Chapter 15, Problem 55P

To determine

Find the time-varying function $y\left(t\right)$ for the given differential equation using Laplace transform method.

Answer to Problem 55P

The time-varying function $y\left(t\right)$ for the given differential equation is $\left(\frac{1}{40}+\frac{1}{20}{e}^{-2t}-\frac{3}{104}{e}^{-4t}-\frac{3}{65}{e}^{-t}\cos \left(2t\right)-\frac{2}{65}{e}^{-t}\sin \left(2t\right)\right)u\left(t\right)$.

Explanation of Solution

Given data:

The differential equation is,

$\frac{d^{3}y}{d{t}^3}+6\frac{d^{2}y}{d{t}^2}+8\frac{dy}{dt}=e^{-t}\cos 2t$ (1)

The initial conditions are zero. That is,

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

Formula used:

Write the general expression for the Laplace transform.

$F\left(s\right)=\mathcal{L}\left[f\left(t\right)\right]$ (2)

Write the general expression for the inverse Laplace transform.

$f\left(t\right)={\mathcal{L}}^{-1}\left[F\left(s\right)\right]$ (3)

Write the general expressions to find the Laplace transform function.

$\mathcal{L}\left[\frac{df}{dt}\right]=sF\left(s\right)-f\left(0^{-}\right)$ (4)

$\mathcal{L}\left[\frac{d^{2}f}{d{t}^2}\right]=s^{2}F\left(s\right)-sf\left(0^{-}\right)-f\text{'}\left(0^{-}\right)$ (5)

$\mathcal{L}\left[\frac{d^{3}f}{d{t}^3}\right]=s^{3}F\left(s\right)-s^{2}f\left(0^{-}\right)-sf\text{'}\left(0^{-}\right)-f\text{'}\text{'}\left(0^{-}\right)$ (6)

$\mathcal{L}\left[e^{-at}\cos \omega t\right]=\frac{s+a}{{\left(s+a\right)}^2+{\omega }^2}$ (7)

Here,

$s$ is a complex variable,

$\omega$ is the angular frequency, and

$s+a$ is the frequency shift or frequency translation.

Write the general expressions to find the inverse Laplace transform function.

${\mathcal{L}}^{-1}\left[\frac{1}{s+a}\right]=e^{-at}u\left(t\right)$ (8)

${\mathcal{L}}^{-1}\left[\frac{s+a}{{\left(s+a\right)}^2+{\omega }^2}\right]=e^{-at}\cos \omega tu\left(t\right)$ (9)

${\mathcal{L}}^{-1}\left[\frac{\omega }{{\left(s+a\right)}^2+{\omega }^2}\right]=e^{-at}\sin \omega tu\left(t\right)$ (10)

${\mathcal{L}}^{-1}\left[\frac{1}{s}\right]=u\left(t\right)$ (11)

Calculation:

Apply Laplace transform function given in equation (2), (4), (5), (6) and (7) to equation (1).

$\left(\begin{array}{l}\left[s^{3}Y\left(s\right)-s^{2}y\left(0^{-}\right)-sy\text{'}\left(0^{-}\right)-y\text{'}\text{'}\left(0^{-}\right)\right]\\ +6\left[s^{2}Y\left(s\right)-sy\left(0^{-}\right)-y\text{'}\left(0^{-}\right)\right]+8\left[sY\left(s\right)-y\left(0^{-}\right)\right]\end{array}\right)=\frac{s+1}{{\left(s+1\right)}^2+2^2}$

$\begin{array}{l}\left(\begin{array}{l}\left[s^{3}Y\left(s\right)-s^{2}y\left(0\right)-sy\text{'}\left(0\right)-y\text{'}\text{'}\left(0\right)\right]\\ +6\left[s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\right]+8\left[sY\left(s\right)-y\left(0\right)\right]\end{array}\right)=\frac{s+1}{{\left(s+1\right)}^2+2^2}\\ \left\{\begin{array}{l}\because y\left(0^{-}\right)=y\left(0\right),\\ y\text{'}\left(0^{-}\right)=y\text{'}\left(0\right)\end{array}\right\}\end{array}$ (12)

Substitute 0 for $y\left(0\right)$, $0$ for $y\text{'}\left(0\right)$ and $0$ for $y\text{'}\text{'}\left(0\right)$ in equation (12).

