Q17RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: Find a general solution to the given differential equation.$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{11}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

he general solution to the given differential equation is:

$\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^t\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{\frac{-1}{2}\mathbf{t}}\mathbf{cos}\left(\frac{\sqrt{43}}{2}\mathbf{t}\right)\mathbf{+}{\mathbf{c}}_3{\mathbf{e}}^{\frac{-1}{2}\mathbf{t}}\mathbf{sin}\left(\frac{\sqrt{43}}{2}\mathbf{t}\right)$

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 $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{10}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{11}\mathbf{y}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation for the above equation ${\mathbf{m}}^3\mathbf{+}\mathbf{10}\mathbf{m}\mathbf{-}\mathbf{11}\mathbf{=}\mathbf{0}.$

Step 3: Now find the roots of the auxiliary equation

Solve the auxiliary equation,

 $\begin{array}{c}{\mathbf{m}}^3\mathbf{+}\mathbf{10}\mathbf{m}\mathbf{-}\mathbf{11}\mathbf{=}\mathbf{0}\\ {\mathbf{m}}^3\mathbf{-}{\mathbf{m}}^2\mathbf{+}{\mathbf{m}}^2\mathbf{-}\mathbf{m}\mathbf{+}\mathbf{11}\mathbf{m}\mathbf{-}\mathbf{11}\mathbf{=}\mathbf{0}\\ \left(m-1\right)\left({\mathbf{m}}^2\mathbf{+}\mathbf{m}\mathbf{+}\mathbf{11}\right)\mathbf{=}\mathbf{0}\\ \mathbf{m}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{m}\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{\pm }\sqrt{1-44}}{2}\\ \mathbf{m}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{m}\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{\pm }\mathbf{i}\sqrt{43}}{2}\end{array}$

The roots of the auxiliary equation are, ${\mathbf{m}}_1\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{m}}_2\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{+}\mathbf{i}\sqrt{43}}{2}\mathbf{,}{\mathbf{m}}_3\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{-}\mathbf{i}\sqrt{43}}{2}.$

 

The general solution of the given equation is, $\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^t\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{\frac{-1}{2}\mathbf{t}}\mathbf{cos}\left(\frac{\sqrt{43}}{2}\mathbf{t}\right)\mathbf{+}{\mathbf{c}}_3{\mathbf{e}}^{\frac{-1}{2}\mathbf{t}}\mathbf{sin}\left(\frac{\sqrt{43}}{2}\mathbf{t}\right)$