Q15E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution for $\mathbf{t}\mathbf{<}\mathbf{0}$.

$\mathbf{y}\text{'}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\frac{\mathbf{1}}{\mathbf{t}}\mathbf{y}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\frac{\mathbf{5}}{{\mathbf{t}}^{\mathbf{2}}}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution of the given equation $\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)-\frac{1}{\mathrm{t}}\mathrm{y}\text{'}\left(\mathrm{t}\right)+\frac{5}{{\mathrm{t}}^2}\mathrm{y}\left(\mathrm{t}\right)=0$ is:

${\mathrm{y}}_{\mathrm{G}}={\mathrm{C}}_1\mathrm{tcos}\left(2\ln \right(-\mathrm{t}\left)\right)+{\mathrm{C}}_2\mathrm{tsin}\left(2\ln \right(-\mathrm{t}\left)\right)$ for $\mathrm{t}<0$.

Step 1: Using Cauchy-Euler equation

Given $\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)-\frac{1}{\mathrm{t}}\mathrm{y}\text{'}\left(\mathrm{t}\right)+\frac{5}{{\mathrm{t}}^2}\mathrm{y}\left(\mathrm{t}\right)=0$

 

Here one has $\mathrm{t}<0$ so we expect to have complex roots, with a $\ln (-\mathrm{t})$  to keep $\ln \left(\mathrm{t}\right)$ defined.

 

One has a hidden Cauchy-Euler; we will multiply by ${\mathrm{t}}^2$ to reveal it.

${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}-\mathrm{ty}\text{'}+5=0$ 

 

Hence, one chooses a solution of the form

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

 

Step 2: Substitute the values

 By Substitution, one sees the function cancels out and we are left with

$\begin{array}{rcl}{\mathrm{r}}^2-\mathrm{r}-\mathrm{r}+5& =& 0\\ {\mathrm{r}}^2-2\mathrm{r}+5& =& 0\end{array}$ 

 

One will complete the square

$\begin{array}{rcl}{\mathrm{r}}^2-2\mathrm{r}+1+4& =& 0\\ {\left({\mathrm{r}}^2-1\right)}^2& =& -4\\ \mathrm{r}-1& =& \pm 2\mathrm{i}\\ \mathrm{r}& =& 1\pm 2\mathrm{i}\end{array}$ 

 

Therefore, ${\mathrm{y}}_{\mathrm{G}}={\mathrm{C}}_1\mathrm{tcos}\left(2\ln \right(-\mathrm{t}\left)\right)+{\mathrm{C}}_2\mathrm{tsin}\left(2\ln \right(-\mathrm{t}\left)\right)$