91Ó°ÊÓ

(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.

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

The general equation is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\cos 4\mathrm{t}+{\mathrm{c}}_2\sin 4\mathrm{t}$.

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

Using the given initial values, we get:

$\begin{array}{l}\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\cos 4\mathrm{t}+{\mathrm{c}}_2\sin 4\mathrm{t}\\ \mathrm{y}\left(0\right)={\mathrm{c}}_1+0\\ 2=c_1\\ \mathrm{y}\text{'}\left(\mathrm{t}\right)=-4{\mathrm{c}}_1\sin 4\mathrm{t}+4{\mathrm{c}}_2\cos 4\mathrm{t}\\ \mathrm{y}\text{'}\left(0\right)=-4{\mathrm{c}}_1\sin 0+4{\mathrm{c}}_2\cos 0\\ 0=4c_2\\ {\mathrm{c}}_2=0\end{array}$

Thus, the solution is y=2cos4t.

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

(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.

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

The general equation is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{(-50+\sqrt{2499})\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{(-50-\sqrt{2499})\mathrm{t}}$.

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

Since the powers of exponential functions tend to $−\mathrm{∞}\mathbf{ast}→\mathrm{∞},\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}→0$.

(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 $−\mathrm{∞}\text{ â¶Ä‰}\mathrm{as}\text{ â¶Ä‰}\mathrm{t}→\mathrm{∞}$.

(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 $−\mathrm{∞}\text{ â¶Ä‰}\mathbf{as}\text{ â¶Ä‰}\mathbf{t}→\mathrm{∞}$.

(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 $\mathrm{∞}\text{ â¶Ä‰}\mathbf{as}\text{ â¶Ä‰}\mathbf{t}→\mathrm{∞}$.

This is the required result.