Q14E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the given Cauchy-Euler equation for $\mathbf{t}\mathbf{>}\mathbf{0}$

${\mathbf{t}}^2\mathbf{y}\text{'}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{3}\mathbf{ty}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$

 

Step-by-Step Solution

Answer

The general solution to the given equation  ${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)-3\mathrm{ty}\text{'}\left(\mathrm{t}\right)+4\mathrm{y}\left(\mathrm{t}\right)=0$ is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{t}}^2+{\mathrm{c}}_2{\mathrm{t}}^2\mathrm{lnt}$.

Step 1: Check the associated characteristic equation.

For a differential equation ${\mathrm{at}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)+\mathrm{bty}\text{'}\left(\mathrm{t}\right)+\mathrm{cy}\left(\mathrm{t}\right)=\mathrm{f}\left(\mathrm{t}\right)$, we have that the associated characteristic equation is ${\mathrm{ar}}^2+(\mathrm{b}-\mathrm{a})\mathrm{r}+\mathrm{c}=0$.

 

In our case, the coefficient of  ${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)$ is $\mathrm{a}=1$, the one multiplying is $\mathrm{ty}\text{'}\left(\mathrm{t}\right)$is $\mathrm{b}=-3$ and the one multiplying $\mathrm{y}\left(\mathrm{t}\right)$ is $\mathrm{c}=4$, therefore the associated characteristic equation is:

$\begin{array}{l}{\mathrm{r}}^2+(-3-1)\mathrm{r}+4=0\\ {\mathrm{r}}^2-4\mathrm{r}+4=0\\ {(\mathrm{r}-2)}^2=0\\ \mathrm{r}=2,2\end{array}$

 

Step 2: Check the differential equation.

When one has a double root to the associated characteristic equation, linearly independent solutions to the given differential equation are ${\mathrm{y}}_1\left(\mathrm{t}\right)={\mathrm{t}}^{\mathrm{r}}$  and ${\mathrm{y}}_2\left(\mathrm{t}\right)={\mathrm{t}}^{\mathrm{r}}\mathrm{lnt}$.

 

So, we have ${\mathrm{y}}_1\left(\mathrm{t}\right)={\mathrm{t}}^2$ , ${\mathrm{y}}_2\left(\mathrm{t}\right)={\mathrm{t}}^2\mathrm{lnt}$ and hence the general solution to the given equation is $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{c}}_1{\mathbf{t}}^2\mathbf{+}{\mathbf{c}}_2{\mathbf{t}}^2\mathbf{lnt}$.