Q15RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: Find a general solution to the given differential equation. $\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution to the given differential equation is:

$\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{-t}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{-2t}\mathbf{+}{\mathbf{c}}_3{\mathbf{e}}^{\frac{-1}{3}\mathbf{t}}$

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

The differential equation is $\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation for the above equation  $\mathbf{3}{\mathbf{m}}^3\mathbf{+}\mathbf{10}{\mathbf{m}}^2\mathbf{+}\mathbf{9}\mathbf{m}\mathbf{+}\mathbf{2}\mathbf{=}\mathbf{0}.$

Step 2: Find the roots of the auxiliary equation.

 

Solve the auxiliary equation,

 $\begin{array}{c}\mathbf{3}{\mathbf{m}}^3\mathbf{+}\mathbf{10}{\mathbf{m}}^2\mathbf{+}\mathbf{9}\mathbf{m}\mathbf{+}\mathbf{2}\mathbf{=}\mathbf{0}\\ \mathbf{3}{\mathbf{m}}^3\mathbf{+}{\mathbf{m}}^2\mathbf{+}\mathbf{9}{\mathbf{m}}^2\mathbf{+}\mathbf{3}\mathbf{m}\mathbf{+}\mathbf{6}\mathbf{m}\mathbf{+}\mathbf{2}\mathbf{=}\mathbf{0}\\ {\mathbf{m}}^2\left(3m+1\right)\mathbf{+}\mathbf{3}\mathbf{m}\left(3m+1\right)\mathbf{+}\mathbf{2}\left(3m+1\right)\mathbf{=}\mathbf{0}\\ \left(3m+1\right)\left({\mathbf{m}}^2\mathbf{+}\mathbf{3}\mathbf{m}\mathbf{+}\mathbf{2}\right)\mathbf{=}\mathbf{0}\\ \left(3m+1\right)\left({\mathbf{m}}^2\mathbf{+}\mathbf{2}\mathbf{m}\mathbf{+}\mathbf{m}\mathbf{+}\mathbf{2}\right)\mathbf{=}\mathbf{0}\\ \left(3m+1\right)\left(\mathbf{m}\left(m+2\right)\mathbf{+}\mathbf{1}\left(m+2\right)\right)\mathbf{=}\mathbf{0}\\ \left(m+1\right)\left(m+2\right)\left(3m+1\right)\mathbf{=}\mathbf{0}\end{array}$

 

The roots of the auxiliary equation are ${\mathbf{m}}_1\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}{\mathbf{m}}_2\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{,}{\mathbf{m}}_3\mathbf{=}\mathbf{-}\frac{1}{3}.$

Thus, the general solution of the given equation is $\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{-t}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{-2t}\mathbf{+}{\mathbf{c}}_3{\mathbf{e}}^{\frac{-1}{3}\mathbf{t}}.$