$\begin{array}{l}\left[s^{3}Y\left(s\right)-s^2\left(0\right)-s\left(0\right)-0\right]+6\left[s^{2}Y\left(s\right)-s\left(0\right)-0\right]+8\left[sY\left(s\right)-0\right]=\frac{s+1}{{\left(s+1\right)}^2+2^2}\\ s^{3}Y\left(s\right)+6s^{2}Y\left(s\right)+8sY\left(s\right)=\frac{s+1}{s^2+2s+1+2^2}\left\{\because {\left(a+b\right)}^2=a^2+2ab+b^2\right\}\\ \left(s^3+6s^2+8s\right)Y\left(s\right)=\frac{s+1}{s^2+2s+1+4}\end{array}$

$\left(s^3+6s^2+8s\right)Y\left(s\right)=\frac{s+1}{s^2+2s+5}$ (13)

Rearrange the equation (13) to find $Y\left(s\right)$.

$\begin{array}{c}Y\left(s\right)=\frac{s+1}{\left(s^3+6s^2+8s\right)\left(s^2+2s+5\right)}\\ =\frac{s+1}{s\left(s^2+6s+8\right)\left(s^2+2s+5\right)}\\ =\frac{s+1}{s\left(s^2+2s+4s+8\right)\left(s^2+2s+5\right)}\\ =\frac{s+1}{s\left(s\left(s+2\right)+4\left(s+2\right)\right)\left(s^2+2s+5\right)}\end{array}$

Reduce the equation as follows,

$Y\left(s\right)=\frac{s+1}{s\left(s+2\right)\left(s+4\right)\left(s^2+2s+5\right)}$ (14)

Expand $Y\left(s\right)$ using partial fraction.

$Y\left(s\right)=\frac{s+1}{s\left(s+2\right)\left(s+4\right)\left(s^2+2s+5\right)}=\frac{A}{s}+\frac{B}{s+2}+\frac{C}{s+4}+\frac{Ds+E}{s^2+2s+5}$ (15)

Here,

A, B, C, D and E are the constants.

Now, to find the constants by using residue and algebraic method.

Constant A:

$A={sY\left(s\right)|}_{s=0}$ (16)

Substitute equation (14) in equation (16) to find the constant A.

$\begin{array}{c}A={s\times \frac{s+1}{s\left(s+2\right)\left(s+4\right)\left(s^2+2s+5\right)}|}_{s=0}\\ ={\frac{s+1}{\left(s+2\right)\left(s+4\right)\left(s^2+2s+5\right)}|}_{s=0}\\ =\frac{0+1}{\left(0+2\right)\left(0+4\right)\left({\left(0\right)}^2+2\left(0\right)+5\right)}\\ =\frac{1}{40}\end{array}$

Constant B:

$B={\left(s+2\right)Y\left(s\right)|}_{s+2=0}$ (17)

Substitute equation (14) in equation (17) to find the constant B.

$\begin{array}{c}B={\left(s+2\right)\times \frac{s+1}{s\left(s+2\right)\left(s+4\right)\left(s^2+2s+5\right)}|}_{s=-2}\\ ={\frac{s+1}{s\left(s+4\right)\left(s^2+2s+5\right)}|}_{s=-2}\\ =\frac{-2+1}{\left(-2\right)\left(-2+4\right)\left({\left(-2\right)}^2+2\left(-2\right)+5\right)}\\ =\frac{1}{20}\end{array}$

Constant C:

$C={\left(s+4\right)Y\left(s\right)|}_{s+4=0}$ (18)

Substitute equation (14) in equation (18) to find the constant C.

$\begin{array}{c}C={\left(s+4\right)\times \frac{s+1}{s\left(s+2\right)\left(s+4\right)\left(s^2+2s+5\right)}|}_{s=-4}\\ ={\frac{s+1}{s\left(s+2\right)\left(s^2+2s+5\right)}|}_{s=-4}\\ =\frac{-4+1}{\left(-4\right)\left(-4+2\right)\left({\left(-4\right)}^2+2\left(-4\right)+5\right)}\\ =-\frac{3}{104}\end{array}$

Consider the partial fraction,

$\begin{array}{l}\frac{s+1}{s\left(s+2\right)\left(s+4\right)\left(s^2+2s+5\right)}=\frac{A}{s}+\frac{B}{s+2}+\frac{C}{s+4}+\frac{Ds+E}{s^2+2s+5}\\ \frac{s+1}{s\left(s+2\right)\left(s+4\right)\left(s^2+2s+5\right)}=\frac{\left(\begin{array}{l}A\left(s+2\right)\left(s+4\right)\left(s^2+2s+5\right)+Bs\left(s+4\right)\left(s^2+2s+5\right)\\ +Cs\left(s+2\right)\left(s^2+2s+5\right)+\left(Ds+E\right)s\left(s+2\right)\left(s+4\right)\end{array}\right)}{s\left(s+2\right)\left(s+4\right)\left(s^2+2s+5\right)}\\ s+1=\left(\begin{array}{l}A\left(s+2\right)\left(s+4\right)\left(s^2+2s+5\right)+Bs\left(s+4\right)\left(s^2+2s+5\right)\\ +Cs\left(s+2\right)\left(s^2+2s+5\right)+\left(Ds+E\right)s\left(s+2\right)\left(s+4\right)\end{array}\right)\\ s+1=\left(\begin{array}{l}A\left(s^2+2s+4s+8\right)\left(s^2+2s+5\right)+B\left(s^2+4s\right)\left(s^2+2s+5\right)\\ +C\left(s^2+2s\right)\left(s^2+2s+5\right)+\left(Ds+E\right)s\left(s^2+2s+4s+8\right)\end{array}\right)\end{array}$

