Q11RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the given differential equation. ${\mathbf{t}}^{\mathbf{2}}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{+}\mathbf{5}\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\mathbf{0}\mathbf{,} \mathbf{t}\mathbf{>}\mathbf{0}$

Step-by-Step Solution

Answer

$\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{t}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{cos}\left(\frac{\sqrt{\mathbf{19}}}{\mathbf{2}}\mathbf{lnt}\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{t}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{sin}\left(\frac{\sqrt{\mathbf{19}}}{\mathbf{2}}\mathbf{lnt}\right)$

Step 1: Firstly, write the auxiliary equation of the given differential equation

The differential equation is,

 ${\mathbf{t}}^{\mathbf{2}}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{t}\right)\mathbf{+}\mathbf{5}\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\mathbf{0} ......\left(\mathbf{1}\right)$

 The auxiliary equation for the above equation,

$$\begin{gathered} \mathbf{x}\mathbf{=}{\mathbf{t}}^{\mathbf{m}} \\ \mathbf{x}\mathbf{\text{'}}\mathbf{=}{\mathbf{mt}}^{\mathbf{m}\mathbf{-}\mathbf{1}} \\ \mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{m}\left(\mathbf{m}\mathbf{-}\mathbf{1}\right){\mathbf{t}}^{\mathbf{m}\mathbf{-}\mathbf{2}} \end{gathered}$$ 

 From equation (1),

$$\begin{gathered} {\mathbf{t}}^{\mathbf{2}}\mathbf{m}\left(\mathbf{m}\mathbf{-}\mathbf{1}\right){\mathbf{t}}^{\mathbf{m}\mathbf{-}\mathbf{2}}\mathbf{+}\mathbf{5}{\mathbf{t}}^{\mathbf{m}}\mathbf{=}\mathbf{0} \\ \mathbf{m}\left(\mathbf{m}\mathbf{-}\mathbf{1}\right){\mathbf{t}}^{\mathbf{m}}\mathbf{+}\mathbf{5}{\mathbf{t}}^{\mathbf{m}}\mathbf{=}\mathbf{0} \\ \mathbf{m}\left(\mathbf{m}\mathbf{-}\mathbf{1}\right)\mathbf{+}\mathbf{5}\mathbf{=}\mathbf{0} \\ {\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{m}\mathbf{+}\mathbf{5}\mathbf{=}\mathbf{0} \end{gathered}$$ 

Step 2: Find roots of the auxiliary equation

Solve the auxiliary equation,

 $$\begin{gathered} {\mathbf{m}}^{\mathbf{2}}\mathbf{-}\mathbf{m}\mathbf{+}\mathbf{5}\mathbf{=}\mathbf{0} \\ \mathbf{m}\mathbf{=}\frac{\mathbf{-}\left(\mathbf{-}\mathbf{1}\right)\mathbf{\pm }\sqrt{\mathbf{1}\mathbf{-}\mathbf{20}}}{\mathbf{2}} \\ \mathbf{m}\mathbf{=}\frac{\mathbf{1}\mathbf{\pm }\sqrt{\mathbf{-}\mathbf{19}}}{\mathbf{2}} \\ \mathbf{m}\mathbf{=}\frac{\mathbf{1}\mathbf{\pm }\mathbf{i}\sqrt{\mathbf{19}}}{\mathbf{2}} \end{gathered}$$

The roots of the auxiliary equation are, 

${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}\mathbf{+}\mathbf{i}\sqrt{\mathbf{19}}}{\mathbf{2}}\mathbf{,} {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}\mathbf{-}\mathbf{i}\sqrt{\mathbf{19}}}{\mathbf{2}}$

$$\begin{gathered} \mathbf{x}\mathbf{=}{\mathbf{t}}^{\mathbf{m}} \\ \mathbf{x}\mathbf{=}{\mathbf{e}}^{\mathbf{m}\left(\mathbf{lnt}\right)} \end{gathered}$$

 The general solution of the given equation is:

 $\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{t}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{cos}\left(\frac{\sqrt{\mathbf{19}}}{\mathbf{2}}\mathbf{lnt}\right)\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{t}}^{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{sin}\left(\frac{\sqrt{\mathbf{19}}}{\mathbf{2}}\mathbf{lnt}\right)$