Q9E - 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.

${\mathbf{t}}^2\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{7}\mathbf{ty}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{7}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general equation is $\mathrm{y}={\mathrm{c}}_1{\mathrm{t}}^{-7}+{\mathrm{c}}_2\mathrm{t}$.

Step 1: Find the auxiliary equation

Given differential equation  ${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)+7\mathrm{ty}\text{'}\left(\mathrm{t}\right)-7\mathrm{y}\left(\mathrm{t}\right)=0 \dots \left(1\right)$         

 

Assume 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), and we get;

 $\begin{array}{c}{\mathrm{t}}^2\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2}+7{\mathrm{trt}}^{\mathrm{r}-1}-7{\mathrm{t}}^{\mathrm{r}}=0\\ \left(\mathrm{r}\right(\mathrm{r}-1)+7\mathrm{r}-7){\mathrm{t}}^{\mathrm{r}}=0\\ {\mathrm{r}}^2+6\mathrm{r}-7=0\end{array}$

Step 2: Determine the general equation.

The roots of the equation are:

 $\begin{array}{l}(\mathrm{r}+7)(\mathrm{r}-1)=0\\ \mathrm{r}=-7,1\end{array}$

Thus, the general solution is $\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{t}}^{-7}\mathbf{+}{\mathbf{c}}_2\mathbf{t}.$