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Expert-verified Found in: Page 186 ### 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 # A mass–spring system is driven by a sinusoidal external force ${\mathbf{g}}\left(\mathbf{t}\right){\mathbf{=}}{\mathbf{5}}{\mathbf{sint}}$. The mass equals 1, the spring constant equals 3, and the damping coefficient equals 4. If the mass is initially located at ${\mathbf{y}}\left(\mathbf{0}\right){\mathbf{=}}\frac{1}{2}$and at rest, i.e., ${\mathbf{y}}{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}{\mathbf{=}}{\mathbf{0}}$, find its equation of motion.

The equation of motion is $\mathrm{y}=2{\mathrm{e}}^{-\mathrm{t}}-\frac{1}{2}{\mathrm{e}}^{-3\mathrm{t}}-\mathrm{cost}+\frac{1}{2}\mathrm{sint}.$

See the step by step solution

## Step 1: Use the given information for writing the differential equation.

Given that,

The mass equals $\mathrm{m}=1$,

The spring constant equals $\mathrm{c}=3$,

And the damping coefficient equals $\mathrm{b}=4$.

The differential equation is,

$\begin{array}{c}\mathrm{my}\text{'}\text{'}+\mathrm{by}\text{'}+\mathrm{cy}=\mathrm{g}\left(\mathrm{t}\right)\\ \mathrm{y}\text{'}\text{'}+4\mathrm{y}\text{'}+3\mathrm{y}=5\mathrm{sint}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(1\right)\end{array}$

Write the homogeneous differential equation of the equation (1),

$\mathrm{y}\text{'}\text{'}+4\mathrm{y}\text{'}+3\mathrm{y}=0$

## Step 2: Now find the complementary solution of the given equation.

The auxiliary equation for the above equation,

${\mathrm{m}}^{2}+4\mathrm{m}+3=0$

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^{2}+4\mathrm{m}+3=0\\ {\mathrm{m}}^{2}+3\mathrm{m}+\mathrm{m}+3=0\\ \mathrm{m}\left(\mathrm{m}+3\right)+1\left(\mathrm{m}+3\right)=0\\ \left(\mathrm{m}+1\right)\left(\mathrm{m}+3\right)=0\end{array}$

The roots of auxiliary equation are,

${\mathrm{m}}_{1}=-1,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{m}}_{2}=-3$

The complimentary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_{1}{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-3\mathrm{t}}$

## Step 3: Now find the particular solution to find a general solution for the equation.

Assume, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acost}+\mathrm{Bsint}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(2\right)$

Now find the first and second derivative of above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=-\mathrm{Asint}+\mathrm{Bcost}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=-\mathrm{Acost}-\mathrm{Bsint}\end{array}$

Substitute the value of ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right),\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)$ and ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$ in the equation (1),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}+4\mathrm{y}\text{'}+3\mathrm{y}=5\mathrm{sint}\\ -\mathrm{Acost}-\mathrm{Bsint}+4\left(-\mathrm{Asint}+\mathrm{Bcost}\right)+3\left(\mathrm{Acost}+\mathrm{Bsint}\right)=5\mathrm{sint}\\ \left(2\mathrm{B}-4\mathrm{A}\right)\mathrm{sint}+\left(2\mathrm{A}+4\mathrm{B}\right)\mathrm{cost}=5\mathrm{sint}\end{array}$

Comparing the all coefficients of the above equation,

$\begin{array}{c}2\mathrm{B}-4\mathrm{A}=5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}.......\left(3\right)\\ 4\mathrm{B}+2\mathrm{A}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(4\right)\end{array}$

Solve the above equations,

$\begin{array}{c}2\left(2\mathrm{B}-4\mathrm{A}\right)=5×2\\ 4\mathrm{B}-8\mathrm{A}=10\\ 4\mathrm{B}+2\mathrm{A}=0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=-1\end{array}$

Substitute the value of A in the equation (4),

role="math" localid="1655115721256" $\begin{array}{c}4\mathrm{B}+2\left(-1\right)=0\\ \mathrm{B}=\frac{1}{2}\end{array}$

