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Expert-verified Found in: Page 173 ### 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 # Using the mass-spring analogy, predict the behavior as ${\mathbf{t}}{\to }{\infty }$ of the solution to the given initial value problem. Then confirm your prediction by actually solving the problem.$\begin{array}{l}\mathbf{\left(}\mathbf{a}\mathbf{\right)}.\text{\hspace{0.17em}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{16}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}.\text{\hspace{0.17em}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{100}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}.\text{\hspace{0.17em}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{8}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\\ \mathbf{\left(}\mathbf{d}\mathbf{\right)}.\text{\hspace{0.17em}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\\ \mathbf{\left(}\mathbf{e}\mathbf{\right)}.\text{\hspace{0.17em}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\end{array}$

1. The solution is y=2cos4t.
2. The general solution is $\mathrm{y}\left(\mathrm{t}\right)=\frac{-50+\sqrt{2499}}{2\sqrt{2499}}{\mathrm{e}}^{\left(-50+\sqrt{2499}\right)\mathrm{t}}+\frac{50+\sqrt{2499}}{2\sqrt{2499}}{\mathrm{e}}^{\left(-50-\sqrt{2499}\right)\mathrm{t}}$
3. The general solution is $\mathrm{y}\left(\mathrm{t}\right)=2{\mathrm{e}}^{2\mathrm{t}}-{\mathrm{e}}^{4\mathrm{t}}$.
4. The general solution is $\mathrm{y}\left(\mathrm{t}\right)=-\frac{1}{2}{\mathrm{e}}^{-3\mathrm{t}}-\frac{3}{2}{\mathrm{e}}^{\mathrm{t}}$.
5. The general solution is $\mathrm{y}\left(\mathrm{t}\right)=-\frac{2}{5}{\mathrm{e}}^{-3\mathrm{t}}+\frac{3}{5}{\mathrm{e}}^{\mathrm{t}}$.
See the step by step solution

## Step 1: Find the general solution.

(a).

The differential equation is $\mathrm{y}\text{'}\text{'}+16\mathrm{y}=0$.

The auxiliary equation is ${r}^{2}+16=0$.

Find the roots of the auxiliary equation.

$\begin{array}{c}{\mathrm{r}}^{2}+16=0\\ \mathrm{r}=±4\mathrm{i}\end{array}$

The general equation is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_{1}\mathrm{cos}4\mathrm{t}+{\mathrm{c}}_{2}\mathrm{sin}4\mathrm{t}$.

Apply initial conditions $\mathrm{y}\left(0\right)=2,\mathrm{y}\text{'}\left(0\right)=0$.

Using the given initial values, we get:

$\begin{array}{l}\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_{1}\mathrm{cos}4\mathrm{t}+{\mathrm{c}}_{2}\mathrm{sin}4\mathrm{t}\\ \mathrm{y}\left(0\right)={\mathrm{c}}_{1}+0\\ 2={c}_{1}\\ \mathrm{y}\text{'}\left(\mathrm{t}\right)=-4{\mathrm{c}}_{1}\mathrm{sin}4\mathrm{t}+4{\mathrm{c}}_{2}\mathrm{cos}4\mathrm{t}\\ \mathrm{y}\text{'}\left(0\right)=-4{\mathrm{c}}_{1}\mathrm{sin}0+4{\mathrm{c}}_{2}\mathrm{cos}0\\ 0=4{c}_{2}\\ {\mathrm{c}}_{2}=0\end{array}$

Thus, the solution is y=2cos4t.

As ${\mathbf{-}}{\mathbf{1}}{\mathbf{\le }}{\mathbf{cos}}{\mathbf{4}}{\mathbf{t}}{\mathbf{\le }}{\mathbf{1}}$therefore the solution oscillates between -2 and 2.

## Step 2: Check the result of the general solution

(b).

Here the differential equation is $\mathrm{y}\text{'}\text{'}+100\mathrm{y}\text{'}+\mathrm{y}=0$.

The auxiliary equation is ${\mathrm{r}}^{2}+100\mathrm{r}+1=0$.

Find the roots of the auxiliary equation.

$\begin{array}{c}{\mathrm{r}}^{2}+100\mathrm{r}+1=0\\ \mathrm{r}=-50±\sqrt{2499}\end{array}$

The general equation is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_{1}{\mathrm{e}}^{\left(-50+\sqrt{2499}\right)\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{e}}^{\left(-50-\sqrt{2499}\right)\mathrm{t}}$.

