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Mathematical Methods of Physics: For Undergraduate

Sunil Karna

Section 6.6 Examples C

Example 6.6.1.

Using Laplace Transforms, find the solution of the initial value problem.
\begin{equation*} y''-4y'+4y =64\sin 2t \end{equation*}
and \(y(0) = 0, y'(0)=1\text{.}\)
Solution.
\begin{equation} y''-4y'+4y =64\sin 2t\tag{6.6.1} \end{equation}
Taking Laplace Transform on both sides of equation (6.6.1), we have -
\begin{equation*} \mathscr{L}[y''-4y'+4y] = \mathscr{L}[64\sin 2t] \end{equation*}
or,
\begin{equation*} \mathscr{L}[y'']-4\mathscr{L}[y']+4\mathscr{L}[y] = 64\mathscr{L}[\sin 2t] \end{equation*}
or,
\begin{equation*} s^{2}\mathscr{L}[y(t)]-sy(0)-y'(0)-4[s\mathscr{L}y(t)-y(0)]+4\mathscr{L}[y(t)]=64\frac{2}{s^{2}+4} \end{equation*}
or,
\begin{equation*} s^{2}Y(s)-sy(0)-y'(0)-4sY(s)+4y(0)+4y(s)=\frac{128}{s^{2}+4} \end{equation*}
putting \(y(0) =0\) and \(y'(0) = 1\text{,}\) we get -
\begin{equation*} s^{2}Y(s)-s.0-1-4sY(s)+40+4Y(s)=\frac{128}{s^{2}+4} \end{equation*}
or,
\begin{equation*} Y(s)[s^{2}-4s+4]=1+ \frac{128}{s^{2}+4} \end{equation*}
or,
\begin{equation*} (s-2)^{2}Y(s) = 1+ \frac{128}{s^{2}+4} \end{equation*}
or,
\begin{equation*} Y(s) = \frac{1}{(s-2)^{2}}+ \frac{128}{(s^{2}+4)(s-2)^{2}} \end{equation*}
\begin{equation*} = \frac{1}{(s-2)^{2}}-\frac{8}{(s-2)}+\frac{16}{(s-2)^{2}}+\frac{8s}{s^{2}+4} \end{equation*}
\begin{equation*} \because \quad Y(s) = \mathscr{L}[y(t)] \end{equation*}
\begin{equation*} \therefore \quad y(t) = -8e^{2t}+17te^{2t}+8\cos 2t \end{equation*}

Example 6.6.2.

Solve \(y''-3y'+2y =2e^{-t}\) and \(y(0)=2, y'(0)=1\text{.}\)
Solution.
\begin{equation} y''-3y'+2y =2e^{-t}\tag{6.6.2} \end{equation}
Taking Laplace Transform on both sides, we have -
\begin{equation*} \mathscr{L}[y''-3y'+2y] =\mathscr{L}[2e^{-t}] \end{equation*}
or,
\begin{equation*} \mathscr{L}[y'']-3\mathscr{L}[y']+2\mathscr{L}[y] =\mathscr{L}[2e^{-t}] \end{equation*}
or,
\begin{equation*} s^{2}\mathscr{L}[y(t)]-sy(0)-y'(0) -3[s\mathscr{L}y(t)-y(0)]+2\mathscr{L}[y(t)] =2\frac{1}{s+1} \end{equation*}
or,
\begin{equation*} s^{2}Y(s) -sy(0)-y'(0)-3sY(s) +3y(0)+2Y(s) = \frac{2}{s+1} \end{equation*}
or,
\begin{equation*} (s^{2}-3s+2)Y(s) -2s -1+3\times 2 =\frac{2}{s+1} \end{equation*}
or,
\begin{equation*} (s^{2}-3s+2)Y(s) = \frac{2}{s+1}+2s-5 \end{equation*}
\begin{equation*} = \frac{2+(2s-5)(s+1)}{5} =\frac{2s^{2}-3s-3}{s+1} \end{equation*}
\begin{equation} \therefore \quad Y(s) = \frac{2s^{2}-3s-3}{(s+1)(s^{2}-3s+2)}\tag{6.6.3} \end{equation}
Now, solving by partial fraction, we get -
\begin{equation*} \frac{2s^{2}-3s-3}{(s+1)(s^{2}-3s+2)} = \frac{A}{s+1}+\frac{Bs}{s^{2}-3s+2} \end{equation*}
or,
\begin{equation*} 2s^{2}-3s-3 =A(s^{2}-3s+2)+Bs(s+1) \end{equation*}
put \(s=-1\text{,}\) we get -
\begin{equation*} 2+3-3 =A(1+3+2)+0 \Rightarrow A =\frac{2}{6}=\frac{1}{3} \end{equation*}
put \(s=+1\text{,}\) we get -
\begin{equation*} 2-3-3 =A(1-3+2)+B.2 \Rightarrow B = -\frac{4}{2} =-2 \end{equation*}
Now, from equation (6.6.3), we get -
\begin{equation*} Y(s) = \frac{1}{3(s+1)}-\frac{2s}{s^{2}-3s+2} \end{equation*}
\begin{equation*} \therefore \quad y(t) = \mathscr{L}^{-1}[Y(s)] \end{equation*}
\begin{equation*} = \mathscr{L}^{-1}[\frac{1}{3(s+1)}] -2\mathscr{L}^{-1}[\frac{s}{(s-1)(s-2)}] \end{equation*}
\begin{equation*} = \frac{1}{3}e^{-t}-2\mathscr{L}^{-1}\left[\frac{-1}{s-1}+\frac{2}{s-2}\right] = \frac{1}{3}e^{-t}+2e^{t}-4e^{2t} \end{equation*}

