ExamplesB

Section 6.4 ExamplesB

Example 6.4.1.

Find the Laplace transform of \((1+\cos2t)\text{.}\)

Solution.

\begin{equation*} \mathscr{L}(1+\cos2t)=\int\limits_{0}^{\infty}e^{-st}(1+\cos2t)\,dt \end{equation*}

\begin{equation*} = \int\limits_{0}^{\infty}e^{-st}.1\,dt + \int\limits_{0}^{\infty}e^{-st}\cos at\,dt \end{equation*}

\begin{equation*} =\frac{1}{s}+ \frac{s}{s^{2}+2^{2}} = \frac{1}{s}+ \frac{s}{s^{2}+4} \end{equation*}

Example 6.4.2.

Find the Laplace transform of \(\cos^{2}t\text{.}\)

Solution.

\begin{equation*} \because \quad \cos2t = 2\cos^{2}t-1; \quad \therefore \cos^{2}t= \frac{1}{2}[\cos 2t+1] \end{equation*}

Now,

\begin{equation*} \mathscr{L}[\cos^{2}t]= \int\limits_{0}^{\infty}e^{-st}\cos^{2}t\,dt \end{equation*}

\begin{equation*} = \frac{1}{2}\int\limits_{0}^{\infty}e^{-st}[\cos 2t+1] =\frac{1}{2}[ \frac{1}{s}+ \frac{s}{s^{2}+4}]. \end{equation*}

(see problem Example 6.4.1)

Example 6.4.3.

Find the Laplace transform of \(t\sin at\text{.}\)

Solution.

\begin{equation*} \mathscr{L}[t\sin at] = \mathscr{L}\left[t\{\frac{e^{iat}-e^{-iat}}{2i}\}\right] =\frac{1}{2i}[\mathscr{L}(te^{iat})-\mathscr{L}(te^{-iat})] \end{equation*}

\begin{equation*} =\frac{1}{2i}\left[\frac{1}{(s-ia)^{2}}-\frac{1}{(s+ia)^{2}}\right]. \end{equation*}

(see formula 8 in Section 6.3)

\begin{equation*} =\frac{1}{2i}\left[\frac{s^{2}+2ias-a^{2}-s^{2}+2ias+a^{2}}{(s^{2}+a^{2})^{2}}\right] \end{equation*}

\begin{equation*} = \frac{1}{2i}\left[\frac{4ias}{(s^{2}+a^{2})^{2}}\right]= \frac{2as}{(s^{2}+a^{2})^{2}} \end{equation*}

Example 6.4.4.

Find the Laplace Transform of \(\cos at \sinh at\text{.}\)

Solution.

\begin{equation*} \mathscr{L}[\cos at \sinh at] = \mathscr{L}\left[\cos at\{\frac{e^{at}-e^{-at}}{2}\}\right] \end{equation*}

\begin{equation*} =\frac{1}{2}[\mathscr{L}(\cos at e^{at})-\mathscr{L}(\cos at e^{-at})] \end{equation*}

\begin{equation*} = \frac{1}{2}\left[\frac{s-a}{(s-a)^{2}+a^{2}}-\frac{s+a}{(s+a)^{2}+a^{2}}\right] \end{equation*}

Example 6.4.5.

Find \(\mathscr{L}[\sin (\omega t + \theta)]\text{.}\)

Solution.

\begin{equation*} \mathscr{L}[\sin (\omega t + \theta)] = \mathscr{L}[\sin \omega t \cos \theta +\cos \omega t \sin \theta] \end{equation*}

\begin{equation*} = \cos \theta \mathscr{L}[\sin \omega t] +\sin \theta \mathscr{L}[\cos \omega t] \end{equation*}

\begin{equation*} = \cos \theta \left[\frac{\omega}{s^{2}+\omega ^{2}}\right] + \sin \theta \left[\frac{s}{s^{2}+\omega ^{2}}\right]. \end{equation*}

Example 6.4.6.

Find the inverse transform of \(\frac{1}{s^{2}-5s+6}\text{.}\)

Solution.

