Q11E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: find a general solution to the given differential equation.

$\mathbf{4}\mathbf{w}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{20}\mathbf{w}\mathbf{\text{'}}\mathbf{+}\mathbf{25}\mathbf{w}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

Answer

 

The general solution of the given equation is $\mathbf{w}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{5}}{\mathbf{2}}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{te}}^{\mathbf{-}\frac{\mathbf{5}}{\mathbf{2}}\mathbf{t}}.$

Step 1: Write the auxiliary equation of the given differential equation.

The differential equation is $\mathbf{4}\mathbf{w}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{20}\mathbf{w}\mathbf{\text{'}}\mathbf{+}\mathbf{25}\mathbf{w}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation for the above equation $\mathbf{4}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{20}\mathbf{m}\mathbf{+}\mathbf{25}\mathbf{=}\mathbf{0}.$

 

Step 2: Now find the roots of the auxiliary equation.

Solve the auxiliary equation,

 

$$\begin{gathered} \mathbf{4}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{20}\mathbf{m}\mathbf{+}\mathbf{25}\mathbf{=}\mathbf{0} \\ {\left(\mathbf{2}\mathbf{m}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{2}\left(\mathbf{2}\right)\left(\mathbf{5}\right)\mathbf{m}\mathbf{+}{\mathbf{5}}^{\mathbf{2}}\mathbf{=}\mathbf{0} \\ {\left(\mathbf{2}\mathbf{m}\mathbf{+}\mathbf{5}\right)}^{\mathbf{2}}\mathbf{=}\mathbf{0} \end{gathered}$$

The roots of the auxiliary equation are  ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\frac{\mathbf{5}}{\mathbf{2}}\mathbf{,} \& {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\frac{\mathbf{5}}{\mathbf{2}}.$

Step 3: Write the general solution.

If the auxiliary equation has repeated real roots, then the general solution is given as;

 $\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{{\mathbf{m}}_{\mathbf{1}}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{te}}^{{\mathbf{m}}_{\mathbf{2}}\mathbf{t}}$

Thus, the general solution of the given equation is $\mathbf{w}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{5}}{\mathbf{2}}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{te}}^{\mathbf{-}\frac{\mathbf{5}}{\mathbf{2}}\mathbf{t}}.$