Reduce the equation as follows,

$\begin{array}{l}s+1=\left(\begin{array}{l}A\left(s^2+6s+8\right)\left(s^2+2s+5\right)+B\left(s^4+2s^3+5s^2+4s^3+8s^2+20s\right)\\ +C\left(s^4+2s^3+5s^2+2s^3+4s^2+10s\right)+\left(Ds+E\right)\left(s^3+6s^2+8s\right)\end{array}\right)\\ s+1=\left(\begin{array}{l}A\left(s^4+2s^3+5s^2+6s^3+12s^2+30s+8s^2+16s+40\right)+B\left(s^4+6s^3+13s^2+20s\right)\\ +C\left(s^4+4s^3+9s^2+10s\right)+D\left(s^4+6s^3+8s^2\right)+E\left(s^3+6s^2+8s\right)\end{array}\right)\\ s+1=\left(\begin{array}{l}A{s}^4+8A{s}^3+25A{s}^2+46As+40A+B{s}^4+6B{s}^3+13B{s}^2+20Bs\\ +C{s}^4+4C{s}^3+9C{s}^2+10Cs+D{s}^4+6D{s}^3+8D{s}^2+E{s}^3+6E{s}^2+8Es\end{array}\right)\end{array}$

$s+1=\left(\begin{array}{l}\left(A+B+C+D\right)s^4+\left(8A+6B+4C+6D+E\right)s^3\\ +\left(25A+13B+9C+8D+6E\right)s^2+\left(46A+20B+10C+8E\right)s\\ +40A\end{array}\right)$ (19)

Equate the coefficients of $s^4$ in equation (19).

$0=A+B+C+D$ (20)

Substitute $\frac{1}{40}$ for A, $\frac{1}{20}$ for B, and $-\frac{3}{104}$ for C in equation (20) to find the constant D.

$\begin{array}{l}0=\frac{1}{40}+\frac{1}{20}-\frac{3}{104}+D\\ D=-\frac{1}{40}-\frac{1}{20}+\frac{3}{104}\\ D=-\frac{3}{65}\end{array}$

Equate the coefficients of $s^3$ in equation (19).

$0=8A+6B+4C+6D+E$ (21)

Substitute $\frac{1}{40}$ for A, $\frac{1}{20}$ for B, $-\frac{3}{104}$ for C, and $-\frac{3}{65}$ for D in equation (21) to find the constant E.

$\begin{array}{l}0=8\left(\frac{1}{40}\right)+6\left(\frac{1}{20}\right)+4\left(-\frac{3}{104}\right)+6\left(-\frac{3}{65}\right)+E\\ E=-\frac{8}{40}-\frac{6}{20}+\frac{12}{104}+\frac{18}{65}\\ E=-\frac{7}{65}\end{array}$

Substitute $\frac{1}{40}$ for A, $\frac{1}{20}$ for B, $-\frac{3}{104}$ for C, $-\frac{3}{65}$ for D, and $-\frac{7}{65}$ for E in equation (15) to find $Y\left(s\right)$.

$\begin{array}{c}Y\left(s\right)=\frac{\left(\frac{1}{40}\right)}{s}+\frac{\left(\frac{1}{20}\right)}{s+2}+\frac{\left(-\frac{3}{104}\right)}{s+4}+\frac{\left(-\frac{3}{65}\right)s+\left(-\frac{7}{65}\right)}{s^2+2s+5}\\ =\frac{\left(\frac{1}{40}\right)}{s}+\frac{\left(\frac{1}{20}\right)}{s+2}+\frac{\left(-\frac{3}{104}\right)}{s+4}+\frac{\left(-\frac{1}{65}\right)\left(3s+7\right)}{s^2+2s+1+4}\\ =\frac{\left(\frac{1}{40}\right)}{s}+\frac{\left(\frac{1}{20}\right)}{s+2}+\frac{\left(-\frac{3}{104}\right)}{s+4}+\frac{\left(-\frac{1}{65}\right)\left(3s+3+4\right)}{{\left(s+1\right)}^2+4}\left\{\because {\left(a+b\right)}^2=a^2+2ab+b^2\right\}\end{array}$

