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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

# The auxiliary equation for the given differential equation has complex roots. Find a general solution. ${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{4}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{7}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}$

The auxiliary equation for the given differential equation $\mathrm{y}\text{'}\text{'}-4\mathrm{y}\text{'}+7\mathrm{y}=0$ has complex roots and its general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{2\mathrm{t}}\left({\mathrm{c}}_{1}\mathrm{cos}\left(\sqrt{3}\mathrm{t}\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(\sqrt{3}\mathrm{t}\right)\right)$ .

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 the roots of the auxiliary equation.

Given differential equation is $\mathrm{y}\text{'}\text{'}-4\mathrm{y}\text{'}+7\mathrm{y}=0.$

Then the auxiliary equation is ${\mathrm{r}}^{2}-4\mathrm{r}+7=0.$

$\mathrm{r}=\frac{4±\sqrt{{4}^{2}-4×1×7}}{2×1}\phantom{\rule{0ex}{0ex}}\mathrm{r}=\frac{4±\sqrt{16-28}}{}\phantom{\rule{0ex}{0ex}}\mathrm{r}=\frac{4±2\mathrm{i}\sqrt{3}}{}\phantom{\rule{0ex}{0ex}}\mathrm{r}=2±\sqrt{3}\mathrm{i}$

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{2×\mathrm{t}}\left({\mathrm{c}}_{1}cos\left(\sqrt{3}\mathrm{t}\right)+{\mathrm{c}}_{2}sin\left(\sqrt{3}\mathrm{t}\right)\right)\phantom{\rule{0ex}{0ex}}\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{2\mathrm{t}}\left({\mathrm{c}}_{1}\mathrm{cos}\left(\sqrt{3}\mathrm{t}\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(\sqrt{3}\mathrm{t}\right)\right)$