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Q4.3-10E

Expert-verified
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 ${\mathbf{y}}{\text{'}}{\text{'}}{\mathbf{+}}{\mathbf{4}}{\mathbf{y}}{\text{'}}{\mathbf{+}}{\mathbf{8}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}$

The general solution of the given equation $\mathrm{y}\text{'}\text{'}+4\mathrm{y}\text{'}+8\mathrm{y}=0$ is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\left({\mathrm{c}}_{1}\mathrm{cos}\left(2\mathrm{t}\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(2\mathrm{t}\right)\right)$.

See the step by step solution

## Step 1: Differentiate the value of y.

Given differential equation is $\mathrm{y}\text{'}\text{'}+4\mathrm{y}\text{'}+8\mathrm{y}=0.$

Let role="math" localid="1654067583036" $\mathrm{y}={\mathrm{e}}^{\mathrm{rt}}$

Therefore, we have:

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=\mathrm{r}{\mathrm{e}}^{\mathrm{rt}}\phantom{\rule{0ex}{0ex}}$

$\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)={\mathrm{r}}^{2}{\mathrm{e}}^{\mathrm{rt}}$

## Step 2: Finding the roots.

Then the auxiliary equation is ${\mathrm{r}}^{2}+4\mathrm{r}+8=0$

The roots of the auxiliary equation are:

$\mathrm{r}=\frac{-4±\sqrt{{4}^{2}-4×1×8}}{2×1}\phantom{\rule{0ex}{0ex}}\mathrm{r}=\frac{-4±\sqrt{16-32}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{r}=\frac{-4±4\mathrm{i}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{r}=-2±2\mathrm{i}$

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-2×\mathrm{t}}\left({\mathrm{c}}_{1}\mathrm{cos}\left(2\mathrm{t}\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(2\mathrm{t}\right)\right)\phantom{\rule{0ex}{0ex}}\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-2\mathrm{t}}\left({\mathrm{c}}_{1}\mathrm{cos}\left(2\mathrm{t}\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(2\mathrm{t}\right)\right)$