Example 6.6.3.

\(y''-2y'+2y =e^{-x}; y(0) =1, y'(0)=0\text{.}\)
Solution.
we have
\begin{equation*} \mathscr{L}[y''-2y'+2y] =\mathscr{L}[e^{-x}] \end{equation*}
or,
\begin{equation*} s^{2}Y(s)-sy(0)-y'(0)-2s\mathscr{L}[y(x)] +2y(0)+2\mathscr{L}y(x) = \frac{1}{s+1} \end{equation*}
or,
\begin{equation*} s^{2}Y(s)-s.1-2sY(s) +2.1+2Y(s) = \frac{1}{s+1} \end{equation*}
or,
\begin{equation*} (s^{2}-2s+2)Y(s) = \frac{1}{s+1}+(s-2) \end{equation*}
or,
\begin{equation*} Y(s) =\frac{1+(s-2)(s+1)}{(s+1)(s^{2}-2s+2)} \end{equation*}
or,
\begin{equation*} \frac{s^{2}-s-1}{(s+1)(s^{2}-2s+2)} = \frac{A}{s+1}+\frac{Bs+C}{s^{2}-2s+2} \end{equation*}
or,
\begin{equation*} =\frac{1/5}{s+1}+\frac{(4/5)s-(7/5)}{s^{2}-2s+2} \end{equation*}
\begin{equation*} = \frac{1}{5(s+1)}+\frac{1}{5}\frac{4s}{(s-1)^{2}+1}-\frac{7}{5}\frac{1}{(s-1)^{2}+1} \end{equation*}
or,
\begin{equation*} y(x) =\mathscr{L}^{-1}\left[\frac{1}{5(s+1)}+\frac{4}{5}\frac{(s-1)}{(s-1)^{2}+1}-\frac{3}{5}\frac{1}{(s-1)^{2}+1}\right] \end{equation*}
or,
\begin{equation*} = \frac{1}{5}e^{-x}+\frac{4}{5}e^{x}\cos x -\frac{3}{5}e^{x}\sin x \end{equation*}
\begin{equation*} =\frac{1}{5}e^{-x}+\frac{e^{x}}{5}(4\cos x -3\sin x) \end{equation*}

Example 6.6.4.

Find the inverse transform of \(\frac{e^{-3s}}{s^{3}}\text{.}\)
Solution.
We know that
\begin{equation*} \mathscr{L}^{-1}[\frac{1}{s^{2}}] = \frac{t^{2}}{2!}= \frac{1}{2}t^{2} \end{equation*}
By the second shifting theorem, we get -
\begin{equation*} \mathscr{L}^{-1}[\frac{e^{-3s}}{s^{2}}] = \frac{1}{2}(t-3)^{2}.u(t-3) = \begin{cases} 0,\quad 0 \lt t \lt 3\\ \frac{1}{2}(t-3)^{2}, \quad t \lt 3 \end{cases} \end{equation*}

Example 6.6.5.

Find
\begin{equation*} \mathscr{L}^{-1}\left[\frac{se^{-2s}}{s^{2}+16}\right]. \end{equation*}
Solution.
\begin{equation*} \mathscr{L}^{-1}\left[\frac{s}{s^{2}+16}\right]=\cos 4t \end{equation*}
or,
\begin{equation*} \mathscr{L}^{-1}\left[\frac{se^{-2s}}{s^{2}+16}\right]=\cos 4(t-2)u(t-2) \end{equation*}

Example 6.6.6.

