Q6RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the given differential equation. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{8}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{14}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

Thus, the general solution of the given differential equation is:

$\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{\left(\mathbf{-}\mathbf{4}\mathbf{-}\sqrt{30}\right)\mathbf{t}}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{\left(\mathbf{-}\mathbf{4}\mathbf{+}\sqrt{30}\right)\mathbf{t}}$

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

The differential equation is$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{8}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{14}\mathbf{y}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation for the above equation is${\mathbf{m}}^2\mathbf{+}\mathbf{8}\mathbf{m}\mathbf{-}\mathbf{14}\mathbf{=}\mathbf{0}.$

 

Step 2: Find the roots of the auxiliary equation.

Solve the auxiliary equation,

 $\begin{array}{c}{\mathbf{m}}^2\mathbf{+}\mathbf{8}\mathbf{m}\mathbf{-}\mathbf{14}\mathbf{=}\mathbf{0}\\ \mathbf{m}\mathbf{=}\frac{\mathbf{-}\mathbf{8}\mathbf{\pm }\sqrt{\mathbf{64}\mathbf{-}\left(-56\right)}}{2}\\ \mathbf{m}\mathbf{=}\frac{\mathbf{-}\mathbf{8}\mathbf{\pm }\sqrt{120}}{2}\\ \mathbf{m}\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{\pm }\sqrt{30}\end{array}$

 

The roots of the auxiliary equation are${\mathbf{m}}_1\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{+}\sqrt{30}\mathbf{,}\&{\mathbf{m}}_2\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{-}\sqrt{30}.$

 

Thus, the general solution of the given equation is$\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{\left(\mathbf{-}\mathbf{4}\mathbf{-}\sqrt{30}\right)\mathbf{t}}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{\left(\mathbf{-}\mathbf{4}\mathbf{+}\sqrt{30}\right)\mathbf{t}}.$