Q31RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find the solution to the given initial value problem.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathbf{y}\mathbf{=}\mathbf{6}\mathbf{cos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{-}\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right)\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{2}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{8}$

Step-by-Step Solution

Answer

The general solution to the given differential equation is;

$\mathbf{y}\mathbf{=}\mathbf{2}{\mathbf{e}}^{\mathbf{t}}\mathbf{cos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{-}\frac{\mathbf{7}}{\mathbf{3}}{\mathbf{e}}^{\mathbf{t}}\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right)\mathbf{-}\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right)$

Write the auxiliary equation of the given differential equation

The given differential equation is,

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathbf{y}\mathbf{=}\mathbf{6}\mathbf{cos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{-}\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right) ......\left(\mathbf{1}\right)$

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

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathbf{y}\mathbf{=}\mathbf{0}$

The auxiliary equation for the above equation,

 ${\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{m}\mathbf{+}\mathbf{10}\mathbf{=}\mathbf{0}$

Now find the complementary solution of the given equation

Solve the above equation,

$\begin{array}{rcl}{\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{m}\mathbf{+}\mathbf{10}& \mathbf{=}& \mathbf{0}\\ \mathbf{m}& \mathbf{=}& \frac{\mathbf{2}\mathbf{\pm }\sqrt{\mathbf{4}\mathbf{-}\mathbf{40}}}{\mathbf{2}}\\ \mathbf{m}& \mathbf{=}& \mathbf{1}\mathbf{\pm }\mathbf{3}\mathbf{i}\end{array}$

The root of an auxiliary equation is ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{3}\mathbf{i}\mathbf{,} {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{3}\mathbf{i}.$

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

Now find the particular solution to find a general solution

Assume, the particular solution of equation (1),

${\mathbf{y}}_{\mathbf{p}}\mathbf{=}\mathbf{Acos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{+}\mathbf{Bsin}\left(\mathbf{3}\mathbf{t}\right) ......\left(\mathbf{2}\right)$

Now find the first and second derivatives of above equation,

$$\begin{gathered} {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{Asin}\left(\mathbf{3}\mathbf{t}\right)\mathbf{+}\mathbf{3}\mathbf{Bcos}\left(\mathbf{3}\mathbf{t}\right) \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{9}\mathbf{Acos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{-}\mathbf{9}\mathbf{Bsin}\left(\mathbf{3}\mathbf{t}\right) \end{gathered}$$

Substitute the value of $\mathbf{y}, \mathbf{y}\mathbf{\text{'}}$and $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}$ the equation (1),

$\begin{array}{rcl}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathbf{y}& \mathbf{=}& \mathbf{6}\mathbf{cos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{-}\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right)\mathbf{-}\mathbf{9}\mathbf{Acos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{-}\mathbf{9}\mathbf{Bsin}\left(\mathbf{3}\mathbf{t}\right)\\ & & \mathbf{-}\mathbf{2}\left[\mathbf{-}\mathbf{3}\mathbf{Asin}\left(\mathbf{3}\mathbf{t}\right)\mathbf{+}\mathbf{3}\mathbf{Bcos}\left(\mathbf{3}\mathbf{t}\right)\right]\mathbf{+}\mathbf{10}\left[\mathbf{Acos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{+}\mathbf{Bsin}\left(\mathbf{3}\mathbf{t}\right)\right]\\ & \mathbf{=}& \mathbf{6}\mathbf{cos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{-}\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right)\\ \left[\mathbf{A}\mathbf{-}\mathbf{6}\mathbf{B}\right]\mathbf{cos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{+}\left[\mathbf{B}\mathbf{+}\mathbf{6}\mathbf{A}\right]\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right)& \mathbf{=}& \mathbf{6}\mathbf{cos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{-}\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right)\\ & & \end{array}$

Comparing the all coefficients of the above equation,

$$\begin{gathered} \mathbf{A}\mathbf{-}\mathbf{6}\mathbf{B}\mathbf{=}\mathbf{6} ......\left(\mathbf{3}\right) \\ \mathbf{B}\mathbf{+}\mathbf{6}\mathbf{A}\mathbf{=}\mathbf{-}\mathbf{1} ......\left(\mathbf{4}\right) \end{gathered}$$

Solve the equation (3) and (4),

$$\begin{gathered} -1\times \left[\mathbf{B}\mathbf{+}\mathbf{6}\mathbf{A}\right]=-1\mathbf{\times }\mathbf{6} \\ \mathbf{36}\mathbf{A}\mathbf{+}\mathbf{6}\mathbf{B}\mathbf{=}\mathbf{-}\mathbf{6} \\ \mathbf{A}\mathbf{-}\mathbf{6}\mathbf{B}\mathbf{=}\mathbf{6} \\ \mathbf{A}\mathbf{=}\mathbf{0} \end{gathered}$$