Express the following function in terms of units step functions and find its Laplace Transform.
\begin{equation*} f(t)= \begin{cases} 8, & t \lt 2\\ 6, & t \gt 2 \end{cases} \end{equation*}
Solution.
we have
\begin{equation*} f(t)= \begin{cases} 8+0, & t \lt 2\\ 8-2, & t \gt 2 \end{cases} = 8+ \begin{cases} 0, & t \lt 2\\ -2, & t \gt 2 \end{cases} \end{equation*}
\begin{equation*} = 8+ (-2) \begin{cases} 0, & t \lt 2\\ 1, & t \gt 2 \end{cases} =8-2u(t-2) \end{equation*}
or,
\begin{equation*} \mathscr{L}[f(t)] =8\mathscr{L}(1) -2\mathscr{L}[u(t-2)] =\frac{8}{s}-2\frac{e^{-2s}}{s}. \end{equation*}

Example 6.6.7.

Express the following in terms of unit step function and find its Laplace Transform.
\begin{equation*} f(t)= \begin{cases} E, & a \lt t \lt b\\ 0, & t \gt b \end{cases} \end{equation*}
Solution.
\begin{equation*} f(t)= E \begin{cases} 1, & a \lt t \lt b\\ 0, & t \gt b \end{cases} = E[u(t-a)-u(t-b)] \end{equation*}
or,
\begin{equation*} \mathscr{L}[f(t)] = E[\frac{e^{-as}}{s}-\frac{e^{-bs}}{s}] \end{equation*}

Example 6.6.8.

Find the Laplace transform of \(t^{2}u(t-3)\text{.}\)
Solution.
\begin{equation*} t^{2}u(t-3) = [(t-3)^{2}+6(t-3) +9]u(t-3) \end{equation*}
\begin{equation*} =(t-3)^{2}u(t-3)+6(t-3)u(t-3) +9u(t-3) \end{equation*}
or,
\begin{equation*} \mathscr{L}[t^{2}u(t-3] \end{equation*}
\begin{equation*} = \mathscr{L}[(t-3)^{2}u(t-3)]+6\mathscr{L}[(t-3)u(t-3)] +9\mathscr{L}[u(t-3)] \end{equation*}
\begin{equation*} = e^{-3s}\left[\frac{2}{s^{3}}+\frac{6}{s^{2}}+\frac{9}{s}\right] = e^{-3s}L[(t+3)^{2}]. \end{equation*}

Example 6.6.9.

Find the current in the given LCR circuit, if the switch is closed at time \(t=0\) and the initial charge on the capacitor is \(Q_{o}\text{.}\)
Solution.
From the given circuit,
we can have the equation as
\begin{equation*} RI+L\frac{\,dI}{\,dt}+\frac{Q}{C}=E_{o} \end{equation*}
using Kirchhoff’s voltage law.
\begin{equation*} \because \frac{\,dQ}{\,dt} = I \end{equation*}
and
\begin{equation*} \,dQ = I\,dt \end{equation*}
or,
\begin{equation*} Q(t) = \int\limits_{0}^{t}I(t)\,dt+c \end{equation*}
where \(c\) is an integration constant. At \(t=0\text{,}\) \(Q=Q_{o}\text{.}\)
\(\therefore c=Q_{o}\text{.}\) Hence,
\begin{equation*} Q(t) = \int\limits_{0}^{t}I(t)\,dt+Q_{o} \end{equation*}
or,
\begin{equation*} RI+L\frac{\,dI}{\,dt}+\frac{1}{C}\{Q_{o}+\int\limits_{0}^{t}I(t)\,dt\}=E_{o} \end{equation*}
Taking Laplace transform on bothe sides, we get-
\begin{equation*} Ri(s)+L[si(s)-i(0)]+\frac{1}{C}\{Q_{o}\frac{1}{s}+\frac{i(s)}{s}\}=E_{o}\frac{1}{s} \end{equation*}
\(\because \quad \mathscr{L}[I(t)] =i(s)]\) or,
\begin{equation*} \left(R+Ls+\frac{1}{Cs}\right)i(s) = \frac{E_{o}-Q_{o}/C}{s} \end{equation*}
or,
\begin{equation*} i(s)=\frac{E_{o}-Q_{o}/C}{\left(Rs+Ls^{2}+\frac{1}{C}\right)}= \frac{E_{o}-Q_{o}/C}{L\left(\frac{Rs}{L}+s^{2}+\frac{1}{LC}\right)} \end{equation*}
or,
\begin{equation*} =\frac{E_{o}-Q_{o}/C}{L\left[ \{s^{2}+2s\frac{R}{2L}+\left(\frac{R}{2L}\right)^{2}\} +\{\frac{1}{LC}-\left(\frac{R}{2L}\right)^{2}\}\right] } \end{equation*}
\begin{equation*} =\frac{E_{o}-Q_{o}/C}{L\left[ \left(s+\frac{R}{2L}\right)^{2}+\{\frac{1}{LC}-\left(\frac{R^{2}}{4L^{2}}\right)\}\right]} = \frac{E_{o}-Q_{o}/C}{L[(s+a)^{2}+b^{2}]} \end{equation*}
where \(a=\frac{R}{2L} \) and \(b= \sqrt{\frac{1}{LC}-\frac{R^{2}}{4L^{2}}}.\) but
\begin{equation*} \mathscr{L}[e^{-at}\sin bt] = \frac{b}{(s+a)^{2}+b^{2}} \end{equation*}
\begin{equation*} \therefore i(t)= \mathscr{L}^{-1}\left[ \frac{E_{o}-Q_{o}/C}{L\{(s+a)^{2}+b^{2}\}}\right] \end{equation*}
\begin{equation*} = \frac{E_{o}-Q_{o}/C}{L} \left(\frac{1}{b}e^{-at}\sin bt\right) \end{equation*}
\begin{equation*} =\left(\frac{E_{o}-Q_{o}/C}{L\sqrt{\frac{1}{LC}-\frac{R^{2}}{4L^{2}}}}\right)\left\{e^{-\frac{R}{2l}t}\sin \left( \sqrt{\frac{1}{LC}-\frac{R^{2}}{4L^{2}}}\right)t \right\} \end{equation*}
which is the required result for the current at any time \(t\text{.}\)

