Q14RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: Find a general solution to the given differential equation. $\mathbf{\nu }\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{v}\mathbf{\text{'}}\mathbf{+}\mathbf{7}\mathbf{v}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution to the given differential equation is:

 $\mathbf{v}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{2t}\mathbf{cos}\left(\sqrt{3}\mathbf{t}\right)\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{2t}\mathbf{sin}\left(\sqrt{3}\mathbf{t}\right)$

Step 1: Complex conjugate roots.

If the auxiliary equation has complex conjugate roots, $\alpha \pm i\beta$ 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: Firstly, write the auxiliary equation of the given differential equation

The differential equation is $\mathbf{\nu }\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{v}\mathbf{\text{'}}\mathbf{+}\mathbf{7}\mathbf{v}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation for the above equation,${\mathbf{m}}^2\mathbf{-}\mathbf{4}\mathbf{m}\mathbf{+}\mathbf{7}\mathbf{=}\mathbf{0}.$

Step 3: Now find the roots of the auxiliary equation

Solve the auxiliary equation,

 $\begin{array}{c}{\mathbf{m}}^2\mathbf{-}\mathbf{4}\mathbf{m}\mathbf{+}\mathbf{7}\mathbf{=}\mathbf{0}\\ \mathbf{m}\mathbf{=}\frac{\mathbf{4}\mathbf{\pm }\sqrt{16-28}}{2}\\ \mathbf{m}\mathbf{=}\frac{\mathbf{4}\mathbf{\pm }\sqrt{-12}}{2}\\ \mathbf{m}\mathbf{=}\frac{\mathbf{4}\mathbf{\pm }\mathbf{i}\sqrt{12}}{2}\\ \mathbf{m}\mathbf{=}\mathbf{2}\mathbf{\pm }\mathbf{i}\sqrt{3}\end{array}$

 

The roots of the auxiliary equation are, ${\mathbf{m}}_1\mathbf{=}\mathbf{2}\mathbf{+}\mathbf{i}\sqrt{3}\mathbf{,}\&{\mathbf{m}}_2\mathbf{=}\mathbf{2}\mathbf{-}\mathbf{i}\sqrt{3}.$

 

The general solution of the given equation is $\mathbf{v}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{2t}\mathbf{cos}\left(\sqrt{3}\mathbf{t}\right)\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{2t}\mathbf{sin}\left(\sqrt{3}\mathbf{t}\right).$