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Expert-verified Found in: Page 288 ### 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 # Referring to the coupled mass-spring system discussed in Example , suppose an external force ${\mathbf{E}}{\mathbf{\left(}}{\mathbf{t}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{37}}{\mathbf{cos}}{\mathbf{3}}{\mathbf{t}}$ is applied to the second object of mass ${\mathbf{1}}{\mathbf{}}{\mathbf{kg}}$. The displacement functions ${\mathbf{x}}\left(\mathbf{t}\right)$ and ${\mathbf{y}}\left(\mathbf{t}\right)$ now satisfy the system$\mathbf{\left(}\mathbf{16}\mathbf{\right)}\mathbf{\left(}\mathbf{2}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{6}\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\phantom{\rule{0ex}{0ex}}\mathbf{\left(}\mathbf{17}\mathbf{\right)}\mathbf{\left(}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{37}\mathbf{cos}\mathbf{3}\mathbf{t}$ (a) Show that ${\mathbf{x}}\left(\mathbf{t}\right)$ satisfies the equation $\mathbf{\left(}\mathbf{18}\mathbf{\right)}{{\mathbf{x}}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{5}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{4}\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{37}\mathbf{cos}\mathbf{3}\mathbf{t}$(b) Find a general solution ${\mathbf{x}}\left(\mathbf{t}\right)$ to the equation (18). [Hint: Use undetermined coefficients with ${{\mathbf{x}}}_{{\mathbf{p}}}{\mathbf{=}}{\mathbf{Acos}}{\mathbf{3}}{\mathbf{t}}{\mathbf{+}}{\mathbf{Bsin}}{\mathbf{3}}{\mathbf{t}}$.](c) Substitute ${\mathbf{x}}\left(\mathbf{t}\right)$ back into (16) to obtain a formula for ${\mathbf{y}}\left(\mathbf{t}\right)$.(d) If both masses are displaced 2m to the right of their equilibrium positions and then released, find the displacement functions ${\mathbf{x}}\left(\mathbf{t}\right)$ and ${\mathbf{y}}\left(\mathbf{t}\right)$.

(a) It has been shown that $\mathrm{x}\left(\mathrm{t}\right)$ satisfies the equation $\left(18\right){\mathrm{x}}^{\left(4\right)}\left(\mathrm{t}\right)+5\mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)+4\mathrm{x}\left(\mathrm{t}\right)=37\mathrm{cos}3\mathrm{t}$.

(b) $\mathrm{x}\left(\mathrm{t}\right)={\mathrm{c}}_{1}\mathrm{cost}+{\mathrm{c}}_{2}\mathrm{sint}+{\mathrm{c}}_{3}\mathrm{cos}2\mathrm{t}+{\mathrm{c}}_{4}\mathrm{sin}2\mathrm{t}+\frac{37}{40}\mathrm{cos}3\mathrm{t}$

(c) $\mathrm{y}\left(\mathrm{t}\right)=2{\mathrm{c}}_{1}\mathrm{cost}+2{\mathrm{c}}_{2}\mathrm{sint}-{\mathrm{c}}_{3}\mathrm{cos}2\mathrm{t}-{\mathrm{c}}_{4}\mathrm{sin}2\mathrm{t}-3×\frac{37}{20}\mathrm{cos}3\mathrm{t}$

(d) The displacement functions x and y are

$\mathrm{x}\left(\mathrm{t}\right)=\frac{23}{8}\mathrm{cost}-\frac{9}{5}\mathrm{cos}2\mathrm{t}+\frac{37}{40}\mathrm{cos}3\mathrm{t}\phantom{\rule{0ex}{0ex}}\mathrm{y}\left(\mathrm{t}\right)=\frac{23}{4}\mathrm{cost}+\frac{9}{5}\mathrm{cos}2\mathrm{t}-3×\frac{37}{40}\mathrm{cos}3\mathrm{t}$

See the step by step solution

## Solving the given equation

Let's rewrite the given system in operator form:

$\left(2{\mathrm{D}}^{2}+6\right)\left[\mathrm{x}\right]-2\left[\mathrm{y}\right]=0\phantom{\rule{0ex}{0ex}}-2\left[\mathrm{x}\right]+\left({\mathrm{D}}^{2}+2\right)\left[\mathrm{y}\right]=37\mathrm{cos}3\mathrm{t}$

One wants to obtain the equation for x so we will eliminate y from the system. To do so one will multiply the first equation by $\left({\mathrm{D}}^{2}+2\right)$ and the second by 2 and then add them together:

$\begin{array}{rcl}& & \left({\mathrm{D}}^{2}+6\right)\left({\mathrm{D}}^{2}+2\right)\left[\mathrm{x}\right]-2\left({\mathrm{D}}^{2}+2\right)\left[\mathrm{y}\right]=0\\ & & -4\left[\mathrm{x}\right]+2\left({\mathrm{D}}^{2}+2\right)\left[\mathrm{y}\right]=2×37\mathrm{cos}3\mathrm{t}\\ & & \left(\left(2{\mathrm{D}}^{2}+6\right)\left({\mathrm{D}}^{2}+2\right)-4\right)\left[\mathrm{x}\right]=2×37\mathrm{cos}3\mathrm{t}\end{array}$

$\left(2{\mathrm{D}}^{4}+4{\mathrm{D}}^{2}+6{\mathrm{D}}^{2}+12-4\right)\left[\mathrm{x}\right]=2×37\mathrm{cos}3\mathrm{t}\phantom{\rule{0ex}{0ex}}2\left({\mathrm{D}}^{4}+5{\mathrm{D}}^{2}+4\right)\left[\mathrm{x}\right]=2×37\mathrm{cos}3\mathrm{t}\phantom{\rule{0ex}{0ex}}\left({\mathrm{D}}^{4}+5{\mathrm{D}}^{2}+4\right)\left[\mathrm{x}\right]=37\mathrm{cos}3\mathrm{t}$

Rewriting this back in standard form we have that $\mathrm{x}\left(\mathrm{t}\right)$ satisfies

${\mathrm{x}}^{\left(4\right)}\left(\mathrm{t}\right)+5\mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)+4\mathrm{x}\left(\mathrm{t}\right)=37\mathrm{cos}3\mathrm{t}$.

## Finding roots

To find a general solution, one will first find a homogeneous solution for $\mathrm{x}\left(\mathrm{t}\right)$.

The auxiliary equation is:

${\mathrm{r}}^{4}+5{\mathrm{r}}^{2}+4=0$ , and its roots are:

${\mathrm{r}}^{4}+5{\mathrm{r}}^{2}+4=\left({\mathrm{r}}^{2}+1\right)\left({\mathrm{r}}^{2}+4\right)=0\phantom{\rule{0ex}{0ex}}{\mathrm{r}}^{2}=-1,{\mathrm{r}}^{2}=-4\phantom{\rule{0ex}{0ex}}{\mathrm{r}}_{1,2}=±\mathrm{i},{\mathrm{r}}_{3,4}=±2\mathrm{i}$

Therefore, the homogeneous solution for x is:

${\mathrm{x}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{c}}_{1}\mathrm{cost}+{\mathrm{c}}_{2}\mathrm{sint}+{\mathrm{c}}_{3}\mathrm{cos}2\mathrm{t}+{\mathrm{c}}_{4}\mathrm{sin}2\mathrm{t}$

## Finding derivatives

One will find a particular solution by using the method of undetermined coefficients and assume that a particular solution has a form of ${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acos}3\mathrm{t}+\mathrm{Bsin}3\mathrm{t}$.

One needs the second and the fourth derivative of ${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)$:

$\mathrm{x}\text{'}=-3\mathrm{Asin}3\mathrm{t}+3\mathrm{Bcos}3\mathrm{t},\mathrm{x}\text{'}\text{'}=-9\mathrm{Acos}3\mathrm{t}-9\mathrm{Bsin}3\mathrm{t}\phantom{\rule{0ex}{0ex}}\mathrm{x}\text{'}\text{'}\text{'}=27\mathrm{Asin}3\mathrm{t}-27\mathrm{Bcos}3\mathrm{t},{\mathrm{x}}^{\left(4\right)}=81\mathrm{Acos}3\mathrm{t}+81\mathrm{Bsin}3\mathrm{t}$