Example 6.6.10.

Find the current in the given circuit, if the switch is connected at \(t=0\) and disconnected at \(t=a\text{.}\)
Solution.
With reference to circuit,
the conditions under which current i flows are \(i=0\) at \(t=0\text{,}\) i.e.,
\begin{equation*} E(t)= \begin{cases} E, & 0 \lt t \lt a\\ 0, & t \gt a \end{cases} \end{equation*}
Take the Laplace on both sides of the given equation, we get -
\begin{equation*} \mathscr{L}\{Li' +Ri\} = \mathscr{L}\{E(t)\} \end{equation*}
or,
\begin{equation*} Lsi(s) -i(0)+Ri(s) = \int\limits_{0}^{\infty}e^{-st}E(t)\,dt \end{equation*}
or,
\begin{equation*} (sL+R)i(s) = \int\limits_{0}^{a}e^{-st}E(t)\,dt+\int\limits_{a}^{\infty}e^{-st}.0\,dt \end{equation*}
\begin{equation*} = E\left(\frac{e^{-st}}{-s}\right)_{0}^{a} = \frac{E}{s}(-e^{-sa}+1) \end{equation*}
or,
\begin{equation*} i(s) =\frac{E(1-e^{-sa})}{s(sL+R)} = \frac{E}{s(sL+R)} -\frac{Ee^{-sa}}{s(sL+R)} \end{equation*}
on inversion,
\begin{equation} i(t) = \mathscr{L}^{-1}\left[\frac{E}{s(sL+R)}\right] - \mathscr{L}^{-1}\left[\frac{Ee^{-sa}}{s(sL+R)}\right] \tag{6.6.4} \end{equation}
here
\begin{equation*} \mathscr{L}^{-1}\left[ \frac{1}{\{s-(-R/L\}}\right]=e^{-(R/L)t} \end{equation*}
or,
\begin{equation*} \mathscr{L}^{-1}\left[ \frac{1}{s\{s+\frac{R}{L}\}}\right] = \int\limits_{0}^{t}e^{-(R/L)t}\,dt \end{equation*}
\begin{equation*} = \left[\frac{e^{-(R/L)t}}{-(R/L)}\right]_{0}^{t} \end{equation*}
\begin{equation*} =\mathscr{L}\left[\frac{e-(R/L)t}{R}+\frac{1}{R} \right] =\frac{L}{R}\left[1-e^{-(R/L)t}\right]. \end{equation*}
Hence,
\begin{equation*} \mathscr{L}^{-1}\left[\frac{E}{s(sL+R)}\right] = \frac{E}{L}L^{-1}\left[\frac{1}{s(s+R/L)}\right] \end{equation*}
\begin{equation*} = \frac{E}{R}\left[1-e^{-(R/L)t}\right] \end{equation*}
and
\begin{equation*} \mathscr{L}^{-1}\left[\frac{Ee^{-sa}}{s(sL+R)}\right] = \frac{E}{R}\left[1-e^{-(R/L)(t-a)}\right]u(t-a) \end{equation*}
(by second shifting theorem.)
Therefore from equation (6.6.4), we get -
\begin{equation*} i(t) = \frac{E}{R}\left[1-e^{-(R/L)t}\right]-\frac{E}{R}\left[1-e^{-(R/L)(t-a)}\right]u(t-a) \end{equation*}
\begin{equation*} i(t) =\frac{E}{R}\left[1-e^{-(R/L)t}\right], \end{equation*}
for \(0 \lt t \lt a; u(t-a) =0\text{,}\) and
\begin{equation*} i(t) = \frac{E}{R}\left[1-e^{-(R/L)t}\right]-\frac{E}{R}\left[1-e^{-(R/L)(t-a)}\right].1, \end{equation*}
for \(t \gt a; u(t-a) = 1 \text{.}\)
\begin{equation*} =\frac{E}{R}\left[e^{-(R/L)(t-a)}-e^{-(R/L)t}\right] \end{equation*}
\begin{equation*} = \frac{E}{R}\left[e^{(Ra/L)}-1\right]e^{-(R/L)t}. \end{equation*}