Apply initial conditions $\mathrm{y}\left(0\right)=1,\mathrm{y}\text{'}\left(0\right)=0$.

$\begin{array}{c}\mathrm{y}\left(0\right)={\mathrm{c}}_{1}=\frac{-50+\sqrt{2499}}{2\sqrt{2499}}\\ \mathrm{y}\text{'}\left(0\right)={\mathrm{c}}_{2}=\frac{50+\sqrt{2499}}{2\sqrt{2499}}\end{array}$

The solution is $\mathrm{y}\left(\mathrm{t}\right)=\frac{-50+\sqrt{2499}}{2\sqrt{2499}}{\mathrm{e}}^{\left(-50+\sqrt{2499}\right)\mathrm{t}}+\frac{50+\sqrt{2499}}{2\sqrt{2499}}{\mathrm{e}}^{\left(-50-\sqrt{2499}\right)\mathrm{t}}$.

Since the powers of exponential functions tend to ${-}{\infty }{\mathbf{ast}}{\to }{\infty }{,}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}{\to }{0}$.

## Step 3: Determine the solution

(c).

Here the differential equation is $\mathrm{y}\text{'}\text{'}-6\mathrm{y}\text{'}+8\mathrm{y}=0$.

The auxiliary equation is:

$\begin{array}{c}{\mathrm{r}}^{2}-6\mathrm{r}+8=0\\ \mathrm{r}=2,4\end{array}$

The general equation is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_{1}{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{e}}^{4\mathrm{t}}$.

Apply initial conditions $\mathrm{y}\left(0\right)=1,\mathrm{y}\text{'}\left(0\right)=0$

$\begin{array}{c}\mathrm{y}\left(0\right)={\mathrm{c}}_{1}=2\\ \mathrm{y}\text{'}\left(0\right)={\mathrm{c}}_{2}=-1\end{array}$

The solution is $\mathrm{y}\left(\mathrm{t}\right)=2{\mathrm{e}}^{2\mathrm{t}}-{\mathrm{e}}^{4\mathrm{t}}$.

The solution approaches to $-\infty \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{as}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{t}\to \infty$.

## Step 4: find the result.

(d).

Here the differential equation is $\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}-3\mathrm{y}=0$.

The auxiliary equation is:

$\begin{array}{c}{\mathrm{r}}^{2}+2\mathrm{r}-3=0\\ \mathrm{r}=3,1\end{array}$

The general equation is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_{1}{\mathrm{e}}^{-3\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{e}}^{\mathrm{t}}$.

Apply initial conditions $\mathrm{y}\left(0\right)=-2,\mathrm{y}\text{'}\left(0\right)=0$

role="math" localid="1654848021100" $\begin{array}{c}\mathrm{y}\left(0\right)={\mathrm{c}}_{1}=-\frac{1}{2}\\ \mathrm{y}\text{'}\left(0\right)={\mathrm{c}}_{2}=-\frac{3}{2}\end{array}$

The general solution is $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{1}{2}{\mathbf{e}}^{-3t}\mathbf{-}\frac{3}{2}{\mathbf{e}}^{t}$.

The solution approaches to ${-}{\infty }{\text{\hspace{0.17em}\hspace{0.17em}}}{\mathbf{as}}{\text{\hspace{0.17em}\hspace{0.17em}}}{\mathbf{t}}{\to }{\infty }$.

## Step 5: evaluate the result

(e).

Here the differential equation is $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{0}$.

The auxiliary equation is:

$\begin{array}{c}{\mathrm{r}}^{2}-\mathrm{r}-6=0\\ \mathrm{r}=-2,3\end{array}$

The general equation is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_{1}{\mathrm{e}}^{-2\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{e}}^{3\mathrm{t}}$.

Apply initial conditions $\mathrm{y}\left(0\right)=1,\mathrm{y}\text{'}\left(0\right)=1$

role="math" localid="1654848262559" $\begin{array}{c}\mathrm{y}\left(0\right)={\mathrm{c}}_{1}=-\frac{2}{5}\\ \mathrm{y}\text{'}\left(0\right)={\mathrm{c}}_{2}=\frac{3}{5}\end{array}$

The solution is $\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{2}{5}{\mathbf{e}}^{-3t}\mathbf{+}\frac{3}{5}{\mathbf{e}}^{t}$.

The solution approaches to $\infty \text{\hspace{0.17em}\hspace{0.17em}}\mathbf{as}\text{\hspace{0.17em}\hspace{0.17em}}\mathbf{t}\to \infty$.

This is the required result. ### Want to see more solutions like these? 