Q16RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: Find a general solution to the given differential equation.$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution to the given differential equation is:

$\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{-t}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{-t}\mathbf{cos}\left(\sqrt{2}\mathbf{t}\right)\mathbf{+}{\mathbf{c}}_3{\mathbf{e}}^{-t}\mathbf{sin}\left(\sqrt{2}\mathbf{t}\right)$

Step 1: Complex conjugate roots.

If the auxiliary equation has complex conjugate roots, then the general solution is given as:

 

 $y\left(t\right)=c_1{e}^{\alpha t}cos\beta t+c_2{e}^{\alpha t}sin\beta t$

Step 2: Write the auxiliary equation of the given differential equation

The differential equation is $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation for the above equation ${\mathbf{m}}^3\mathbf{+}\mathbf{3}{\mathbf{m}}^2\mathbf{+}\mathbf{5}\mathbf{m}\mathbf{+}\mathbf{3}\mathbf{=}\mathbf{0}.$

 

Step 3: Now find the roots of the auxiliary equation

Solve the auxiliary equation,

 $\begin{array}{c}{\mathbf{m}}^3\mathbf{+}\mathbf{3}{\mathbf{m}}^2\mathbf{+}\mathbf{5}\mathbf{m}\mathbf{+}\mathbf{3}\mathbf{=}\mathbf{0}\\ {\mathbf{m}}^3\mathbf{+}{\mathbf{m}}^2\mathbf{+}\mathbf{2}{\mathbf{m}}^2\mathbf{+}\mathbf{2}\mathbf{m}\mathbf{+}\mathbf{3}\mathbf{m}\mathbf{+}\mathbf{3}\mathbf{=}\mathbf{0}\\ \left(m+1\right)\left({\mathbf{m}}^2\mathbf{+}\mathbf{2}\mathbf{m}\mathbf{+}\mathbf{3}\right)\mathbf{=}\mathbf{0}\\ \mathbf{m}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{m}\mathbf{=}\frac{\mathbf{-}\mathbf{2}\mathbf{\pm }\sqrt{4-12}}{2}\\ \mathbf{m}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{m}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{\pm }\mathbf{i}\sqrt{2}\end{array}$

 

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

 

The general solution of the given equation is,

$\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{-t}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{-t}\mathbf{cos}\left(\sqrt{2}\mathbf{t}\right)\mathbf{+}{\mathbf{c}}_3{\mathbf{e}}^{-t}\mathbf{sin}\left(\sqrt{2}\mathbf{t}\right)$