Solution. The two facts needed are \(\mathcal {L}\{t^n\} = \dfrac {n!}{s^{n+1}}\) for \(s>0\) and \(\mathcal {L}\{e^{at}\} = \dfrac {1}{s-a}\) for \(s>a\); both follow from the defining integral, and each converges only where the exponential decays.

(a)

\[\mathcal {L}\big \{9t^3-e^{-t}\big \} = \frac {54}{s^4} - \frac {1}{s+1}, \qquad s>0 .\]

(b)

\(\dfrac {3}{s-4}\), for \(s>4\).

(c)

Using \(\cosh 7t = \dfrac {e^{7t}+e^{-7t}}{2}\) and the corresponding form for \(\sinh \), \[\frac {2s}{s^2-49} + \frac {21}{s^2-49} = \frac {2s+21}{s^2-49}, \qquad s>7 .\]

(d)

Expand first: \(\big (3e^{-4t}+4\big )^2 = 9e^{-8t}+24e^{-4t}+16\), so \[\frac {9}{s+8} + \frac {24}{s+4} + \frac {16}{s},\qquad s>0 ,\] the binding condition coming from the constant term.

Solution.

(a)

This is not a shift in \(t\) of the kind the second shifting theorem handles, since there is no unit step; expand instead: \[\sin \Big (3t-\frac 12\Big ) = \sin 3t\cos \tfrac 12 - \cos 3t\sin \tfrac 12 ,\] so the transform is \[\frac {3\cos \frac 12 - s\sin \frac 12}{s^2+9}.\]

(b)

Use the product-to-sum identity \(\cos A\sin B = \dfrac {\sin (A+B)-\sin (A-B)}{2}\), giving \(\cos 3t\sin 4t = \dfrac {\sin 7t+\sin t}{2}\), so the transform is \[\frac 12\Big (\frac {7}{s^2+49} + \frac {1}{s^2+1}\Big ).\]

(c)

Term by term, \[\frac {5}{s+6} + \frac {72}{s^5} + \frac {3s}{s^2-16} - \frac {10}{s^2-25}.\]

Solution. The second shifting theorem is \(\mathcal {L}\{g(t)u(t-a)\} = e^{-as}\mathcal {L}\{g(t+a)\}\), so the function must first be rewritten in terms of \(\tau = t-a\).

(a)

Here \(a=7\), and with \(t = \tau +7\), \[(t-3)^3 = (\tau +4)^3 = \tau ^3+12\tau ^2+48\tau +64 .\] Hence \[\mathcal {L} = e^{-7s}\Big (\frac {6}{s^4} + \frac {24}{s^3} + \frac {48}{s^2} + \frac {64}{s}\Big ).\] Note the shift in the step is \(7\) while the shift inside the cube is \(3\); taking them to be the same is the usual error.

(b)

Here \(a=2\), and with \(t = \tau +2\), \[t^3e^{-4t} = (\tau +2)^3e^{-4\tau -8} = e^{-8}\big (\tau ^3+6\tau ^2+12\tau +8\big )e^{-4\tau }.\] Using the first shifting theorem on each term, \[\mathcal {L} = e^{-2s-8}\Big (\frac {6}{(s+4)^4} + \frac {12}{(s+4)^3} + \frac {12}{(s+4)^2} + \frac {8}{s+4}\Big ).\]

Solution.

(a)

Let \(f = \sinh ^2\omega t\), so \(f(0)=0\) and \(f' = 2\omega \sinh \omega t\cosh \omega t = \omega \sinh 2\omega t\). Then \[\mathcal {L}\{f'\} = sF(s) - 0 \implies \omega \cdot \frac {2\omega }{s^2-4\omega ^2} = sF(s),\] so \[F(s) = \frac {2\omega ^2}{s\big (s^2-4\omega ^2\big )}.\] The identity \(\sinh ^2\omega t = \dfrac {\cosh 2\omega t-1}{2}\) gives the same answer directly, and agreement is worth checking.

