Q4RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the given differential equation.$\mathbf{9}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{30}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{25}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution to the given differential equation is $\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{\frac{5}{3}\mathbf{t}}\mathbf{+}{\mathbf{c}}_2{\mathbf{te}}^{\frac{5}{3}\mathbf{t}}.$

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

The given differential equation is$\mathbf{9}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{30}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{25}\mathbf{y}\mathbf{=}\mathbf{0}.$

 

The auxiliary equation is$\mathbf{9}{\mathbf{m}}^2\mathbf{-}\mathbf{30}\mathbf{m}\mathbf{+}\mathbf{25}\mathbf{=}\mathbf{0}.$

Step 2: Find the roots of the auxiliary equation

Solve the auxiliary equation,

 $\begin{array}{c}\mathbf{9}{\mathbf{m}}^2\mathbf{-}\mathbf{30}\mathbf{m}\mathbf{+}\mathbf{25}\mathbf{=}\mathbf{0}\\ {\left(3m-5\right)}^2\mathbf{=}\mathbf{0}\\ \mathbf{m}\mathbf{=}\frac{5}{3}\mathbf{,}\frac{5}{3}\end{array}$

The roots of the auxiliary equation are ${\mathbf{m}}_1\mathbf{=}\frac{5}{3}\mathbf{,}\&{\mathbf{m}}_2\mathbf{=}\frac{5}{3}.$

 

The general solution of the given equation is$\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^{\frac{5}{3}\mathbf{t}}\mathbf{+}{\mathbf{c}}_2{\mathbf{te}}^{\frac{5}{3}\mathbf{t}}.$