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Q4.3-14E

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. ${\mathbf{y}}{\text{'}}{\text{'}}{\mathbf{+}}{\mathbf{2}}{\mathbf{y}}{\text{'}}{\mathbf{+}}{\mathbf{5}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}$

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

See the step by step solution

## Step 1: Complex conjugate roots.

If the auxiliary equation has complex conjugate roots ${\alpha }{±}{i}{\beta }$, then the general solution is given as:

${y}\left(t\right){=}{{c}}_{{1}}{{e}}^{\alpha t}{c}{o}{s}{\beta }{t}{+}{{c}}_{{2}}{{e}}^{\alpha t}{s}{i}{n}{\beta }{t}$.

## Step 2: Finding roots of the auxiliary equation.

Given differential equation is $\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+5\mathrm{y}=0.$

Then the auxiliary equation is ${\mathrm{r}}^{2}+2\mathrm{r}+5=0.$

The roots of the auxiliary equation are:

role="math" localid="1654070409803" $\mathrm{r}=\frac{-2±\sqrt{{2}^{2}-4×1×5}}{2×1}\phantom{\rule{0ex}{0ex}}\mathrm{r}=\frac{-2±\sqrt{4-20}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{r}=\frac{-2±\sqrt{-16}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{r}=\frac{-2±4\mathrm{i}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{r}=-1±2\mathrm{i}$

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