Let us convert the given function into partial fractions.

\begin{equation*} \mathscr{L}^{-1}\left(\frac{1}{s^{2}-5s+6}\right) = \mathscr{L}^{-1}\left(\frac{1}{s-3}\right)-L^{-1}\left(\frac{1}{s-2}\right) \end{equation*}

\begin{equation*} = e^{3t}-e^{2t} \end{equation*}

Example 6.4.7.

Find

\begin{equation*} \mathscr{L}^{-1}\left(\frac{s-1}{s^{2}-6s+25}\right). \end{equation*}

Solution.

Hint.

\begin{equation*} \frac{s-1}{s^{2}-6s+25} = \frac{s-1}{(s-3)^{2}+4^{2}} \end{equation*}

\begin{equation*} =\frac{s-3+2}{(s-3)^{2}+4^{2}} =\frac{s-3}{(s-3)^{2}+4^{2}}+\frac{1}{2} \frac{4}{(s-3)^{2}+4^{2}} \end{equation*}

Example 6.4.8.

Find

\begin{equation*} \mathscr{L}^{-1}\left(\frac{s+4}{s(s-1)(s^{2}+4)}\right). \end{equation*}

Hint.

\begin{equation*} \frac{s+4}{s(s-1)(s^{2}+4)} =\frac{A}{s}+\frac{B}{s-1}+\frac{Cs+D}{s^{2}+4} \end{equation*}

or,

\begin{equation*} s+4 \end{equation*}

\begin{equation} = A(s-1)(s^{2}+4) +Bs(s^{2}+4) +(Cs +D) -s(s-1)\tag{6.4.1} \end{equation}

Solution.

putting, \(s=1\text{,}\) we get - \(4=-4A; \quad \therefore A = -1\text{,}\) again, put \(s=1\text{,}\) we get - \(5=B.5; \quad \therefore B = 1\text{,}\) equating the coefficient of \(s^{3}\) on both sides of equation (6.4.1), we have, \(0=A+B+C\) or,

\begin{equation*} 0=-1+1+C; \quad \therefore C = 0 \end{equation*}

equating the coefficient of \(s\) on both sides of equation (6.4.1), we get- \(1=4A+4B-D =-4+4-D\) or, \(D=-1\text{.}\) On putting \(A,B,C,D\) in equation (6.4.1), we get -

\begin{equation*} \frac{s+4}{s(s-1)(s^{2}+4)} =-\frac{1}{s}+\frac{1}{s-1}-\frac{1}{s^{2}+4} \end{equation*}

Example 6.4.9.

Find

\begin{equation*} \mathscr{L}^{-1}\left(\frac{s^{2}}{(s^{2}+a^{2})(s^{2}+b^{2})}\right). \end{equation*}

Solution.

Here,

\begin{equation*} \left(\frac{s^{2}}{(s^{2}+a^{2})(s^{2}+b^{2})}\right) =\frac{A}{(s^{2}+a^{2})}+ \frac{B}{(s^{2}+b^{2})} \end{equation*}

or,

\begin{equation*} s^{2}=A(s^{2}+b^{2})+B(s^{2}+a^{2}) =s^{2}(A+B)+Ab^{2}+Ba^{2} \end{equation*}

now, \(A+B=1 \) and \(Ab^{2}+Ba^{2}=0 \quad \Rightarrow Ab^{2}=-Ba^{2}\)

\begin{equation*} \therefore \quad -\frac{Ba^{2}}{b^{2}}+B=1 \end{equation*}

or,

\begin{equation*} B\left(1-\frac{a^{2}}{b^{2}}\right)=1 \end{equation*}

or,

\begin{equation*} B\left(\frac{b^{2}-a^{2}}{b2}\right)=1 \end{equation*}

\begin{equation*} \therefore \quad B= \frac{b2}{b^{2}-a^{2}} \end{equation*}

again,

\begin{equation*} Ab^{2}=- \left(\frac{b2}{b^{2}-a^{2}}\right)a^{2} \end{equation*}

or,

\begin{equation*} A=\frac{a2}{a^{2}-b^{2}} \end{equation*}

\begin{equation*} \therefore \quad \frac{s^{2}}{(s^{2}+a^{2})(s^{2}+b^{2})} = \frac{a2}{a^{2}-b^{2}}\frac{1}{s^{2}+a^{2}}-\frac{b2}{a^{2}-b^{2}}\frac{1}{s^{2}+b^{2}} \end{equation*}

\begin{equation*} = \frac{1}{a^{2}-b^{2}}\left[\frac{a^{2}}{s^{2}+a^{2}}-\frac{b^{2}}{s^{2}+b^{2}}\right] \end{equation*}

