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Answers without the blur. Sign up and see all textbooks for free! Q5.6-7E

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 # Suppose the displacement functions ${\mathbf{x}}\left(\mathbf{t}\right)$ and ${\mathbf{y}}\left(\mathbf{t}\right)$ for a coupled mass-spring system (similar to the one discussed in Problem 6) satisfy the initial value problem$\begin{array}{rcl}& & \mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{5}\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}\\ & & \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{sin}\mathbf{2}\mathbf{t}\\ & & \mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\\ & & \mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}{}{}{}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\end{array}$ Solve for ${\mathbf{x}}\left(\mathbf{t}\right)$ and ${\mathbf{y}}\left(\mathbf{t}\right)$

The solution for $\mathrm{x}\left(\mathrm{t}\right)=\frac{2}{5}\mathrm{cost}+\frac{4}{5}\mathrm{sint}-\frac{2}{5}\mathrm{cos}\sqrt{6}\mathrm{t}+\frac{\sqrt{6}}{5}\mathrm{sin}\sqrt{6}\mathrm{t}-\mathrm{sin}2\mathrm{t}$ and

$\mathrm{y}\left(\mathrm{t}\right)=\frac{4}{5}\mathrm{cost}+\frac{8}{5}\mathrm{sint}+\frac{1}{5}\mathrm{cos}\sqrt{6}\mathrm{t}-\frac{\sqrt{6}}{10}\mathrm{sin}\sqrt{6}\mathrm{t}-\frac{1}{2}\mathrm{sin}2\mathrm{t}$.

See the step by step solution

## Using the elimination procedure

Given system can be written in the following form:

$\begin{array}{rcl}& & \left({\mathrm{D}}^{2}+5\right)\left[\mathrm{x}\right]-2\left[\mathrm{y}\right]=0\\ & & -2\left[\mathrm{x}\right]+\left({\mathrm{D}}^{2}+2\right)\left[\mathrm{y}\right]=3\mathrm{sin}2\mathrm{t}\end{array}$

By ${\mathrm{L}}_{1}={\mathrm{D}}^{2}+5,{\mathrm{L}}_{2}=-2,{\mathrm{L}}_{3}=-2,{\mathrm{L}}_{4}={\mathrm{D}}^{2}+2,{\mathrm{f}}_{1}=0,{\mathrm{f}}_{2}=\mathrm{sin}2\mathrm{t}$, using the elimination procedure,

One obtains $\left({\mathrm{L}}_{1}{\mathrm{L}}_{4}-{\mathrm{L}}_{2}{\mathrm{L}}_{3}\right)\left[\mathrm{x}\right]={\mathrm{L}}_{4}\left[{\mathrm{f}}_{1}\right]-{\mathrm{L}}_{2}\left[{\mathrm{f}}_{2}\right]$

i.e.

$\left(\left({\mathrm{D}}^{2}+5\right)\left({\mathrm{D}}^{2}+2\right)-4\right)\left[\mathrm{x}\right]=\left({\mathrm{D}}^{2}+2\right)\left[0\right]-\left(-2\right)\left[3\mathrm{sin}2\mathrm{t}\right]\phantom{\rule{0ex}{0ex}}\left({\mathrm{D}}^{4}+7{\mathrm{D}}^{2}+6\right)\left[\mathrm{x}\right]=0+6\mathrm{sin}2\mathrm{t}\phantom{\rule{0ex}{0ex}}\left(\left({\mathrm{D}}^{2}+1\right)\left({\mathrm{D}}^{2}+6\right)\right)\left[\mathrm{x}\right]=6\mathrm{sin}2\mathrm{t}\cdots \left(1\right)$

## Finding derivatives

The auxiliary equation is $\left({\mathrm{r}}^{2}+1\right)\left({\mathrm{r}}^{2}+6\right)=0$ and its roots ${\mathrm{r}}_{1,2}=±\mathrm{i},{\mathrm{r}}_{3,4}=±\mathrm{i}\sqrt{6}$.