Example 6.6.11.

Solve the given equation of simple harmonic motion using Laplace Transform
\begin{equation*} m\ddot{x}+kx =0 \end{equation*}
at \(x(0) =0\) and \(x'(0) =1\text{.}\)
Solution.
Taking Laplace transform, we have -
\begin{equation*} \mathscr{L}[\ddot{x}+(k/m)x]=0 \end{equation*}
or,
\begin{equation*} s^{2}x(s) -sx(0) -x'(0)+(k/m)x(s) =0 \end{equation*}
or,
\begin{equation*} \{s^{2}+(k/m)\}x(s) =1 \end{equation*}
\begin{equation*} \therefore x(s) = \frac{1}{\{s^{2}+(k/m)\}}=\frac{1}{s^{2}+a^{2}} \end{equation*}
we know that
\begin{equation*} \mathscr{L}[\sin at] =\frac{a}{s^{2}+a^{2}} \end{equation*}
Hence,
\begin{equation*} \mathscr{L}^{-1}[x(s)] =x(t)=\mathscr{L}^{-1}[\frac{1}{s^{2}+a^{2}}] = \frac{1}{a}\sin at \end{equation*}
\begin{equation*} \therefore x(t) = \frac{1}{\sqrt{k/m}}\sin (k/m)t. \end{equation*}

Example 6.6.12.

Find the Laplace transform of the waveform
\begin{equation*} f(t) = \left(\frac{2t}{3}\right) \end{equation*}
if \(0 \leq t \leq 3\text{.}\)
Solution.
\begin{equation*} \mathscr{L}\{f(t)\} = \frac{\int\limits_{0}^{T}e^{-st} f(t)\,dt}{1-e^{-sT}} \end{equation*}
or,
\begin{equation*} \mathscr{L}\{\frac{2t}{3}\} = \frac{\int\limits_{0}^{3}e^{-st} \frac{2t}{3}\,dt}{1-e^{-3s}} = \frac{1}{1-e^{-3s}}\frac{2}{3}\left[t.\frac{e^{-st}}{-s}-1.\frac{-st}{+s^{2}}\right]_{0}^{3} \end{equation*}
as \(T = 3\text{.}\) Or,
\begin{equation*} =\frac{1}{1-e^{-3s}}\frac{2}{3}\left[3.\frac{e^{-3s}}{-s}-\frac{e^{-3s}}{+s^{2}}+\frac{1}{s^{2}}\right] \end{equation*}
\begin{equation*} = \frac{2e^{-3s}}{-s(1-e^{-3s})}+\frac{2}{3s^{2}}. \end{equation*}

Example 6.6.13.