Hence,

\begin{equation*} \mathscr{L}^{-1}\left(\frac{s^{2}}{(s^{2}+a^{2})(s^{2}+b^{2})}\right) \end{equation*}

\begin{equation*} = \frac{1}{a^{2}-b^{2}}\left[a^{2}\left(\frac{1}{a}\sin at\right)-b^{2}\left(\frac{1}{b}\sin bt\right)\right] \end{equation*}

\begin{equation*} =\frac{1}{a^{2}-b^{2}}[a\sin at-b\sin bt] \end{equation*}

Example 6.4.10.

Find the inverse Laplace Transform of \(\frac{1}{s(s^{2}+1)}\text{.}\)

Solution.

We have

\begin{equation*} \mathscr{L}^{-1}\left[\frac{1}{(s^{2}+1)}\right] = \sin t \end{equation*}

or,

\begin{equation*} \mathscr{L}^{-1}\left[\frac{1}{s(s^{2}+1)}\right] = \int\limits_{0}^{t}1.\sin t\,dt \end{equation*}

or,

\begin{equation*} [-\cos t]_{0}^{t} = (-\cos t +1) \end{equation*}

i.e.,

\begin{equation*} \mathscr{L}^{-1}\left[\frac{1}{s(s^{2}+1)}\right] = 1-\cos t \end{equation*}

Example 6.4.11.

Find the inverse Laplace transform of \(\frac{1}{s^{2}(s^{2}+a^{2})}\text{.}\)

Solution.

We have

\begin{equation*} \mathscr{L}^{-1}\left[\frac{1}{(s^{2}+a^{2})}\right] = \frac{1}{a}\sin at \end{equation*}

or,

\begin{equation*} \mathscr{L}^{-1}\left[\frac{1}{s(s^{2}+a^{2})}\right] =\frac{1}{a}\int\limits_{0}^{t}\sin at\,dt \end{equation*}

\begin{equation*} = \frac{1}{a}\left[\frac{-\cos at}{a}\right]_{0}^{t} \end{equation*}

\begin{equation*} =\frac{1}{a^{2}}[1-\cos at +1] = \frac{1}{a^{2}}[1-\cos at] \end{equation*}

or,

\begin{equation*} \mathscr{L}^{-1}\left[\frac{1}{s^{2}(s^{2}+a^{2})}\right] = \frac{1}{a^{2}}\int\limits_{0}^{t}(1-\cos at)\,dt \end{equation*}

\begin{equation*} = \frac{1}{a^{2}}\left[\left.t\right\vert_{0}^{t}-\left(\frac{\sin at}{a}\right)_{0}^{t}\right] \end{equation*}

\begin{equation*} = \frac{1}{a^{2}}\left[t-\frac{1}{a}\sin at\right] \end{equation*}

Example 6.4.12.

Find

\begin{equation*} \mathscr{L}^{-1}\left[\frac{1}{s^{2}+4s}\right] \end{equation*}

Hint.

Find

\begin{equation*} \mathscr{L}^{-1}\left[\frac{1}{s(s^{2}+2^{2})}\right] \end{equation*}

Solution.

\begin{equation*} \frac{1}{4}(1-\cos 2t) \end{equation*}

Example 6.4.13.

Find the Laplace Transform of \(t\sinh at\text{.}\)

Solution.

\begin{equation*} \mathscr{L}[\sinh at] = \frac{a}{s^{2}-a^{2}} \end{equation*}

or,

\begin{equation*} \mathscr{L}[t\sinh at] =-\frac{d}{\,ds}\left[\frac{a}{s^{2}-a^{2}}\right] = \frac{2as}{(s^{2}-a^{2})^{2}} \end{equation*}

Example 6.4.14.