Reduce the equation as follows,

$\begin{array}{c}Y\left(s\right)=\frac{\left(\frac{1}{40}\right)}{s}+\frac{\left(\frac{1}{20}\right)}{s+2}+\frac{\left(-\frac{3}{104}\right)}{s+4}+\frac{\left(-\frac{1}{65}\right)\left(3s+3\right)+\left(-\frac{1}{65}\right)4}{{\left(s+1\right)}^2+4}\\ =\frac{\left(\frac{1}{40}\right)}{s}+\frac{\left(\frac{1}{20}\right)}{s+2}+\frac{\left(-\frac{3}{104}\right)}{s+4}+\frac{\left(-\frac{3}{65}\right)\left(s+1\right)}{{\left(s+1\right)}^2+4}+\frac{\left(-\frac{1}{65}\right)4}{{\left(s+1\right)}^2+4}\end{array}$

$Y\left(s\right)=\frac{\left(\frac{1}{40}\right)}{s}+\frac{\left(\frac{1}{20}\right)}{s+2}+\frac{\left(-\frac{3}{104}\right)}{s+4}+\frac{\left(-\frac{3}{65}\right)\left(s+1\right)}{{\left(s+1\right)}^2+2^2}+\frac{\left(-\frac{2}{65}\right)2}{{\left(s+1\right)}^2+2^2}$ (22)

Apply inverse Laplace transform given in equation (3) to equation (22). Therefore,

$\begin{array}{c}y\left(t\right)={\mathcal{L}}^{-1}\left[Y\left(s\right)\right]\\ ={\mathcal{L}}^{-1}\left[\frac{\left(\frac{1}{40}\right)}{s}+\frac{\left(\frac{1}{20}\right)}{s+2}+\frac{\left(-\frac{3}{104}\right)}{s+4}+\frac{\left(-\frac{3}{65}\right)\left(s+1\right)}{{\left(s+1\right)}^2+2^2}+\frac{\left(-\frac{2}{65}\right)2}{{\left(s+1\right)}^2+2^2}\right]\\ =\left\{\begin{array}{l}{\mathcal{L}}^{-1}\left[\frac{\left(\frac{1}{40}\right)}{s}\right]+{\mathcal{L}}^{-1}\left[\frac{\left(\frac{1}{20}\right)}{s+2}\right]+{\mathcal{L}}^{-1}\left[\frac{\left(-\frac{3}{104}\right)}{s+4}\right]+{\mathcal{L}}^{-1}\left[\frac{\left(-\frac{3}{65}\right)\left(s+1\right)}{{\left(s+1\right)}^2+2^2}\right]\\ +{\mathcal{L}}^{-1}\left[\frac{\left(-\frac{2}{65}\right)2}{{\left(s+1\right)}^2+2^2}\right]\end{array}\right\}\end{array}$

$y\left(t\right)=\left\{\begin{array}{l}\frac{1}{40}{\mathcal{L}}^{-1}\left[\frac{1}{s}\right]+\frac{1}{20}{\mathcal{L}}^{-1}\left[\frac{1}{s+2}\right]-\frac{3}{104}{\mathcal{L}}^{-1}\left[\frac{1}{s+4}\right]-\frac{3}{65}{\mathcal{L}}^{-1}\left[\frac{\left(s+1\right)}{{\left(s+1\right)}^2+2^2}\right]\\ -\frac{2}{65}{\mathcal{L}}^{-1}\left[\frac{2}{{\left(s+1\right)}^2+2^2}\right]\end{array}\right\}$ (23)

Apply inverse Laplace transform function given in equation (8), (9), (10) and (11) to equation (23).

$\begin{array}{c}y\left(t\right)=\frac{1}{40}u\left(t\right)+\frac{1}{20}{e}^{-2t}u\left(t\right)-\frac{3}{104}{e}^{-4t}u\left(t\right)-\frac{3}{65}{e}^{-t}\cos \left(2t\right)u\left(t\right)-\frac{2}{65}{e}^{-t}\sin \left(2t\right)u\left(t\right)\\ =\left(\frac{1}{40}+\frac{1}{20}{e}^{-2t}-\frac{3}{104}{e}^{-4t}-\frac{3}{65}{e}^{-t}\cos \left(2t\right)-\frac{2}{65}{e}^{-t}\sin \left(2t\right)\right)u\left(t\right)\end{array}$

Conclusion:

Thus, the time-varying function $y\left(t\right)$ for the given differential equation is $\left(\frac{1}{40}+\frac{1}{20}{e}^{-2t}-\frac{3}{104}{e}^{-4t}-\frac{3}{65}{e}^{-t}\cos \left(2t\right)-\frac{2}{65}{e}^{-t}\sin \left(2t\right)\right)u\left(t\right)$.