Therefore, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{Acost}+\mathrm{Bsint}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=-\mathrm{cost}+\frac{1}{2}\mathrm{sint}\end{array}$

## Step 4: Find the general solution and use the given initial condition,

Therefore, the general solution is,

$\begin{array}{l}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}={\mathrm{c}}_{1}{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-3\mathrm{t}}-\mathrm{cost}+\frac{1}{2}\mathrm{sint}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(5\right)\end{array}$

Given initial condition,

$\mathrm{y}\left(0\right)=\frac{1}{2},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{y}\text{'}\left(0\right)=0$

Substitute the value of $\mathrm{y}=\frac{1}{2}$ and t = 0 in the equation (3),

$\begin{array}{c}\frac{1}{2}={\mathrm{c}}_{1}{\mathrm{e}}^{-\left(0\right)}+{\mathrm{c}}_{2}{\mathrm{e}}^{-3\left(0\right)}-\mathrm{cos}\left(0\right)+\frac{1}{2}\mathrm{sin}\left(0\right)\\ \frac{1}{2}={\mathrm{c}}_{1}+{\mathrm{c}}_{2}-1\\ {\mathrm{c}}_{1}+{\mathrm{c}}_{2}=\frac{3}{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}.....\left(6\right)\end{array}$

Now find the derivative of above equation,

$\mathrm{y}\text{'}=-{\mathrm{c}}_{1}{\mathrm{e}}^{-\mathrm{t}}-3{\mathrm{c}}_{2}{\mathrm{e}}^{-3\mathrm{t}}+\mathrm{sint}+\frac{1}{2}\mathrm{cost}$

Substitute the value of y’ = 0 and t = 0 in the above equation,

$\begin{array}{c}0=-{\mathrm{c}}_{1}{\mathrm{e}}^{-\left(0\right)}-3{\mathrm{c}}_{2}{\mathrm{e}}^{-3\left(0\right)}+\mathrm{sin}\left(0\right)+\frac{1}{2}\mathrm{cos}\left(0\right)\\ 0=-{\mathrm{c}}_{1}-3{\mathrm{c}}_{2}+\frac{1}{2}\\ {\mathrm{c}}_{1}+3{\mathrm{c}}_{2}=\frac{1}{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(7\right)\end{array}$

Solve the (6) and (7) equations,

$\begin{array}{c}{\mathrm{c}}_{1}+{\mathrm{c}}_{2}=\frac{3}{2}\\ {\mathrm{c}}_{1}+3{\mathrm{c}}_{2}=\frac{1}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}_{2}=-\frac{1}{2}\end{array}$

Substitute the value of ${\mathrm{c}}_{2}$ in the equation (6),

role="math" localid="1655116289545" $\begin{array}{c}{\mathrm{c}}_{1}+{\mathrm{c}}_{2}=\frac{3}{2}\\ {\mathrm{c}}_{1}+\left(\frac{-1}{2}\right)=\frac{3}{2}\\ {\mathrm{c}}_{1}=2\end{array}$

Substitute the value of ${\mathrm{c}}_{1}$ and ${\mathrm{c}}_{2}$ in the equation (5),

role="math" localid="1655116374807" $\begin{array}{c}\mathrm{y}={\mathrm{c}}_{1}{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-3\mathrm{t}}-\mathrm{cost}+\frac{1}{2}\mathrm{sint}\\ \mathrm{y}=2{\mathrm{e}}^{-\mathrm{t}}-\frac{1}{2}{\mathrm{e}}^{-3\mathrm{t}}-\mathrm{cost}+\frac{1}{2}\mathrm{sint}\end{array}$

Thus, the equation of motion is:

$\mathrm{y}=2{\mathrm{e}}^{-\mathrm{t}}-\frac{1}{2}{\mathrm{e}}^{-3\mathrm{t}}-\mathrm{cost}+\frac{1}{2}\mathrm{sint}$ ### Want to see more solutions like these? 