Therefore, the solution to the corresponding homogeneous equation is

${\mathrm{x}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{c}}_{1}\mathrm{cost}+{\mathrm{c}}_{2}\mathrm{sint}+{\mathrm{c}}_{3}\mathrm{cos}\sqrt{6}\mathrm{t}+{\mathrm{c}}_{4}\mathrm{sin}\sqrt{6}\mathrm{t}$.

Let's check if $\mathrm{x}\left(\mathrm{t}\right)=-\mathrm{sin}2\mathrm{t}$ is a particular solution for (1).

First, one will find derivatives

$\begin{array}{rcl}& & \mathrm{x}\text{'}\left(\mathrm{t}\right)=-2\mathrm{cos}2\mathrm{t}\\ & & \mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)=4\mathrm{sin}2\mathrm{t}\\ & & \mathrm{x}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)=8\mathrm{cos}2\mathrm{t}\\ & & {\mathrm{x}}^{\left(4\right)}\left(\mathrm{t}\right)=-16\mathrm{sin}2\mathrm{t}\end{array}$

## Finding x

Since

${\mathrm{x}}^{\left(4\right)}\left(\mathrm{t}\right)+7\mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)+6\mathrm{x}\left(\mathrm{t}\right)=-16\mathrm{sin}2\mathrm{t}+7×4\mathrm{sin}2\mathrm{t}+6\left(-\mathrm{sin}2\mathrm{t}\right)\phantom{\rule{0ex}{0ex}}=6\mathrm{sin}2\mathrm{t}$

${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=-\mathrm{sin}2\mathrm{t}$ is a particular solution to 1 and the general solution is

$\mathrm{x}\left(\mathrm{t}\right)={\mathrm{x}}_{\mathrm{h}}\left(\mathrm{t}\right)+{\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{c}}_{1}\mathrm{cost}+{\mathrm{c}}_{2}\mathrm{sint}+{\mathrm{c}}_{3}\mathrm{cos}\sqrt{6}\mathrm{t}+{\mathrm{c}}_{4}\mathrm{sin}\sqrt{6}\mathrm{t}-\mathrm{sin}2\mathrm{t}$ .

From the first equation of the system, $\mathrm{x}\text{'}\text{'}+5\mathrm{x}-2\mathrm{y}=0$ one can find a general solution for $\mathrm{y}\left(\mathrm{t}\right)$ but the first one needs to find $\mathrm{x}\text{'}\text{'}$

$\mathrm{x}\text{'}\left(\mathrm{t}\right)=-{\mathrm{c}}_{1}\mathrm{sint}+{\mathrm{c}}_{2}\mathrm{cost}-\sqrt{6}{\mathrm{c}}_{3}\mathrm{sin}\sqrt{6}\mathrm{t}+\sqrt{6}{\mathrm{c}}_{4}\mathrm{cos}\sqrt{6}\mathrm{t}-2\mathrm{cos}2\mathrm{t}\phantom{\rule{0ex}{0ex}}\mathrm{x}\text{'}\text{'}\left(\mathrm{t}\right)=-{\mathrm{c}}_{1}\mathrm{cost}-{\mathrm{c}}_{2}\mathrm{sint}-6{\mathrm{c}}_{3}\mathrm{cos}\sqrt{6}\mathrm{t}-6{\mathrm{c}}_{4}\mathrm{sin}\sqrt{6}\mathrm{t}+4\mathrm{sin}2\mathrm{t}$

