Q32RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find the solution to the given initial value problem.

$\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{11}}{\mathbf{2}}$

Step-by-Step Solution

Answer

The general solution to the given initial value problem is $\mathrm{y}={\mathrm{e}}^{\frac{\mathrm{t}}{2}}\cos \left(\mathrm{t}\right)-6{\mathrm{e}}^{\frac{\mathrm{t}}{2}}\sin \left(\mathrm{t}\right)$

Write the auxiliary equation of the given differential equation

The given differential equation is,

 $\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{0} ......\left(\mathbf{1}\right)$

The auxiliary equation for the above equation,

$\begin{array}{rcl}\mathbf{4}{\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{m}\mathbf{+}\mathbf{5}& \mathbf{=}& \mathbf{0}\\ \mathbf{m}& \mathbf{=}& \frac{\mathbf{4}\mathbf{\pm }\sqrt{\mathbf{16}\mathbf{-}\mathbf{80}}}{\mathbf{2}\left(4\right)}\\ \mathbf{m}& \mathbf{=}& \frac{\mathbf{4}\mathbf{\pm }\sqrt{\mathbf{-}\mathbf{64}}}{\mathbf{8}}\\ \mathbf{m}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{2}}\mathbf{\pm }\mathbf{i}\\ & & \end{array}$

Now find the general solution of the given equation

The root of an auxiliary equation is, 

${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{+}\mathbf{i}\mathbf{,} \& {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{-}\mathbf{i}$

The general solution of the given equation is,

$\mathbf{y}\mathbf{=}{\mathbf{Ae}}^{\frac{\mathbf{t}}{\mathbf{2}}}\mathbf{cos}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{Be}}^{\frac{\mathbf{t}}{\mathbf{2}}}\mathbf{sin}\left(\mathbf{t}\right) ......\left(\mathbf{2}\right)$

Use the given initial condition,

Given the initial condition,

$\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{11}}{\mathbf{2}}$

Substitute the value of $y=1$ and $t=0$ in the equation (2),

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{Ae}}^{\frac{\mathbf{t}}{\mathbf{2}}}\mathbf{cos}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{Be}}^{\frac{\mathbf{t}}{\mathbf{2}}}\mathbf{sin}\left(\mathbf{t}\right) \\ \mathbf{1}\mathbf{=}{\mathbf{Ae}}^{\frac{\mathbf{0}}{\mathbf{2}}}\mathbf{cos}\left(\mathbf{0}\right)\mathbf{+}{\mathbf{Be}}^{\frac{\mathbf{0}}{\mathbf{2}}}\mathbf{sin}\left(\mathbf{0}\right) \\ \mathbf{A}\mathbf{=}\mathbf{1} \end{gathered}$$

Now find the derivative of the equation (2),

$\mathbf{y}\mathbf{\text{'}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Ae}}^{\frac{\mathbf{t}}{\mathbf{2}}}\mathbf{cos}\left(\mathbf{t}\right)\mathbf{-}{\mathbf{Ae}}^{\frac{\mathbf{t}}{\mathbf{2}}}\mathbf{sin}\left(\mathbf{t}\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Be}}^{\frac{\mathbf{t}}{\mathbf{2}}}\mathbf{sin}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{Be}}^{\frac{\mathbf{t}}{\mathbf{2}}}\mathbf{cos}\left(\mathbf{t}\right)$

Substitute the value of $\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\frac{\mathbf{11}}{\mathbf{2}}$and $t=0$ in the above equation,

$\begin{array}{rcl}\mathbf{-}\frac{\mathbf{11}}{\mathbf{2}}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Ae}}^{\frac{\mathbf{0}}{\mathbf{2}}}\mathbf{cos}\left(\mathbf{0}\right)\mathbf{-}{\mathbf{Ae}}^{\frac{\mathbf{0}}{\mathbf{2}}}\mathbf{sin}\left(\mathbf{0}\right)\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Be}}^{\frac{\mathbf{0}}{\mathbf{2}}}\mathbf{sin}\left(\mathbf{0}\right)\mathbf{+}{\mathbf{Be}}^{\frac{\mathbf{0}}{\mathbf{2}}}\mathbf{cos}\left(\mathbf{0}\right)\\ \frac{\mathbf{1}}{\mathbf{2}}\mathbf{A}\mathbf{+}\mathbf{B}& \mathbf{=}& \mathbf{-}\frac{\mathbf{11}}{\mathbf{2}} ......\left(\mathbf{3}\right)\\ & & \end{array}$

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

$\begin{array}{rcl}\frac{\mathbf{1}}{\mathbf{2}}\left(\mathbf{1}\right)\mathbf{+}\mathbf{B}& \mathbf{=}& \mathbf{-}\frac{\mathbf{11}}{\mathbf{2}}\\ \mathbf{B}& \mathbf{=}& \mathbf{-}\frac{\mathbf{11}}{\mathbf{2}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\\ \mathbf{B}& \mathbf{=}& \mathbf{-}\mathbf{6}\\ & & \end{array}$

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

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{Ae}}^{\frac{\mathbf{t}}{\mathbf{2}}}\mathbf{cos}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{Be}}^{\frac{\mathbf{t}}{\mathbf{2}}}\mathbf{sin}\left(\mathbf{t}\right) \\ \mathbf{y}\mathbf{=}{\mathbf{e}}^{\frac{\mathbf{t}}{\mathbf{2}}}\mathbf{cos}\left(\mathbf{t}\right)\mathbf{-}\mathbf{6}{\mathbf{e}}^{\frac{\mathbf{t}}{\mathbf{2}}}\mathbf{sin}\left(\mathbf{t}\right) \end{gathered}$$

Thus, the general solution to the given initial value problem is $\mathbf{y}\mathbf{=}{\mathbf{e}}^{\frac{\mathbf{t}}{\mathbf{2}}}\mathbf{cos}\left(\mathbf{t}\right)\mathbf{-}\mathbf{6}{\mathbf{e}}^{\frac{\mathbf{t}}{\mathbf{2}}}\mathbf{sin}\left(\mathbf{t}\right)$.