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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{y}}{\mathbf{=}}{\mathbf{0}}{\mathbf{;}}$${\mathbf{y}}\left(\mathbf{0}\right){\mathbf{=}}{\mathbf{1}}{\mathbf{,}}$${\mathbf{y}}{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}$

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

See the step by step solution

## Step 1: Differentiate the value of y.

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

Let $\mathrm{y}={\mathrm{e}}^{\mathrm{rt}}$

Therefore,

$\begin{array}{c}\mathrm{y}\text{'}\left(\mathrm{t}\right)=\mathrm{r}{\mathrm{e}}^{\mathrm{rt}}\\ \mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)={\mathrm{r}}^{2}{\mathrm{e}}^{\mathrm{rt}}\end{array}$

## Step 2: Finding the general solution.

Then the auxiliary equation is ${\mathrm{r}}^{2}-2\mathrm{r}+1=0$

$\begin{array}{c}{\left(\mathrm{r}-1\right)}^{2}=0\\ \mathrm{r}-1=0\\ \mathrm{r}=1\end{array}$

Therefore, the general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{te}}^{\mathrm{t}}$ .

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

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

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

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

Then,

$\begin{array}{l}\mathrm{y}\text{'}\left(0\right)={\mathrm{c}}_{1}{\mathrm{e}}^{0}+{\mathrm{c}}_{2}{\mathrm{e}}^{0}+\mathrm{c}.0×{\mathrm{e}}^{0}\\ {\mathrm{c}}_{1}+{\mathrm{c}}_{2}=-2\end{array}$

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

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

Therefore, the solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}-3{\mathrm{te}}^{\mathrm{t}}$ . ### Want to see more solutions like these? 