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Q4.3-19E

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Found in: Page 172

### 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

# Find a general solution ${\mathbf{y}}{\text{'}}{\text{'}}{\text{'}}{\mathbf{+}}{\mathbf{y}}{\text{'}}{\text{'}}{\mathbf{+}}{\mathbf{3}}{\mathbf{y}}{\text{'}}{\mathbf{-}}{\mathbf{5}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}$

The general solution of the given equation $\mathrm{y}\text{'}\text{'}\text{'}+\mathrm{y}\text{'}\text{'}+3\mathrm{y}\text{'}-5\mathrm{y}=0$ is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{C}}_{1}{\mathrm{e}}^{\mathrm{t}}+{\mathrm{C}}_{2}{\mathrm{e}}^{-\mathrm{t}}\mathrm{cos}\left(2\mathrm{t}\right)+{\mathrm{C}}_{3}{\mathrm{e}}^{-\mathrm{t}}\mathrm{sin}\left(2\mathrm{t}\right)$.

See the step by step solution

## Step 1: Using rational root theorem.

First, one needs to find the auxiliary equation and solve it. One has ${\mathrm{r}}^{3}+{\mathrm{r}}^{2}+3\mathrm{r}-5=0$.

You can use the rational root theorem.

The first divisor of role="math" localid="1654073691473" $5$ is role="math" localid="1654073695156" ${\mathbf{1}}$ if will be one solution of the equation and $\mathbf{\left(}\mathbf{r}\mathbf{-}\mathbf{1}\mathbf{\right)}$ will be a factor.

Indeed, you have ${1}^{3}+{1}^{2}+3-5=0$

Now you can divide role="math" localid="1654073800144" ${\mathrm{r}}^{3}+{\mathrm{r}}^{2}+3\mathrm{r}-5$ by role="math" localid="1654073795414" $\mathrm{r}-1$ to get ${\mathrm{r}}^{2}+2\mathrm{r}+5$.

## Step 2: Finding factors and roots.

The equation can be factored as $\left(\mathrm{r}-1\right)\left({\mathrm{r}}^{2}+2\mathrm{r}+5\right)=0.$

Since ${\mathrm{r}}^{2}+2\mathrm{r}+5=0$

$\mathrm{r}=\frac{-2±\sqrt{{2}^{2}-4×1×5}}{}\phantom{\rule{0ex}{0ex}}\mathrm{r}=-1±2\mathrm{i}$

The roots of the auxiliary equation are $\mathrm{r}=1,\mathrm{r}=-1+2\mathrm{i}$ and $\mathrm{r}=-1-2\mathrm{i}$

Thus, the general solution of the differential equation is: $y\left(t\right)={\mathrm{C}}_{1}{\mathrm{e}}^{\mathrm{t}}+{\mathrm{C}}_{2}{\mathrm{e}}^{-\mathrm{t}}\mathrm{cos}\left(2\mathrm{t}\right)+{\mathrm{C}}_{3}{\mathrm{e}}^{-\mathrm{t}}\mathrm{sin}\left(2\mathrm{t}\right)$