Find the Laplace transform of the half wave rectifier.
\begin{equation*} f(t) =\begin{cases} \sin \omega, \quad 0 \lt t \lt \pi/\omega\\ 0, \quad \pi/\omega \lt t \lt 2\pi/\omega \end{cases} \end{equation*}
Solution.
\begin{equation*} \mathscr{L}\{f(t)\}=\frac{1}{1-e^{-st}}\int\limits_{0}^{T}e^{-st} f(t)\,dt \end{equation*}
\begin{equation*} = \frac{1}{1-e^{-s(2\pi/\omega )s}}\int\limits_{0}^{2\pi/\omega}e^{-st} f(t)\,dt \end{equation*}
\begin{equation*} =\frac{1}{1-e^{-s(2\pi/\omega )s}}\left[\int\limits_{0}^{\pi/\omega}e^{-st} \sin \omega t\,dt+ \int\limits_{\pi/\omega}^{2\pi/\omega}e^{-st}.0 \,dt\right] \end{equation*}
we know that,
\begin{equation*} \int e^{ax}\sin bx\,dx = e^{ax}\left(\frac{a\sin bx-b\cos bx}{a^{2}+b^{2}}\right) \end{equation*}
\begin{equation*} =\frac{1}{1-e^{-(2\pi s/\omega)}}\left[\frac{\omega e^{-\pi s/\omega}+\omega}{s^{2}+\omega^{2}}\right] \end{equation*}
\begin{equation*} = \frac{\omega [1+e^{-\pi s/\omega}]}{(s^{2}+\omega^{2})[[1-e^{-2\pi s/\omega}]} \end{equation*}
\begin{equation*} = \frac{\omega}{(s^{2}+\omega^{2})[1-e^{-\pi s/\omega}]} \end{equation*}

Example 6.6.14.

Find the Laplace transform of the saw tooth wave.
\begin{equation*} f(t) = \frac{kt}{T} \end{equation*}
for \(0 \lt t \lt T \) and \(f(t+T) =f(t)\text{.}\)
Solution.
\begin{equation*} \mathscr{L}\{f(t)\}=\frac{1}{1-e^{-sT}}\int\limits_{0}^{T}e^{-st}f(t)\,dt \end{equation*}
\begin{equation*} = \frac{1}{1-e^{-sT}}\int\limits_{0}^{T}e^{-st}.\frac{kt}{T}\,dt = \frac{1}{1-e^{-sT}}\frac{k}{T}\int\limits_{0}^{T}e^{-st}.t\,dt \end{equation*}
\begin{equation*} = -\frac{ke^{-sT}}{s(1-e^{-sT})}+\frac{k}{Ts^{2}}. \end{equation*}

Example 6.6.15.

A periodic square wave function \(f(t)\) in terms of unit step function is given as
\begin{equation*} f(t) =k[u_{o}(t)-2u_{a}(t)+2u_{2a}(t)-2u_{3a}(t)+\cdots]. \end{equation*}
Show that the Laplce Transform of \(f(t)\) is given by
\begin{equation*} \mathscr{L}\{f(t)\}=\frac{k}{s}\tanh (\frac{as}{2}). \end{equation*}
Solution.
\begin{equation*} f(t) =k[u_{o}(t)-2u_{a}(t)+2u_{2a}(t)-2u_{3a}(t)+\cdots] \end{equation*}
\begin{equation*} \text{or,}\quad \mathscr{L}\{f(t)\}=k[\mathscr{L}u_{o}(t)-2\mathscr{L} u_{a}(t)+\cdots] \end{equation*}
\begin{equation*} = \frac{k}{s}\left[1-2\frac{e^{-as}}{1+e^{-as}}\right]= \frac{k}{s}\left[\frac{1-e^{-as}}{1+e^{-as}}\right] \end{equation*}
\begin{equation*} = \frac{k}{s}\left[\frac{e^{as/2}-e^{-as/2}}{e^{as/2}+e^{-as/2}}\right] = \frac{k}{s}\tanh \left(\frac{as}{2}\right) \end{equation*}

Example 6.6.16.