## Simplification

Then simplify,

$2\mathrm{y}=\mathrm{x}\text{'}\text{'}+5\mathrm{x}\phantom{\rule{0ex}{0ex}}=-{\mathrm{c}}_{1}\mathrm{cost}-{\mathrm{c}}_{2}\mathrm{sint}-6{\mathrm{c}}_{3}\mathrm{cos}\sqrt{6}\mathrm{t}-6{\mathrm{c}}_{4}\mathrm{sin}\sqrt{6}\mathrm{t}+4\mathrm{sin}2\mathrm{t}+5{\mathrm{c}}_{1}\mathrm{cost}+5{\mathrm{c}}_{2}\mathrm{sint}\phantom{\rule{0ex}{0ex}}+5{\mathrm{c}}_{3}\mathrm{cos}\sqrt{6}\mathrm{t}+5{\mathrm{c}}_{4}\mathrm{sin}\sqrt{6}\mathrm{t}-5\mathrm{sin}2\mathrm{t}\phantom{\rule{0ex}{0ex}}=4{\mathrm{c}}_{1}\mathrm{cost}+4{\mathrm{c}}_{2}\mathrm{sint}-{\mathrm{c}}_{3}\mathrm{cos}\sqrt{6}\mathrm{t}-{\mathrm{c}}_{4}\mathrm{sin}\sqrt{6}\mathrm{t}-\mathrm{sin}2\mathrm{t}$

Therefore, $\mathrm{y}\left(\mathrm{t}\right)=2{\mathrm{c}}_{1}\mathrm{cost}+2{\mathrm{c}}_{2}\mathrm{sint}-\frac{{\mathrm{c}}_{3}}{2}\mathrm{cos}\sqrt{6}\mathrm{t}-\frac{{\mathrm{c}}_{4}}{2}\mathrm{sin}\sqrt{6}\mathrm{t}-\frac{1}{2}\mathrm{sin}2\mathrm{t}$

It remains to find constants from initial conditions.

## Finding  xt,yt

Let's compute $\mathrm{y}\text{'}$

$\mathrm{y}\left(\mathrm{t}\right)=-2{\mathrm{c}}_{1}\mathrm{sint}+2{\mathrm{c}}_{2}\mathrm{cost}+\frac{{\mathrm{c}}_{3}\sqrt{6}}{2}\mathrm{sin}\sqrt{6}\mathrm{t}-\frac{{\mathrm{c}}_{4}\sqrt{6}}{2}\mathrm{cos}\sqrt{6}\mathrm{t}-\mathrm{cos}2\mathrm{t}$

$0=\mathrm{x}\left(0\right)={\mathrm{c}}_{1}+{\mathrm{c}}_{3}\phantom{\rule{0ex}{0ex}}0=\mathrm{x}\text{'}\left(0\right)={\mathrm{c}}_{2}+\sqrt{6}{\mathrm{c}}_{4}-2\phantom{\rule{0ex}{0ex}}1=\mathrm{y}\left(0\right)=2{\mathrm{c}}_{1}-\frac{{\mathrm{c}}_{3}}{2}\phantom{\rule{0ex}{0ex}}0=\mathrm{y}\text{'}\left(0\right)=2{\mathrm{c}}_{2}-\frac{\sqrt{6}}{2}{\mathrm{c}}_{4}-1$

By solving this system, one obtains ${\mathrm{c}}_{1}=\frac{2}{5},{\mathrm{c}}_{2}=\frac{4}{5},{\mathrm{c}}_{3}=-\frac{2}{5},{\mathrm{c}}_{4}=\frac{\sqrt{6}}{5}$

Finally, $\mathrm{x}\left(\mathrm{t}\right)=\frac{2}{5}\mathrm{cost}+\frac{4}{5}\mathrm{sint}-\frac{2}{5}\mathrm{cos}\sqrt{6}\mathrm{t}+\frac{\sqrt{6}}{5}\mathrm{sin}\sqrt{6}\mathrm{t}-\mathrm{sin}2\mathrm{t}$

And $\mathrm{y}\left(\mathrm{t}\right)=\frac{4}{5}\mathrm{cost}+\frac{8}{5}\mathrm{sint}+\frac{1}{5}\mathrm{cos}\sqrt{6}\mathrm{t}-\frac{\sqrt{6}}{10}\mathrm{sin}\sqrt{6}\mathrm{t}-\frac{1}{2}\mathrm{sin}2\mathrm{t}$ ### Want to see more solutions like these? 