Q37E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

The auxiliary equations for the following differential equations have repeated complex roots. Adapt the "repeated root" procedure of Section $\mathbf{4}\mathbf{.}\mathbf{2}$  to find their general solutions:

$\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{y}\text{'}\text{'}\text{'}\text{'}\mathbf{+}\mathbf{2}\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}$

$\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{y}\text{'}\text{'}\text{'}\text{'}\mathbf{+}\mathbf{4}\mathbf{y}\text{'}\text{'}\text{'}\mathbf{+}\mathbf{12}\mathbf{y}\text{'}\text{'}\mathbf{+}\mathbf{16}\mathbf{y}\text{'}\mathbf{+}\mathbf{16}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

1. The general solution of the given differential equation is:$\mathrm{y}\left(\mathrm{t}\right)=({\mathrm{c}}_1+{\mathrm{c}}_2\mathrm{t})\mathrm{cost}+({\mathrm{c}}_3+{\mathrm{c}}_4\mathrm{t})\mathrm{sint}$ 
2. The general solution of the given differential equation is:$\mathrm{y}\left(\mathrm{t}\right)=({\mathrm{c}}_1+{\mathrm{c}}_2\mathrm{t}){\mathrm{e}}^{-\mathrm{t}}\cos \sqrt{3}\mathrm{t}+({\mathrm{c}}_3+{\mathrm{c}}_4\mathrm{t}){\mathrm{e}}^{-\mathrm{t}}\sin \sqrt{3}\mathrm{t}$

Step 1: Finding the roots and general solution

The auxiliary equation is: $r^4+2{\mathrm{r}}^2+1=0$

 

Now one will find the roots of this equation: 

${\mathrm{r}}^4+2{\mathrm{r}}^2+1=0\iff {({\mathrm{r}}^2+1)}^2=0$

$\begin{array}{c}{\mathrm{r}}^2+1=0\\ {\mathrm{r}}^2=-1\\ {\mathrm{r}}_{1,2}=\pm \mathrm{i}\end{array}$

These roots are both repeated. Similarly, to the procedure when repeated roots are not complex, one has that the general solution is: 

 

 $\begin{array}{c}\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+{\mathrm{c}}_3{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}+\mathrm{t}({\mathrm{c}}_2{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+{\mathrm{c}}_4{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t})\\ \mathrm{y}\left(\mathrm{t}\right)=({\mathrm{c}}_1+{\mathrm{c}}_2\mathrm{t}){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+({\mathrm{c}}_3+{\mathrm{c}}_4\mathrm{t}){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}\end{array}$

 

Where ${\mathrm{r}}_{1,2}=\mathrm{\alpha }\pm \mathrm{\beta i}$. In this case $\mathrm{\alpha }=0$  and $\mathrm{\beta }=1$ , so the general solution of the given differential equation is $\mathrm{y}\left(\mathrm{t}\right)=({\mathrm{c}}_1+{\mathrm{c}}_2\mathrm{t})\mathrm{cost}+({\mathrm{c}}_3+{\mathrm{c}}_4\mathrm{t})\mathrm{sint}$ .

 

Step 2: Finding the roots and general solution.

The differential equation is $\mathrm{y}\text{'}\text{'}\text{'}\text{'}+4\mathrm{y}\text{'}\text{'}\text{'}+12\mathrm{y}\text{'}\text{'}+16\mathrm{y}\text{'}+16\mathrm{y}=0.$

 

The auxiliary equation is: ${\mathrm{r}}^4+4{\mathrm{r}}^3+12{\mathrm{r}}^2+16\mathrm{r}+16=0$ 

 

Let’s solve this: 

$\begin{array}{c}{\mathrm{r}}^4+4{\mathrm{r}}^3+12{\mathrm{r}}^2+16\mathrm{r}+16=0\\ {({\mathrm{r}}^2+2\mathrm{r}+4)}^2=0\end{array}$

$\begin{array}{c}{\mathrm{r}}^2+2\mathrm{r}+4=0\\ {\mathrm{r}}_{1,2}=\frac{-2\pm \sqrt{4-16}}{2}\\ {\mathrm{r}}_{1,2}=-1\pm \sqrt{3}\mathrm{i}\end{array}$

As before, those roots are repeated, so the general solution is: $\mathrm{y}\left(\mathrm{t}\right)=({\mathrm{c}}_1+{\mathrm{c}}_2\mathrm{t}){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+({\mathrm{c}}_3+{\mathrm{c}}_4\mathrm{t}){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}$

 

Where ${\mathrm{r}}_{1,2}=\mathrm{\alpha }\pm \mathrm{\beta i}$. In this case $\mathrm{\alpha }=-1$  and $\mathrm{\beta }=\sqrt{3}$ , so the general solution of the given differential equation is $\mathrm{y}\left(\mathrm{t}\right)=({\mathrm{c}}_1+{\mathrm{c}}_2\mathrm{t}){\mathrm{e}}^{-\mathrm{t}}\cos \sqrt{3}\mathrm{t}+({\mathrm{c}}_3+{\mathrm{c}}_4\mathrm{t}){\mathrm{e}}^{-\mathrm{t}}\sin \sqrt{3}\mathrm{t}$ .