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Q4.3-21E

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

# Solve the given initial value problem. ${\mathbf{y}}{\text{'}}{\text{'}}{\mathbf{+}}{\mathbf{2}}{\mathbf{y}}{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

The solution of the given initial value $\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+2\mathrm{y}=0$ is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-\mathrm{t}}\left(2\mathrm{cost}+3\mathrm{sint}\right)$ when $\mathrm{y}\left(0\right)=2$ and $\mathrm{y}\text{'}\left(0\right)=1$.

See the step by step solution

## Step 1: Initial value problem.

An initial value problem is an ordinary differential equation with a given initial condition. The solution of an ordinary differential equation is known as a general solution which consists of an arbitrary constant. The value of an arbitrary constant can be obtained by using the initial condition.

## Step 2: Finding the general solution.

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

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

role="math" localid="1654075580000" $\mathrm{r}=\frac{-2±\sqrt{{2}^{2}-4×1×2}}{2×1}\phantom{\rule{0ex}{0ex}}\mathrm{r}=\frac{-2±\sqrt{4-8}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{r}=\frac{-2±\sqrt{-4}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{r}=\frac{-2±2\mathrm{i}}{2}\phantom{\rule{0ex}{0ex}}\mathrm{r}=-1±\mathrm{i}$

Therefore, the general solution is:

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

## Step 3: Finding the values of c1 and c2

Given initial conditions are $\mathrm{y}\left(0\right)=2$ and $\mathrm{y}\text{'}\left(0\right)=1$

$\begin{array}{c}\mathrm{y}\left(0\right)={\mathrm{e}}^{-0}\left({\mathrm{c}}_{1}\mathrm{cos}\left(0\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(0\right)\right)\\ {\mathrm{c}}_{1}=2\end{array}$

And

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=-{\mathrm{e}}^{-\mathrm{t}}\left({\mathrm{c}}_{1}\mathrm{cost}+{\mathrm{c}}_{2}\mathrm{sint}\right)+{\mathrm{e}}^{-\mathrm{t}}\left(-{\mathrm{c}}_{1}\mathrm{sint}+{\mathrm{c}}_{2}\mathrm{cost}\right)$

Then we have:

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

Substitute ${\mathrm{c}}_{1}$ in the above equation

$\begin{array}{l}-2+{\mathrm{c}}_{2}=1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}_{2}=3\end{array}$

Therefore, the solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{-\mathrm{t}}\left(2\mathrm{cost}+3\mathrm{sint}\right)$.