Q13E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 9 through 14, find a general solution to the given Cauchy–Euler equation for t>0.$\mathbf{13}\mathbf{.}\mathbf{9}{\mathbf{t}}^2\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{15}\mathbf{ty}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general equation is  $\mathrm{y}={\mathrm{c}}_1{\mathrm{t}}^{-3}+{\mathrm{c}}_2{\mathrm{t}}^{-3}\ln \left(\mathrm{t}\right)$.

Step 1: Find auxiliary equation.

The given differential equation is $9{\mathrm{t}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)+15\mathrm{ty}\text{'}\left(\mathrm{t}\right)+\mathrm{y}\left(\mathrm{t}\right)=0\dots \left(1\right)$

 

Assume  $\mathrm{y}={\mathrm{t}}^{\mathrm{r}}$ then we have:

 $\begin{array}{l} \mathrm{y}\text{'}={\mathrm{rt}}^{\mathrm{r}-1}\\ \mathrm{y}\text{'}\text{'}=\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2}\end{array}$

 

Substitute all values in equation (1), we get:

$\begin{array}{c}9{\mathrm{t}}^2\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2}+15{\mathrm{trt}}^{\mathrm{r}-1}+{\mathrm{t}}^{\mathrm{r}}=0\\ 9{\mathrm{t}}^2\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2}+15{\mathrm{trt}}^{\mathrm{r}-1}+{\mathrm{t}}^{\mathrm{r}}=0\\ \left(9\mathrm{r}\right(\mathrm{r}-1)+15\mathrm{r}+1){\mathrm{t}}^{\mathrm{r}}=0\\ 9{\mathrm{r}}^2+6\mathrm{r}+1=0\end{array}$ 

Step 2: Determine general equation.

The roots of the equation are:

 $\begin{array}{l}{\mathrm{r}}^2+3\mathrm{r}+3\mathrm{r}+1=0\\ (\mathrm{r}+3)(\mathrm{r}+3)=0\\ \mathrm{r}=-3,-3\end{array}$

 

The general equation is $\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{t}}^{-3}\mathbf{+}{\mathbf{c}}_2{\mathbf{t}}^{-3}\mathbf{ln}\left(\mathbf{t}\right)$.