Q12E - 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.

$\mathbf{1}2.\frac{{\mathbf{d}}^2\mathbf{w}}{{\mathbf{dt}}^2}\mathbf{+}\frac{6}{t}\frac{\mathrm{dw}}{\mathrm{dt}}\mathbf{+}\frac{4}{{\mathbf{t}}^2}\mathbf{w}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general equation is $\mathrm{w}={\mathrm{c}}_1{\mathrm{t}}^{-4}+{\mathrm{c}}_2{\mathrm{t}}^{-1}$. 

Step 1: Find the auxiliary equation.

Given differential equation $\frac{{\mathrm{d}}^2\mathrm{w}}{{\mathrm{dt}}^2}+\frac{6}{\mathrm{t}}\frac{\mathrm{dw}}{\mathrm{dt}}+\frac{4}{{\mathrm{t}}^2}\mathrm{w}=0\dots \left(1\right)$

Assume $\mathrm{w}={\mathrm{t}}^{\mathrm{r}}$ then we have;

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

 

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

 

$\begin{array}{c}\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2}+\frac{6}{\mathrm{t}}{\mathrm{rt}}^{\mathrm{r}-1}+\frac{4}{{\mathrm{t}}^2}{\mathrm{t}}^{\mathrm{r}}=0\\ {\mathrm{t}}^2\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2}+6{\mathrm{trt}}^{\mathrm{r}-1}+4{\mathrm{t}}^{\mathrm{r}}=0\\ \left(\mathrm{r}\right(\mathrm{r}-1)+6\mathrm{r}+4){\mathrm{t}}^{\mathrm{r}}=0\\ {\mathrm{r}}^2+5\mathrm{r}+4=0\end{array}$

Step 2: Determine the general equation.

The roots of the equation are:

 

$\begin{array}{l}{\mathrm{r}}^2+4\mathrm{r}+\mathrm{r}+4=0\\ (\mathrm{r}+4)(\mathrm{r}+1)=0\\ \mathrm{r}=-4,-1\end{array}$

 

The general equation is $\mathbf{w}\mathbf{=}{\mathbf{c}}_1{\mathbf{t}}^{-4}\mathbf{+}{\mathbf{c}}_2{\mathbf{t}}^{-1}$.