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

$\begin{array}{rcl}\mathbf{A}\mathbf{-}\mathbf{6}\mathbf{B}& \mathbf{=}& \mathbf{6}\\ \mathbf{0}\mathbf{-}\mathbf{6}\mathbf{B}& \mathbf{=}& \mathbf{6}\\ \mathbf{B}& \mathbf{=}& \mathbf{-}\mathbf{1}\\ & & \end{array}$

Substitute the value of A and Bin the equation (2),

$$\begin{gathered} {\mathbf{y}}_{\mathbf{p}}\mathbf{=}\mathbf{Acos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{+}\mathbf{Bsin}\left(\mathbf{3}\mathbf{t}\right) \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{=}\left(\mathbf{0}\right)\mathbf{cos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{+}\left(\mathbf{-}\mathbf{1}\right)\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right) \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{=}\mathbf{-}\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right) \end{gathered}$$

Find the general solution and use the given initial condition

Therefore, the general solution is,

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{y}}_{\mathbf{c}}\mathbf{+}{\mathbf{y}}_{\mathbf{p}} \\ \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{cos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right)\mathbf{-}\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right) ......\left(\mathbf{5}\right) \end{gathered}$$

Given initial condition,

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

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

$$\begin{gathered} \mathbf{2}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{0}}\mathbf{cos}\left(\mathbf{3}\left(\mathbf{0}\right)\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{0}}\mathbf{sin}\left(\mathbf{3}\left(\mathbf{0}\right)\right)\mathbf{-}\mathbf{sin}\left(\mathbf{3}\left(\mathbf{0}\right)\right) \\ {\mathbf{c}}_{\mathbf{1}}\mathbf{=}\mathbf{2} \end{gathered}$$

Now find the derivative of the equation (5),

$\mathbf{y}\mathbf{\text{'}}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{cos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{-}\mathbf{3}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right)\mathbf{+}\mathbf{3}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{cos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{-}\mathbf{3}\mathbf{cos}\left(\mathbf{3}\mathbf{t}\right)$

Substitute the value of $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{8}$and $t=0$ in the above equation,

 $\begin{array}{rcl}\mathbf{-}\mathbf{8}& \mathbf{=}& {\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{0}}\mathbf{cos}\left(\mathbf{3}\left(\mathbf{0}\right)\right)\mathbf{-}\mathbf{3}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{0}}\mathbf{sin}\left(\mathbf{3}\left(\mathbf{0}\right)\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{0}}\mathbf{sin}\left(\mathbf{3}\left(\mathbf{0}\right)\right)\mathbf{+}\mathbf{3}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{0}}\mathbf{cos}\left(\mathbf{3}\left(\mathbf{0}\right)\right)\mathbf{-}\mathbf{3}\mathbf{cos}\left(\mathbf{3}\left(\mathbf{0}\right)\right)\\ \mathbf{-}\mathbf{8}& \mathbf{=}& {\mathbf{c}}_{\mathbf{1}}\mathbf{+}\mathbf{3}{\mathbf{c}}_{\mathbf{2}}\mathbf{-}\mathbf{3}\\ {\mathbf{c}}_{\mathbf{1}}\mathbf{+}\mathbf{3}{\mathbf{c}}_{\mathbf{2}}& \mathbf{=}& \mathbf{-}\mathbf{5} ......\left(\mathbf{6}\right)\end{array}$

Substitute the value of $c_1$ in the equation (6),

$\begin{array}{rcl}{\mathbf{c}}_{\mathbf{1}}\mathbf{+}\mathbf{3}{\mathbf{c}}_{\mathbf{2}}& \mathbf{=}& \mathbf{-}\mathbf{5}\\ \mathbf{2}\mathbf{+}\mathbf{3}{\mathbf{c}}_{\mathbf{2}}& \mathbf{=}& \mathbf{-}\mathbf{5}\\ \mathbf{3}{\mathbf{c}}_{\mathbf{2}}& \mathbf{=}& \mathbf{-}\mathbf{7}\\ {\mathbf{c}}_{\mathbf{2}}& \mathbf{=}& \frac{\mathbf{-}\mathbf{7}}{\mathbf{3}}\end{array}$

Substitute the value of $c_1$and $c_2$ in the equation (5),

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{t}}\mathbf{cos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right)\mathbf{-}\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right) \\ \mathbf{y}\mathbf{=}\mathbf{2}{\mathbf{e}}^{\mathbf{t}}\mathbf{cos}\left(\mathbf{3}\mathbf{t}\right)\mathbf{-}\frac{\mathbf{7}}{\mathbf{3}}{\mathbf{e}}^{\mathbf{t}}\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right)\mathbf{-}\mathbf{sin}\left(\mathbf{3}\mathbf{t}\right) \end{gathered}$$