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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

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 $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}\left(-\mathrm{cost}+\mathrm{sint}\right)$