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Expert-verified Found in: Page 413 ### 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 # Recompute the coupled mass–spring oscillator motion in Problem 1, Exercises 5.6 (page 287), using Laplace transforms.

$x\left(t\right)=-\frac{8}{17}\mathrm{cos}t-\frac{9}{17}\mathrm{cos}\sqrt{\frac{20}{3}t}\phantom{\rule{0ex}{0ex}}y\left(t\right)=-\frac{6}{17}\mathrm{cos}t+\frac{6}{17}\mathrm{cos}\sqrt{\frac{20}{3}}t$

See the step by step solution

## Step 1:Given Information

The differential equation of a coupled mass spring oscillator is

$\left\{\begin{array}{cc}{x}^{\text{'}\text{'}}+4x-4y=0,& x\left(0\right)=-1,{x}^{\text{'}}\left(0\right)=0\\ 3{y}^{\text{'}\text{'}}-6x+11y=0& y\left(0\right)=0,{y}^{\text{'}}\left(0\right)=0\end{array}$

## Step 2: Determining the coupled mass–spring oscillator motion

The mechanism in Exercises 5.6 Problem 1 is

$\left\{\begin{array}{cc}{x}^{\text{'}\text{'}}+4x-4y=0,& x\left(0\right)=-1,{x}^{\text{'}}\left(0\right)=0\\ 3{y}^{\text{'}\text{'}}-6x+11y=0& y\left(0\right)=0,{y}^{\text{'}}\left(0\right)=0\end{array}$

Using the Laplace transform, we can obtain

$\left\{\begin{array}{l}L\left\{x"+4x-4y\right\}=0\\ L\left\{3y"-6x+11y\right\}=0\end{array}$

$⇒\left\{\begin{array}{l}{s}^{2}X\left(s\right)-sx\left(0\right)-{x}^{\text{'}}\left(0\right)+4X\left(s\right)-4Y\left(s\right)=0\\ 3\left[{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\text{'}}\left(0\right)\right]-6X\left(s\right)+11Y\left(s\right)=0\end{array}$

$⇒\left\{\begin{array}{l}{s}^{2}X\left(s\right)+s+4X\left(s\right)-4Y\left(s\right)=0\\ 3{s}^{2}Y\left(s\right)-6X\left(s\right)+11Y\left(s\right)=0\end{array}$

$⇒\left\{\begin{array}{l}\left({s}^{2}+4\right)X\left(s\right)-4Y\left(s\right)=-s\\ \left(3{s}^{2}+11\right)Y\left(s\right)-6X\left(s\right)=0\end{array}$

$⇒\left\{\begin{array}{l}X\left(s\right)=\frac{4}{{s}^{2}+1}Y\left(s\right)-\frac{s}{{s}^{2}+1}\\ \left(3{s}^{2}+11\right)Y\left(s\right)-6X\left(s\right)=0\end{array}$

We now have

$\begin{array}{c}\left(3{s}^{2}+11\right)Y\left(s\right)-6\left(\frac{4}{{s}^{2}+1}Y\left(s\right)-\frac{s}{{s}^{2}+1}\right)=0\\ \left(3{s}^{2}+11\right)Y\left(s\right)-\frac{24}{{s}^{2}+1}Y\left(s\right)+\frac{6s}{{s}^{2}+1}=0\\ \left(3{s}^{2}+11-\frac{24}{{s}^{2}+1}\right)Y\left(s\right)=-\frac{6s}{{s}^{2}+1}\\ \frac{\left(3{s}^{2}+11\right)\left({s}^{2}+1\right)-24}{{s}^{2}+1}Y\left(s\right)=-\frac{6s}{{s}^{2}+1}\\ \left(3{s}^{4}+23{s}^{2}+20\right)Y\left(s\right)=6s\\ Y\left(s\right)=-\frac{6s}{3{s}^{4}+23{s}^{2}+20}\end{array}$

We can get partial fractions by using partial fractions.

$Y\left(s\right)=-\frac{6s}{17\left({s}^{2}+1\right)}+\frac{6s}{17\left({s}^{2}+\frac{20}{3}\right)}$

As a result, if we use the inverse Laplace, we get

$y\left(t\right)=-\frac{6}{17}\mathrm{cos}t+\frac{6}{17}\mathrm{cos}\sqrt{\frac{20}{3}t}$

We now have X(s):

$\begin{array}{rcl}\left(3{s}^{2}+11\right)\frac{\left(-6s\right)}{3{s}^{4}+23{s}^{2}+20}-6X\left(s\right)& =& 0\\ X\left(s\right)& =& \frac{-s\left(3{s}^{2}+11\right)}{\left({s}^{2}+1\right)\left({s}^{2}+\frac{20}{3}\right)}\end{array}$

We can get partial fractions by using partial fractions.

$X\left(s\right)=-\frac{8s}{17\left({s}^{2}+1\right)}-\frac{9s}{17\left({s}^{2}+\frac{20}{3}\right)}$

As a result, if we use the inverse Laplace, we get

$x\left(t\right)=-\frac{8}{17}\mathrm{cos}t-\frac{9}{17}\mathrm{cos}\sqrt{\frac{20}{3}}t$

## Step 3: Determining the Result

The solution is obtained as:

$x\left(t\right)=-\frac{8}{17}\mathrm{cos}t-\frac{9}{17}\mathrm{cos}\sqrt{\frac{20}{3}t}$

$y\left(t\right)=-\frac{6}{17}\mathrm{cos}t+\frac{6}{17}\mathrm{cos}\sqrt{\frac{20}{3}}t$ ### Want to see more solutions like these? 