Q31E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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.

$\begin{array}{l}\mathbf{\left(}\mathbf{a}\mathbf{\right)}.\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{16}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}.\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{100}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}.\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{8}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\\ \mathbf{\left(}\mathbf{d}\mathbf{\right)}.\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\\ \mathbf{\left(}\mathbf{e}\mathbf{\right)}.\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\end{array}$

Step-by-Step Solution

Answer

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

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.

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

 

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 $\mathbf{-}\mathbf{1}\mathbf{\le }\mathbf{cos}\mathbf{4}\mathbf{t}\mathbf{\le }\mathbf{1}$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.

 

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

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 $-\infty \mathbf{ast}\to \infty ,\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\to 0$.

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 $-\infty \mathrm{as}\mathrm{t}\to \infty$.

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

$\begin{array}{c}\mathrm{y}\left(0\right)={\mathrm{c}}_1=-\frac{1}{2}\\ \mathrm{y}\text{'}\left(0\right)={\mathrm{c}}_2=-\frac{3}{2}\end{array}$

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 $-\infty \mathbf{as}\mathbf{t}\to \infty$.

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

$\begin{array}{c}\mathrm{y}\left(0\right)={\mathrm{c}}_1=-\frac{2}{5}\\ \mathrm{y}\text{'}\left(0\right)={\mathrm{c}}_2=\frac{3}{5}\end{array}$

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 $\infty \mathbf{as}\mathbf{t}\to \infty$.

 

This is the required result.