Use convolution theorem to find
\begin{equation*} \mathscr{L}^{-1}\{\frac{s^{2}}{(s^{2}+a^{2})(s^{2}+b^{2})}\}, \text{for} \quad a \neq b. \end{equation*}
Solution.
We have,
\begin{equation*} \{\frac{s^{2}}{s^{2}+a^{2}}\} = \mathscr{L}\{\cos at \} \end{equation*}
and
\begin{equation*} \{\frac{s^{2}}{s^{2}+b^{2}}\} = \mathscr{L}\{\cos bt \} \end{equation*}
Hence by convolution theorem,
\begin{equation*} \mathscr{L}\{\int\limits_{0}^{t}\cos ax \cos b(t-x) \,dx \} = \frac{s^{2}}{(s^{2}+a^{2})(s^{2}+b^{2})} \end{equation*}
\begin{equation*} \therefore \quad \mathscr{L}^{-1}\{\frac{s^{2}}{(s^{2}+a^{2})(s^{2}+b^{2})}\} = \int\limits_{0}^{t}\cos ax \cos b(t-x) \,dx \end{equation*}
\begin{equation*} = \frac{1}{2} \int\limits_{0}^{t}[\cos ax+bt-bx) +\cos(ax-bt+bx)]\,dx \end{equation*}
\begin{equation*} =\frac{1}{2} \int\limits_{0}^{t}\cos [(a-b)x+bt]\,dx +\frac{1}{2} \int\limits_{0}^{t}\cos[(a+b)x-bt]\,dx \end{equation*}
\begin{equation*} =\left[\frac{\sin [(a-b)x+bt]}{2(a-b)}\right]_{0}^{t}+\left[\frac{\sin [(a+b)x-bt]}{2(a+b)}\right]_{0}^{t} \end{equation*}
\begin{equation*} =\frac{\sin at -\sin bt}{2(a-b)}+\frac{\sin at +\sin bt}{2(a+b)}= \frac{a\sin at -b\sin bt}{a^{2}-b^{2}} \end{equation*}

Example 6.6.17.

Solve the equation,
\begin{equation*} y''(t) +ty'(t)-y(t) =0, \end{equation*}
such that \(y(0) =0, y'(0) =1\text{.}\)
Solution.
we have the equation
\begin{equation} y''(t) +ty'(t)-y(t) =0\tag{6.6.5} \end{equation}
Taking Laplace transform of equation (6.6.5) we have
\begin{equation*} \mathscr{L}\{y''(t) +ty'(t)-y(t)\} =\mathscr{L}\{0\} \end{equation*}
or,
\begin{equation*} \mathscr{L}\{y''(t)\} +\mathscr{L}\{ty'(t)\}-\mathscr{L}\{y(t)\} =\mathscr{L}\{0\} \end{equation*}
or,
\begin{equation*} s^{2}Y(s) -sy(0) -y'(0) +(-1)^{1}\frac{\,d}{\,ds} [sY(s)-y(0)] -Y(s) =0 \end{equation*}
\begin{equation*} [\because \mathscr{L}[y'(t)] = sY(s) -y(0)] \end{equation*}
or,
\begin{equation*} s^{2}Y(s)-1-\frac{\,d}{\,ds}\{sY(s)\}-Y(s) =0 \end{equation*}
or,
\begin{equation*} s^{2}Y(s)-[sY'(s)+Y(s)]-Y(s) =1 \end{equation*}
or,
\begin{equation*} s^{2}Y(s)-sY'(s)-Y(s)]-Y(s) =1 \end{equation*}
or,
\begin{equation*} (s^{2}-2)Y(s)-sY'(s) =1 \end{equation*}
or,
\begin{equation*} Y'(s)-(s-2/s)Y(s) =-1/s \end{equation*}
or,
\begin{equation*} Y'(s)+(\frac{2}{s}-s)Y(s) =-\frac{1}{s} \end{equation*}
which is a linear equation of the first order, like \([y' +P(x)y =Q(x)]\text{.}\) The integrating factor of which is
\begin{equation*} I = \int P(x)\,dx =\int(\frac{2}{s}-s)\,ds =2\log s-\frac{s^{2}}{2} \end{equation*}
or,
\begin{equation*} e^{I}=e^{2\log s-\frac{s^{2}}{2}} =e^{\log s^{2}-\frac{s^{2}}{2}} =s^{2}e^{-s^{2}/2} \end{equation*}
\begin{equation*} \therefore \frac{\,d}{\,ds}[ye^{I}]=Qe^{I} \end{equation*}
or,
\begin{equation*} \frac{\,d}{\,ds} \left[Y(s)s^{2}e^{-s^{2}/2}\right] =-\frac{1}{s} s^{2}e^{-s^{2}/2} \end{equation*}
or,
\begin{equation*} Y(s)s^{2}e^{-s^{2}/2}=\int-se^{-s^{2}/2}\,ds =+\frac{1}{2}\frac{e^{-s^{2}/2}}{1/2}+C \end{equation*}
\begin{equation*} \therefore Y(s) = \frac{1}{s^{2}}+\frac{C}{s^{2}}e^{s^{2}/2} \end{equation*}
Since \(Y(s)\) vanishes as \(x \to \infty\text{,}\) hence C must vanish.
\begin{equation*} \therefore \quad Y(s) = \frac{1}{s^{2}} \end{equation*}
or,
\begin{equation*} y(t) = \mathscr{L}^{-1}\{\frac{1}{s^{2}}\} =\frac{t}{1!}=t \end{equation*}

Example 6.6.18.

