Q1RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the given differential equation $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{8}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{0}$.

Step-by-Step Solution

Answer

The general solution to the given differential equation is$\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{-9t}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^t.$

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

The given differential equation is$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{8}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation for the above equation${\mathbf{m}}^2\mathbf{+}\mathbf{8}\mathbf{m}\mathbf{-}\mathbf{9}\mathbf{=}\mathbf{0}.$

Step 2: Find the roots of the auxiliary equation

Solve the auxiliary equation, 

$\begin{array}{c}{\mathbf{m}}^2\mathbf{+}\mathbf{8}\mathbf{m}\mathbf{-}\mathbf{9}\mathbf{=}\mathbf{0}\\ {\mathbf{m}}^2\mathbf{+}\mathbf{9}\mathbf{m}\mathbf{-}\mathbf{m}\mathbf{-}\mathbf{9}\mathbf{=}\mathbf{0}\\ \mathbf{m}\left(m+9\right)\mathbf{-}\mathbf{1}\left(m+9\right)\mathbf{=}\mathbf{0}\\ \left(m+9\right)\left(m-1\right)\mathbf{=}\mathbf{0}\end{array}$

 

The roots of the auxiliary equation are ${\mathbf{m}}_1\mathbf{=}\mathbf{-}\mathbf{9}\&{\mathbf{m}}_2\mathbf{=}\mathbf{1}.$

 

The general solution of the given equation is$\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{-9t}\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^t.$