 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q25E

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{;}}$$\mathbf{y}\mathbf{\left(}\mathbf{\pi }\mathbf{\right)}\mathbf{=}{{\mathbf{e}}}^{\pi }{\mathbf{,}}$${\mathbf{y}}{\text{'}}\mathbf{\left(}\mathbf{\pi }\mathbf{\right)}\mathbf{=}\mathbf{0}$

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

See the step by step solution

## Step 1: Complex conjugate roots.

If the auxiliary equation has complex conjugate roots ${\mathbf{\alpha }}{\mathbf{±}}{\mathbf{i\beta }}$, then the general solution is given as: ${\mathrm{y}}\left(\mathrm{t}\right){=}{{\mathrm{c}}}_{{1}}{{\mathrm{e}}}^{{\mathrm{\alpha t}}}{\mathrm{cos\beta t}}{+}{{\mathrm{c}}}_{{2}}{{\mathrm{e}}}^{{\mathrm{\alpha t}}}{\mathrm{sin\beta t}}$

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

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

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

$\begin{array}{l}\mathrm{r}=\frac{2±\sqrt{{2}^{2}-4×1×2}}{2×1}\\ \mathrm{r}=\frac{2±\sqrt{4-8}}{}\\ \mathrm{r}=\frac{2±\sqrt{-4}}{}\\ \mathrm{r}=\frac{2±2\mathrm{i}}{}\\ \mathrm{r}=1±\mathrm{i}\end{array}$

Therefore, the general solution is:

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

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

Given initial conditions are $\mathrm{y}\left(\mathrm{\pi }\right)={\mathrm{e}}^{\mathrm{\pi }}$ and $\mathrm{y}\text{'}\left(\mathrm{\pi }\right)=0$

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

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

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

Then ${\mathrm{c}}_{2}=1$

Therefore, the solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}\left(-\mathrm{cost}+\mathrm{sint}\right)$ ### Want to see more solutions like these? 