Prove that
\begin{equation*} \int\limits_{0}^{\infty}e^{-x^{2}} \,dx = \frac{\sqrt{\pi}}{2}. \end{equation*}
Solution.
Let
\begin{equation*} f(t) = \int\limits_{0}^{\infty}e^{-tx^{2}}\,dx \end{equation*}
But,
\begin{equation*} \mathscr{L}\{f(t)\} = \int\limits_{0}^{\infty}e^{-st}f(t)\,dt \end{equation*}
\begin{equation*} = \int\limits_{0}^{\infty} e^{-st}\left[\int\limits_{0}^{\infty} e^{-tx^{2}}\,dx\right]\,dt \end{equation*}
\begin{equation*} = \int\limits_{0}^{\infty}\left[ \int\limits_{0}^{\infty} e^{-st} e^{-tx^{2}}\,dt\right]\,dx \end{equation*}
\begin{equation*} = \int\limits_{0}^{\infty}\mathscr{L}\{e^{-tx^{2}}\}\,dx = \int\limits_{0}^{\infty}\frac{1}{s+x^{2}}\,dx \end{equation*}
\begin{equation*} =\left[\frac{1}{\sqrt{s}}\tan^{-1}\frac{x}{\sqrt{s}}\right]_{0}^{\infty} \end{equation*}
\begin{equation*} = \frac{\pi}{2\sqrt{s}} \left[\because \quad \mathscr{L}\{e^{at}\} =\frac{1}{s-a}\right] \end{equation*}
\begin{equation*} \therefore \quad f(t)=\frac{\pi}{2} \mathscr{L}^{-1}\{\frac{1}{\sqrt{s}}\} \end{equation*}
\begin{equation*} =\frac{\pi}{2}\frac{1}{\sqrt{\pi}t}=\frac{1}{2}\sqrt{\frac{\pi}{t}} \end{equation*}
or,
\begin{equation*} \int\limits_{0}^{\infty}e^{-tx^{2}}\,dx = \frac{1}{2}\sqrt{\frac{\pi}{t}} \end{equation*}
putting \(t=1\text{,}\) we get the result.

Example 6.6.19.

Solve \(ty''(t)+y'(t) +ty(t) =0\)at \(y(0)=1\) and \(y'(0)\) is constant.
Solution.
Taking Laplace transform of the given equation, we have
\begin{equation*} \mathscr{L}[ty''(t)+y'(t) +ty(t)] = 0 \end{equation*}
or,
\begin{equation*} (-1)\frac{\,d}{\,ds}\mathscr{L}[y''(t)] +\mathscr{L}[y'(t)-\frac{\,d}{\,ds}\mathscr{L}[y(t)]=0 \end{equation*}
or,
\begin{equation*} -\frac{\,d}{\,ds}[s^{2}Y(s)-s-k]+sY(s) -1-Y'(s) =0 \end{equation*}
or,
\begin{equation*} -[2sY(s)+s^{2}Y'(s)-1]+sY(s) -1-Y'(s) = 0 \end{equation*}
or,
\begin{equation*} -2sY(s) -s^{2}Y'(s)+1+sY(s)-1-Y'(s) =0 \end{equation*}
or,
\begin{equation*} (s-2s)Y(s)-(s^{2}+1)Y'(s) =0 \end{equation*}
or,
\begin{equation*} \frac{Y'(s)}{Y(s)}=-\frac{s}{s^{2}+1}=-\frac{1}{2}\frac{2s}{s^{2}+1} \end{equation*}
Integrating, we get -
\begin{equation*} \log Y(s) = -\frac{1}{2}\log (s^{2}+1)+\log C \end{equation*}
\begin{equation*} \therefore \quad Y(s) = \frac{C}{\sqrt{s^{2}+1}} \end{equation*}
Taking inverse Laplace transform,
\begin{equation*} \mathscr{L}^{-1}\{Y(s)\}=C\mathscr{L}^{-1}\{\frac{C}{\sqrt{s^{2}+1}}\} \end{equation*}
\begin{equation*} \therefore \quad y(t) = CJ_{o}(t) \end{equation*}
\(J_{o}(t)\) is Bessel’s function when \(t=0\text{,}\) \(y(t) =y(0) = 1.\) \(\therefore \quad 1= CJ_{o}(0) =C,\) \(\because \quad J_{o}(0)=1.\) which is the required solution.
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