## Substituting The values

Now, one has that,

${\mathrm{x}}_{\mathrm{p}}^{\left(4\right)}{\left(\mathrm{t}\right)+5\mathrm{x}}_{\mathrm{p}}^{}\text{'}\text{'}\left(\mathrm{t}\right)+4{\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=81\mathrm{Acos}3\mathrm{t}+81\mathrm{Bsin}3\mathrm{t}+5\left(-9\mathrm{Acos}3\mathrm{t}-9\mathrm{Bsin}3\mathrm{t}\right)+4\left(\mathrm{Acos}3\mathrm{t}+\mathrm{Bsin}3\mathrm{t}\right)\phantom{\rule{0ex}{0ex}}=40\mathrm{Acos}3\mathrm{t}+40\mathrm{Bsin}3\mathrm{t}=37\mathrm{cos}3\mathrm{t}\phantom{\rule{0ex}{0ex}}⇒40\mathrm{A}=37,40\mathrm{B}=0\phantom{\rule{0ex}{0ex}}⇔\mathrm{A}=\frac{37}{40},\mathrm{B}=0$

So, the particular solution for x is ${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=37/40\mathrm{cos}3\mathrm{t}$ and the general solution for x is $\mathrm{x}\left(\mathrm{t}\right)={\mathrm{x}}_{\mathrm{h}}+{\mathrm{x}}_{\mathrm{p}}={\mathrm{c}}_{1}\mathrm{cost}+{\mathrm{c}}_{2}\mathrm{sint}+{\mathrm{c}}_{3}\mathrm{cos}2\mathrm{t}+{\mathrm{c}}_{4}\mathrm{sin}2\mathrm{t}+\frac{37}{40}\mathrm{cos}3\mathrm{t}$.

## Finding derivatives

The first equation of the given system gives us that $\mathrm{y}\left(\mathrm{t}\right)=\mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)+3\mathrm{x}\left(\mathrm{t}\right)$

So, one need to find the second derivative of x and substitute it into the previous equation.

$\begin{array}{rcl}& & \mathrm{x}\text{'}\left(\mathrm{t}\right)=-{\mathrm{c}}_{1}\mathrm{sint}+{\mathrm{c}}_{2}\mathrm{cost}-2{\mathrm{c}}_{3}\mathrm{sin}2\mathrm{t}+2{\mathrm{c}}_{4}\mathrm{cos}2\mathrm{t}-3×\frac{37}{40}\mathrm{sin}3\mathrm{t}\hfill\\ & & \mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)=-{\mathrm{c}}_{1}\mathrm{cost}-{\mathrm{c}}_{2}\mathrm{sint}-4{\mathrm{c}}_{3}\mathrm{cos}2\mathrm{t}-4{\mathrm{c}}_{4}\mathrm{sin}2\mathrm{t}-9×\frac{37}{40}\mathrm{cos}3\mathrm{t}\end{array}$

Substituting this into $\mathrm{y}\left(\mathrm{t}\right)=\mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)+3\mathrm{x}\left(\mathrm{t}\right)$

One has,

$\mathrm{y}\left(\mathrm{t}\right)=-{\mathrm{c}}_{1}\mathrm{cost}-{\mathrm{c}}_{2}\mathrm{sint}-4{\mathrm{c}}_{3}\mathrm{cos}2\mathrm{t}-4{\mathrm{c}}_{4}\mathrm{sin}2\mathrm{t}-9×\frac{37}{40}\mathrm{cos}3\mathrm{t}\phantom{\rule{0ex}{0ex}}+3\left({\mathrm{c}}_{1}\mathrm{cost}+{\mathrm{c}}_{2}\mathrm{sint}+{\mathrm{c}}_{3}\mathrm{cos}2\mathrm{t}+{\mathrm{c}}_{4}\mathrm{sin}2\mathrm{t}+\frac{37}{40}\mathrm{cos}3\mathrm{t}\right)\phantom{\rule{0ex}{0ex}}\mathrm{y}\left(\mathrm{t}\right)=2{\mathrm{c}}_{1}\mathrm{cost}+2{\mathrm{c}}_{2}\mathrm{sint}-{\mathrm{c}}_{3}\mathrm{cos}2\mathrm{t}-{\mathrm{c}}_{4}\mathrm{sin}2\mathrm{t}-3×\frac{37}{20}\mathrm{cos}3\mathrm{t}$

## Finding derivatives

Both masses are displaced 2m to the right of their equilibrium position, so $\mathrm{x}\left(0\right)=\mathrm{y}\left(0\right)=2$, and since they are released, one has that $\mathrm{x}\text{'}\left(0\right)=\mathrm{y}\text{'}\left(0\right)=0$.

