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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{'}}{\text{'}}{\mathbf{-}}{\mathbf{4}}{\mathbf{y}}{\text{'}}{\text{'}}{\mathbf{+}}{\mathbf{7}}{\mathbf{y}}{\text{'}}{\mathbf{-}}{\mathbf{6}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}{\mathbf{;}}$$\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{1}{\mathbf{,}}{\mathbf{y}}{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}$${\mathbf{y}}{\text{'}}{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

The solution of the given initial value $\mathrm{y}\text{'}\text{'}\text{'}-4\mathrm{y}\text{'}\text{'}+7\mathrm{y}\text{'}-6\mathrm{y}=0$ is $\mathrm{y}\left(\mathrm{t}\right)=\sqrt{2}{\mathrm{e}}^{\mathrm{t}}\left(\mathrm{sin}\left(\sqrt{2}\mathrm{t}\right)\right)+{\mathrm{e}}^{2\mathrm{t}}$ when $\mathrm{y}\left(0\right)=1,\mathrm{y}\text{'}\left(0\right)=0$ and $\mathrm{y}\text{'}\text{'}\left(0\right)=0$ .

See the step by step solution

## Step 1: Differentiate the value of y.

Given differential equation is $\mathrm{y}\text{'}\text{'}\text{'}-4\mathrm{y}\text{'}\text{'}+7\mathrm{y}\text{'}-6\mathrm{y}=0$

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

Therefore,

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=\mathrm{r}{\mathrm{e}}^{\mathrm{rt}}$ , $\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)={\mathrm{r}}^{2}{\mathrm{e}}^{\mathrm{rt}}$ and $\mathrm{y}\text{'}\text{'}\text{'}={\mathrm{r}}^{3}{\mathrm{e}}^{\mathrm{rt}}$

Then the auxiliary equation is ${\mathrm{r}}^{3}-4{\mathrm{r}}^{2}+7\mathrm{r}-6=0$.

Now ${\mathrm{r}}^{3}-4{\mathrm{r}}^{2}+7\mathrm{r}-6=\left(\mathrm{r}-2\right)\left({\mathrm{r}}^{2}-2\mathrm{r}+3\right)$

## Step 2: Finding the general solution

Now we have to find the roots of ${\mathrm{r}}^{2}-2\mathrm{r}+3.$

$\begin{array}{c}\mathrm{r}=\frac{2±\sqrt{{2}^{2}-4×1×3}}{2×1}\\ \mathrm{r}=\frac{2±\sqrt{4-12}}{}\\ \mathrm{r}=\frac{2±\sqrt{-8}}{}\\ \mathrm{r}=\frac{2±2\sqrt{2}\mathrm{i}}{}\\ \mathrm{r}=1±\sqrt{2}\mathrm{i}\end{array}$

Therefore, the general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}\left({\mathrm{c}}_{1}\mathrm{cos}\left(\sqrt{2}\mathrm{t}\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(\sqrt{2}\mathrm{t}\right)\right)+{\mathrm{c}}_{3}{\mathrm{e}}^{2\mathrm{t}}$.

## Step 3: Substituting the values of y(0)=1,y'(0)=0 and   y''(0)=0

$\begin{array}{l}\mathrm{y}\left(0\right)={\mathrm{e}}^{0}\left({\mathrm{c}}_{1}\mathrm{cos}\left(\sqrt{2}×0\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(\sqrt{2}×0\right)\right)+{\mathrm{c}}_{3}{\mathrm{e}}^{2×0}\\ {\mathrm{c}}_{1}+{\mathrm{c}}_{3}=1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(1\right)\end{array}$

And

$\mathrm{y}\text{'}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}\left({\mathrm{c}}_{1}\mathrm{cos}\sqrt{2}\mathrm{t}+{\mathrm{c}}_{2}\mathrm{sin}\sqrt{2}\mathrm{t}\right)+\sqrt{2}{\mathrm{e}}^{\mathrm{t}}\left(-{\mathrm{c}}_{1}\mathrm{sin}\sqrt{\mathrm{t}}+{\mathrm{c}}_{2}\mathrm{cos}\sqrt{\mathrm{t}}\right)+2{\mathrm{c}}_{3}{\mathrm{e}}^{2\mathrm{t}}$

