Q2E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: Find a geneQuestion: Find a general solution to the given differential equation.ral solution to the given differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

Answer

 

The general solution of the given equation is $\mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{3}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{te}}^{\mathbf{-}\mathbf{3}\mathbf{t}}.$

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

The differential equation is $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{0}.$$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation for the above equation ${\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathbf{m}\mathbf{+}\mathbf{9}\mathbf{=}\mathbf{0}.$

 

Step 2: Find the roots of the auxiliary equation.

Solve the auxiliary equation,

 $$\begin{gathered} {\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathbf{m}\mathbf{+}\mathbf{9}\mathbf{=}\mathbf{0} \\ {\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{m}\mathbf{+}\mathbf{3}\mathbf{m}\mathbf{+}\mathbf{9}\mathbf{=}\mathbf{0} \\ \mathbf{m}\left(\mathbf{m}\mathbf{+}\mathbf{3}\right)\mathbf{+}\mathbf{3}\left(\mathbf{m}\mathbf{+}\mathbf{3}\right)\mathbf{=}\mathbf{0} \\ \left(\mathbf{m}\mathbf{+}\mathbf{3}\right)\left(\mathbf{m}\mathbf{+}\mathbf{3}\right)\mathbf{=}\mathbf{0} \end{gathered}$$

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

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

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