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Found in: Page 173

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# 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}}$

1. The general solution of the given differential equation is:$\mathrm{y}\left(\mathrm{t}\right)=\left({\mathrm{c}}_{1}+{\mathrm{c}}_{2}\mathrm{t}\right)\mathrm{cost}+\left({\mathrm{c}}_{3}+{\mathrm{c}}_{4}\mathrm{t}\right)\mathrm{sint}$
2. The general solution of the given differential equation is:$\mathrm{y}\left(\mathrm{t}\right)=\left({\mathrm{c}}_{1}+{\mathrm{c}}_{2}\mathrm{t}\right){\mathrm{e}}^{-\mathrm{t}}\mathrm{cos}\sqrt{3}\mathrm{t}+\left({\mathrm{c}}_{3}+{\mathrm{c}}_{4}\mathrm{t}\right){\mathrm{e}}^{-\mathrm{t}}\mathrm{sin}\sqrt{3}\mathrm{t}$
See the step by step solution

## 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:

${\mathrm{r}}^{4}+2{\mathrm{r}}^{2}+1=0⇔{\left({\mathrm{r}}^{2}+1\right)}^{2}=0$

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

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

$\begin{array}{c}\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+{\mathrm{c}}_{3}{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}+\mathrm{t}\left({\mathrm{c}}_{2}{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+{\mathrm{c}}_{4}{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}\right)\\ \mathrm{y}\left(\mathrm{t}\right)=\left({\mathrm{c}}_{1}+{\mathrm{c}}_{2}\mathrm{t}\right){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+\left({\mathrm{c}}_{3}+{\mathrm{c}}_{4}\mathrm{t}\right){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}\end{array}$

Where ${\mathrm{r}}_{1,2}=\mathrm{\alpha }±\mathrm{\beta i}$. In this case $\mathrm{\alpha }=0$ and $\mathrm{\beta }=1$ , so the general solution of the given differential equation is $\mathrm{y}\left(\mathrm{t}\right)=\left({\mathrm{c}}_{1}+{\mathrm{c}}_{2}\mathrm{t}\right)\mathrm{cost}+\left({\mathrm{c}}_{3}+{\mathrm{c}}_{4}\mathrm{t}\right)\mathrm{sint}$ .

## 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:

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

role="math" localid="1654854846964" $\begin{array}{c}{\mathrm{r}}^{2}+2\mathrm{r}+4=0\\ {\mathrm{r}}_{1,2}=\frac{-2±\sqrt{4-16}}{2}\\ {\mathrm{r}}_{1,2}=-1±\sqrt{3}\mathrm{i}\end{array}$

As before, those roots are repeated, so the general solution is: $\mathrm{y}\left(\mathrm{t}\right)=\left({\mathrm{c}}_{1}+{\mathrm{c}}_{2}\mathrm{t}\right){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+\left({\mathrm{c}}_{3}+{\mathrm{c}}_{4}\mathrm{t}\right){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}$

Where ${\mathrm{r}}_{1,2}=\mathrm{\alpha }±\mathrm{\beta i}$. In this case $\mathrm{\alpha }=-1$ and $\mathrm{\beta }=\sqrt{3}$ , so the general solution of the given differential equation is $\mathrm{y}\left(\mathrm{t}\right)=\left({\mathrm{c}}_{1}+{\mathrm{c}}_{2}\mathrm{t}\right){\mathrm{e}}^{-\mathrm{t}}\mathrm{cos}\sqrt{3}\mathrm{t}+\left({\mathrm{c}}_{3}+{\mathrm{c}}_{4}\mathrm{t}\right){\mathrm{e}}^{-\mathrm{t}}\mathrm{sin}\sqrt{3}\mathrm{t}$ .