Find \(\mathscr{L}[t^{2}e^{t}\sin 4t]\text{.}\)

Solution.

\begin{equation*} \mathscr{L}[\sin 4t] = \frac{4}{s^{2}+16} \end{equation*}

or,

\begin{equation*} \mathscr{L}[e^{t}\sin 4t] = \frac{4}{(s-1)^{2}+16} \end{equation*}

or,

\begin{equation*} \mathscr{L}[t^{2}e^{t}\sin 4t] =(-1)^{2}\frac{\,d^{2}}{\,ds^{2}}\left[ \frac{4}{(s-1)^{2}+16}\right] \end{equation*}

\begin{equation*} = \frac{\,d^{2}}{\,ds^{2}}\left[\frac{4}{s^{2}-2s+17}\right] \end{equation*}

\begin{equation*} =\frac{\,d}{\,ds}\left[\frac{-4(2s-2)}{(s^{2}-2s +17)^{2}}\right] \end{equation*}

\begin{equation*} =-4\left[\frac{(s^{2}-2s +17)^{2}\{2s^{2}-4s+34-8s^{2}+16s-8\}}{(s^{2}-2s +17)^{4}}\right] \end{equation*}

\begin{equation*} = \frac{-4\{-6s^{2}+12s+26\}}{(s^{2}-2s +17)^{3}} \end{equation*}

Example 6.4.15.

Find the inverse Laplace Transfoem of the function,

\begin{equation*} \log \left(1+\frac{\omega^{2}}{s^{2}}\right). \end{equation*}

Solution.

Let

\begin{equation*} F(s)= \log (1+\frac{\omega^{2}}{s^{2}}) \end{equation*}

we know that,

\begin{equation*} -\frac{\,d}{\,ds}\log \left(\frac{s^{2}+\omega^{2}}{s^{2}}\right) =-\frac{\,d}{\,ds}[\log (s^{2}+\omega^{2}) -\log s^{2}] \end{equation*}

\begin{equation*} =-\left[\frac{1\cdot2s}{s^{2}+\omega^{2}}-\frac{2s}{s^{2}}\right] = \left[\frac{2}{s}-\frac{2s}{s^{2}+\omega^{2}}\right] \end{equation*}

or,

\begin{equation*} tf(t) =\mathscr{L}^{-1}\left[\frac{2}{s}-\frac{2s}{s^{2}+\omega^{2}}\right] = 2-2\cos \omega t \end{equation*}

or,

\begin{equation*} f(t)=\frac{2}{t}[1-\cos \omega t] \end{equation*}

Example 6.4.16.

Find \(\mathscr{L}^{-1}[\cot^{-1}(1+s)]\text{.}\)

Solution.

Let \(F(s) = \cot^{-1}(1+s)\) we know that

\begin{equation*} \mathscr{L}[tf(t)] =-\frac{\,d}{\,ds}F(s)=-\frac{\,d}{\,ds}\left[\cot^{-1}(1+s)\right] \end{equation*}

\begin{equation*} = -\left[\frac{-1}{1+(1+s)^{2}}\right] = \frac{1}{(s+1)^{2}+1} \end{equation*}

or,

\begin{equation*} tf(t) = \mathscr{L}^{-1}\left[\frac{1}{(s+1)^{2}+1}\right] =e^{-t}\sin t \end{equation*}

\begin{equation*} \therefore f(t) = \frac{1}{t}\left[e^{-t}\sin t\right]. \end{equation*}

Example 6.4.17.

