Q17E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution for $\mathbf{t}\mathbf{<}\mathbf{0}$ . ${\mathbf{t}}^2\mathbf{y}\text{'}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{9}\mathbf{ty}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{17}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The general solution of the given equation ${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)+9\mathrm{ty}\text{'}\left(\mathrm{t}\right)+17\mathrm{y}\left(\mathrm{t}\right)=0$ is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{t}}^{-4}\cos \left(\ln \right(-\mathrm{t}\left)\right)+{\mathrm{c}}_2{\mathrm{t}}^{-4}\sin \left(\ln \right(-\mathrm{t}\left)\right)$ for $\mathrm{t}<0$.

Step 1: Find the associated characteristic equation.

The associated characteristic equation to ${\mathrm{at}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)+\mathrm{bty}\text{'}\left(\mathrm{t}\right)+\mathrm{cy}\left(\mathrm{t}\right)=0$ be equation ${\mathrm{r}}^2+(\mathrm{b}-\mathrm{a})\mathrm{r}+\mathrm{c}=0.$

 

The coefficient of ${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)$ is $\mathrm{a}=1$, the one multiplying $\mathrm{ty}\text{'}\left(\mathrm{t}\right)$ is and the one multiplying $\mathrm{y}\left(\mathrm{t}\right)$ is $\mathrm{c}=17$, and therefore the associated characteristic equation to the given differential equation is:

  $\begin{array}{l}{\mathrm{r}}^2+(9-1)\mathrm{r}+17=0\\ {\mathrm{r}}^2+8\mathrm{r}+17=0\\ \mathrm{r}=\frac{-8\pm \sqrt{64-68}}{2}\\ \mathrm{r}=-4\pm \mathrm{i}\end{array}$

Step 2: Finding the general solution.

When the roots to the characteristic equation are complex,$\mathrm{r}=\mathrm{\alpha }\pm \mathrm{\beta i}$ and if $\mathrm{t}>0$then the linearly independent solutions are ${\mathrm{y}}_1\left(\mathrm{t}\right)={\mathrm{t}}^{\mathrm{\alpha }}\cos \left(\mathrm{\beta lnt}\right)$ and ${\mathrm{y}}_2\left(\mathrm{t}\right)={\mathrm{t}}^{\mathrm{\alpha }}\sin \left(\mathrm{\beta lnt}\right).$ But if we have that $\mathrm{t}<0$, then the linearly independent solutions are given as:

 

${\mathrm{y}}_1\left(\mathrm{t}\right)=(-\mathrm{t}{)}^{\mathrm{\alpha }}\cos \left(\mathrm{\beta ln}\right(-\mathrm{t})\mathrm{and}{\mathrm{y}}_2\left(\mathrm{t}\right)=(-\mathrm{t}{)}^{\mathrm{\alpha }}\sin \left(\mathrm{\beta ln}\right(-\mathrm{t}\left)\right)$

 

Hence, the general solution to the given differential equation is:

$\begin{array}{c}\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{(-\mathrm{t})}^{-4}\cos \left(\ln \right(-\mathrm{t}\left)\right)+{\mathrm{c}}_2{(-\mathrm{t})}^{-4}\sin \left(\ln \right(-\mathrm{t}\left)\right)\\ \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{t}}^{-4}\cos \left(\ln \right(-\mathrm{t}\left)\right)+{\mathrm{c}}_2{\mathrm{t}}^{-4}\sin \left(\ln \right(-\mathrm{t}\left)\right)\end{array}$