Then,

$\begin{array}{l}\mathrm{y}\text{'}\text{'}\left(0\right)={\mathrm{e}}^{0}{\mathrm{c}}_{1}\mathrm{cos}\left(\sqrt{2}×0\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(\sqrt{2}×0\right)+\sqrt{2}{\mathrm{e}}^{0}\left(-{\mathrm{c}}_{1}\mathrm{sin}\left(\sqrt{2}×0\right)+{\mathrm{c}}_{2}\mathrm{cos}\left(\sqrt{2}×0\right)\right)+2{\mathrm{c}}_{3}{\mathrm{e}}^{2×0}\\ {\mathrm{c}}_{1}+\sqrt{2}{\mathrm{c}}_{2}+2{\mathrm{c}}_{3}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(2\right)\end{array}$

And

${\mathrm{y}}^{\text{'}\text{'}\text{'}}={\mathrm{e}}^{\mathrm{t}}{\mathrm{c}}_{1}\mathrm{cos}\left(\sqrt{2}\mathrm{t}\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(\sqrt{2}\mathrm{t}\right)+\sqrt{2}{\mathrm{e}}^{\mathrm{t}}\left(-{\mathrm{c}}_{1}\mathrm{sin}\sqrt{\mathrm{t}}+{\mathrm{c}}_{2}\mathrm{cos}\sqrt{\mathrm{t}}\right)+\sqrt{2}{\mathrm{e}}^{\mathrm{t}}\left(-{\mathrm{c}}_{1}\mathrm{sin}\sqrt{\mathrm{t}}+{\mathrm{c}}_{2}\mathrm{cos}\sqrt{\mathrm{t}}\right)+2{\mathrm{e}}^{\mathrm{t}}\left(-{\mathrm{c}}_{1}\mathrm{cos}\left(\sqrt{2}\mathrm{t}\right)-{\mathrm{c}}_{2}\mathrm{sin}\left(\sqrt{2}\mathrm{t}\right)\right)+4{\mathrm{c}}_{3}{\mathrm{e}}^{2\mathrm{t}}$

Then,

$\begin{array}{l}{\mathrm{y}}^{\text{'}\text{'}\text{'}}\left(0\right)={\mathrm{e}}^{0}\left({\mathrm{c}}_{1}\mathrm{cos}\left(0\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(0\right)+\sqrt{2}{\mathrm{e}}^{0}\left(-{\mathrm{c}}_{1}\mathrm{sin}0+{\mathrm{c}}_{2}\mathrm{cos}0\right)+\sqrt{2}{\mathrm{e}}^{0}\left(-{\mathrm{c}}_{1}\mathrm{sin}0+{\mathrm{c}}_{2}\mathrm{cos}0\right)\\ +2{\mathrm{e}}^{0}\left(-{\mathrm{c}}_{1}\mathrm{cos}\left(0\right)-{\mathrm{c}}_{2}\mathrm{sin}\left(0\right)\right)+4{\mathrm{c}}_{3}{\mathrm{e}}^{0}\right)\\ -{\mathrm{c}}_{1}+2\sqrt{2}{\mathrm{c}}_{2}+4{\mathrm{c}}_{3}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(3\right)\end{array}$

On solving the equations, we get:

${\mathrm{c}}_{1}=0,{\mathrm{c}}_{2}=-\sqrt{2}$ and ${\mathrm{c}}_{3}=1$

Therefore, the solution is $\mathrm{y}\left(\mathrm{t}\right)=\sqrt{2}{\mathrm{e}}^{\mathrm{t}}\left(\mathrm{sin}\left(\sqrt{2}\mathrm{t}\right)\right)+{\mathrm{e}}^{2\mathrm{t}}$. ### Want to see more solutions like these? 