Find the Laplace Transform of \(\frac{\sin 2t}{t}\text{.}\)

Solution.

\begin{equation*} \mathscr{L}[\sin 2t] = \frac{2}{s^{2}+4} \end{equation*}

Now,

\begin{equation*} \mathscr{L}\left[\frac{\sin2t}{t}\right] = \int\limits_{s}^{\infty}\frac{2}{s^{2}+4}\,ds \end{equation*}

\begin{equation*} =2.\frac{1}{2}\left[\tan^{-1}\frac{s}{2}\right]_{s}^{\infty} = \left[\tan^{-1}\infty -\tan^{-1}\frac{s}{2}\right] \end{equation*}

\begin{equation*} = \frac{\pi}{2}-\tan^{-1}\frac{s}{2} = \cot^{-1}\frac{s}{2}. \end{equation*}

Example 6.4.18.

Find the Laplace Transform of \(\frac{1-\cos t}{t^{2}}\text{.}\)

Solution.

\begin{equation*} \mathscr{L}[1-\cos t] = L[1]-L[\cos t] = \frac{1}{s}-\frac{s}{s^{2}+1} \end{equation*}

or,

\begin{equation*} \mathscr{L}\left[\frac{1-\cos t}{t}\right] = \int\limits_{s}^{\infty}\left(\frac{1}{s}-\frac{s}{s^{2}+1}\right)\,ds \end{equation*}

\begin{equation*} = \left[\log s-\frac{1}{2}\log (s^{2}+1)\right]_{s}^{\infty} =\frac{1}{2}\left[\log s^{2}-\log (s^{2}+1)\right]_{s}^{\infty} \end{equation*}

\begin{equation*} =\frac{1}{2}\left[\log \frac{s^{2}}{s^{2}+1}\right]_{s}^{\infty} =\frac{1}{2}\left[\log \frac{s^{2}}{s^{2}(1+1/s^{2})}\right]_{s}^{\infty} \end{equation*}

\begin{equation*} = \frac{1}{2}\left[0-\log \frac{s^{2}}{s^{2}+1}\right] = -\frac{1}{2}\left[\log \frac{s^{2}}{s^{2}+1}\right] \end{equation*}

again

\begin{equation*} \mathscr{L}\left[\frac{1-\cos t}{t^{2}}\right] =- \frac{1}{2} \int\limits_{s}^{\infty}\log \frac{s^{2}}{s^{2}+1} \,ds \end{equation*}

\begin{equation*} = - \frac{1}{2} \int\limits_{s}^{\infty}\log - \frac{1}{2} \int\limits_{s}^{\infty}\log \left(\frac{s^{2}}{s^{2}+1}.1\right)\,ds \end{equation*}

Integrating by parts, we have

\begin{equation*} =- \frac{1}{2}\left[\log \frac{s^{2}}{s^{2}+1}.s-\int\limits_{s}^{\infty}\frac{s^{2}+1}{s^{2}}.\frac{(s^{2}+1).2s -s^{2}(2s)s}{(s^{2}+1)^{2}} \,ds\right]_{s}^{\infty} \end{equation*}

\begin{equation*} =-\frac{1}{2}\left[s \log \frac{s^{2}}{s^{2}+1}-2\int\limits_{s}^{\infty}\frac{1}{s^{2}+1}\,ds\right]_{s}^{\infty} \end{equation*}

\begin{equation*} = -\frac{1}{2}\left[s \log \frac{s^{2}}{s^{2}+1}-2\tan ^{-1}s\right]_{s}^{\infty} \end{equation*}

\begin{equation*} =-\frac{1}{2}\left[0-2(\pi/2) -s \log \frac{s^{2}}{s^{2}+1}+2\tan ^{-1}s\right] \end{equation*}

\begin{equation*} = -\frac{1}{2}\left[-\pi-s \log \frac{s^{2}}{s^{2}+1}+2\tan ^{-1}s\right] \end{equation*}

\begin{equation*} = \frac{\pi}{2}+\frac{s}{2} \log \frac{s^{2}}{s^{2}+1}-\tan^{-1}s \end{equation*}

\begin{equation*} = \left(\frac{\pi}{2}-\tan^{-1}s \right) +\frac{s}{2} \log \frac{s^{2}}{s^{2}+1} \end{equation*}

\begin{equation*} =\cot^{-1}s + \frac{s}{2} \log \frac{s^{2}}{s^{2}+1}. \end{equation*}

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