c}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{11}\left(\mathbf{t}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{3}\left(\mathbf{t}\right)}$.

Find the particular solution

Assume, the particular solution of equation (1),

 ${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{Acost}\mathbf{+}\mathbf{Bsint} ......\left(\mathbf{2}\right)$

Now find the derivative of the above equation,

$$\begin{gathered} {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\mathbf{Asint}\mathbf{+}\mathbf{Bcost} \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\mathbf{Acost}\mathbf{-}\mathbf{Bsint} \end{gathered}$$

Substitute the value of ${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{,} {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\left(\mathbf{t}\right)$ and ${\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{t}\right)$  in the equation (1),

$\begin{array}{rcl}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{8}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{33}\mathbf{y}& \mathbf{=}& \mathbf{546}\mathbf{sint}\\ \mathbf{-}\mathbf{Acost}\mathbf{-}\mathbf{Bsint}\mathbf{-}\mathbf{8}\left(\mathbf{-}\mathbf{Asint}\mathbf{+}\mathbf{Bcost}\right)\mathbf{-}\mathbf{33}\left(\mathbf{Acost}\mathbf{+}\mathbf{Bsint}\right)& \mathbf{=}& \mathbf{546}\mathbf{sint}\\ \left(\mathbf{-}\mathbf{8}\mathbf{B}\mathbf{-}\mathbf{34}\mathbf{A}\right)\mathbf{cost}\mathbf{+}\left(\mathbf{8}\mathbf{A}\mathbf{-}\mathbf{34}\mathbf{B}\right)\mathbf{sint}& \mathbf{=}& \mathbf{546}\mathbf{sint}\end{array}$

Comparing all coefficients of the above equation,

$\begin{array}{rcl}\mathbf{-}\mathbf{8}\mathbf{B}\mathbf{-}\mathbf{34}\mathbf{A}& \mathbf{=}& \mathbf{0}\\ \mathbf{8}\mathbf{A}\mathbf{-}\mathbf{34}\mathbf{B}& \mathbf{=}& \mathbf{546}\\ & & \end{array}$

Solve the above equations,

$\begin{array}{rcl}\mathbf{8}\mathbf{B}\mathbf{+}\mathbf{34}\mathbf{A}& \mathbf{=}& \mathbf{0} \Rightarrow \left(\mathbf{4}\mathbf{B}\mathbf{+}\mathbf{17}\mathbf{A}\mathbf{=}\mathbf{0}\right)\mathbf{\times }\mathbf{17}\\ \mathbf{8}\mathbf{A}\mathbf{-}\mathbf{34}\mathbf{B}& \mathbf{=}& \mathbf{546} \Rightarrow \left(\mathbf{4}\mathbf{A}\mathbf{-}\mathbf{17}\mathbf{B}\mathbf{=}\mathbf{273}\right)\mathbf{\times }\mathbf{4}\\ \mathbf{68}\mathbf{B}\mathbf{+}\mathbf{289}\mathbf{A}& \mathbf{=}& \mathbf{0} ......\left(\mathbf{3}\right)\\ \mathbf{-}\mathbf{68}\mathbf{B}\mathbf{+}\mathbf{16}\mathbf{A}& \mathbf{=}& \mathbf{1092} ......\left(\mathbf{4}\right)\\ \mathbf{A}& \mathbf{=}& \frac{\mathbf{1092}}{\mathbf{305}}\end{array}$

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

$\begin{array}{rcl}\mathbf{68}\mathbf{B}\mathbf{+}\mathbf{289}\left(\frac{\mathbf{1092}}{\mathbf{305}}\right)& \mathbf{=}& \mathbf{0}\\ \mathbf{B}& \mathbf{=}& \mathbf{-}\frac{\mathbf{4641}}{\mathbf{305}}\end{array}$

Therefore, the particular solution of equation (1),

$$\begin{gathered} {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{Acost}\mathbf{+}\mathbf{Bsint} \\ {\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\frac{\mathbf{1092}}{\mathbf{305}}\mathbf{cost}\mathbf{-}\frac{\mathbf{4641}}{\mathbf{305}}\mathbf{sint} \end{gathered}$$

Conclusion, the general solution

Therefore, the general solution is,

 $$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{y}}_{\mathbf{c}}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right) \\ \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{11}\left(\mathbf{t}\right)}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{3}\left(\mathbf{t}\right)}\mathbf{+}\frac{\mathbf{1092}}{\mathbf{305}}\mathbf{cost}\mathbf{-}\frac{\mathbf{4641}}{\mathbf{305}}\mathbf{sint} \end{gathered}$$