First, one will find the first derivative of y:

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=-2{\mathrm{c}}_{2}\mathrm{sint}+2{\mathrm{c}}_{2}\mathrm{cost}+2{\mathrm{c}}_{3}\mathrm{sin}2\mathrm{t}-2{\mathrm{c}}_{4}\mathrm{cos}2\mathrm{t}-9×\frac{37}{40}\mathrm{sin}3\mathrm{t}$

So, from this initial condition,one has that

$\begin{array}{rcl}& & \mathrm{x}\left(0\right)={\mathrm{c}}_{1}×1+{\mathrm{c}}_{2}×0+{\mathrm{c}}_{3}×1+{\mathrm{c}}_{4}×0+\frac{37}{40}×1\\ & =& {\mathrm{c}}_{1}+{\mathrm{c}}_{3}+\frac{37}{40}\\ & =& 2\\ & & \mathrm{y}\left(0\right)=2{\mathrm{c}}_{1}×1+2{\mathrm{c}}_{2}×0-{\mathrm{c}}_{3}×1-{\mathrm{c}}_{4}×0-3×\frac{37}{20}×1\\ & =& 2{\mathrm{c}}_{1}-{\mathrm{c}}_{3}-6×\frac{37}{40}\\ & =& 2\end{array}$

## Substituting the values

Adding the previous two equations together one will get

$\begin{array}{rcl}3{\mathrm{c}}_{1}-5×\frac{37}{40}& =& 4\\ {\mathrm{c}}_{1}& =& \frac{23}{8}\end{array}$

Substituting this into the second equation one will have ${\mathrm{c}}_{2}=2{\mathrm{c}}_{1}-6×\frac{37}{40}-2⇔{\mathrm{c}}_{3}=-\frac{9}{5}$

The third and the fourth initial condition give us

$\mathrm{x}\text{'}\left(0\right)=-{\mathrm{c}}_{1}×0+{\mathrm{c}}_{2}×1-2{\mathrm{c}}_{3}×0+2{\mathrm{c}}_{4}×1-3×\frac{37}{40}×0\phantom{\rule{0ex}{0ex}}={\mathrm{c}}_{2}+2{\mathrm{c}}_{4}\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\mathrm{y}\text{'}\left(0\right)=-2{\mathrm{c}}_{1}×0+2{\mathrm{c}}_{2}×1+2{\mathrm{c}}_{3}×0-2{\mathrm{c}}_{4}×1-9×\frac{37}{40}×0\phantom{\rule{0ex}{0ex}}=2{\mathrm{c}}_{2}-2{\mathrm{c}}_{4}\phantom{\rule{0ex}{0ex}}=0$

## Finding c2,c4

Adding those equations together one gets that $3{\mathrm{c}}_{2}=0⇒{\mathrm{c}}_{2}=0$

Substituting this into the first equation one has that $2{\mathrm{c}}_{4}=0⇒{\mathrm{c}}_{4}=0$

Finally, one has that the displacement functions x and y is:

$\mathrm{x}\left(\mathrm{t}\right)=\frac{23}{8}\mathrm{cost}-\frac{9}{5}\mathrm{cos}2\mathrm{t}+\frac{37}{40}\mathrm{cos}3\mathrm{t}\phantom{\rule{0ex}{0ex}}\mathrm{y}\left(\mathrm{t}\right)=\frac{23}{4}\mathrm{cost}+\frac{9}{5}\mathrm{cos}2\mathrm{t}-3×\frac{37}{40}\mathrm{cos}3\mathrm{t}$

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