(b)

Here the quicker route is the derivative-of-transform rule \(\mathcal {L}\{tf(t)\} = -F'(s)\) with \(f = \sin \frac {\pi t}{2}\), whose transform is \(\dfrac {\pi /2}{s^2+\pi ^2/4}\). Differentiating, \[\mathcal {L}\Big \{t\sin \frac {\pi t}{2}\Big \} = \frac {\pi s}{\big (s^2+\frac {\pi ^2}{4}\big )^2}.\]

Solution.

(i)

\[\int _0^t\big (x^2-\cosh 3x\big )dx = \frac {t^3}{3} - \frac {\sinh 3t}{3},\] whose transform is \[\frac 13\cdot \frac {6}{s^4} - \frac 13\cdot \frac {3}{s^2-9} = \frac {2}{s^4} - \frac {1}{s^2-9}.\]

(ii)

\(F(s) = \mathcal {L}\{t^2-\cosh 3t\} = \dfrac {2}{s^3} - \dfrac {s}{s^2-9}\), so \[\frac {F(s)}{s} = \frac {2}{s^4} - \frac {1}{s^2-9},\] the same. The second route avoids the integration entirely, which is the point of setting the question both ways.

Solution.

(a)

With \(F = \dfrac {1}{s+\alpha }\), differentiating \(n\) times gives \(\dfrac {(-1)^nn!}{(s+\alpha )^{n+1}}\), so \[\mathcal {L}\big \{t^ne^{-\alpha t}\big \} = \frac {n!}{(s+\alpha )^{n+1}}.\]

(b)

By (a) with \(\alpha = -5\) and \(n=4\), \[\frac {3\cdot 4!}{(s-5)^5} + \frac {1}{s^2} = \frac {72}{(s-5)^5} + \frac {1}{s^2}.\]

(c)

Expand and use the first shifting theorem: \((t-4)^2e^{-3t} = \big (t^2-8t+16\big )e^{-3t}\), so \[\frac {2}{(s+3)^3} - \frac {8}{(s+3)^2} + \frac {16}{s+3}.\]

(d)

This is not a transform but a number: it is \(\mathcal {L}\{t\cos 2t\}\) evaluated at \(s=2\). Now \[\mathcal {L}\{t\cos 2t\} = -\frac {d}{ds}\frac {s}{s^2+4} = \frac {s^2-4}{\big (s^2+4\big )^2},\] and at \(s=2\) the numerator vanishes, so \[\int _0^{\infty }te^{-2t}\cos 2t\,dt = 0 .\]

Solution. Neither transform exists. The rule \(\mathcal {L}\Big \{\dfrac {f(t)}{t}\Big \} = \int _s^{\infty }F(u)\,du\) requires \(\dfrac {f(t)}{t}\) to remain bounded as \(t\rightarrow 0\), and in both cases it does not.

(a)

\(\cosh ^2 0 = 1\), so near the origin \(\dfrac {\cosh ^2t}{t}\sim \dfrac 1t\) and \[\int _0 \frac {\cosh ^2t}{t}e^{-st}dt\] diverges logarithmically at the lower limit. Numerically, cutting the integral off at \(t=\varepsilon \) with \(s=5\) gives \(4.77\), \(7.07\) and \(9.37\) for \(\varepsilon = 10^{-3}, 10^{-4}, 10^{-5}\) — growing without bound, as \(\ln \frac 1\varepsilon \) does.

(b)

Here \(3-3e^{-3t} = 9t - \frac {27}{2}t^2+\cdots \), so the quotient behaves like \(\dfrac {9}{t^2}\) near the origin, and the divergence is worse.

The versions that do converge

Each becomes a standard exercise if the power of \(t\) matches the order of vanishing of the numerator. Since \(\sinh ^2 t\) vanishes to second order, \[\mathcal {L}\Big \{\frac {\sinh ^2t}{t}\Big \} = \int _s^{\infty }\frac {2\,du}{u\big (u^2-4\big )} = \frac 14\ln \frac {s^2}{s^2-4},\] and with a single power of \(t\), \[\mathcal {L}\Big \{\frac {3-3e^{-3t}}{t}\Big \} = 3\int _s^{\infty }\Big (\frac 1u-\frac {1}{u+3}\Big )du = 3\ln \frac {s+3}{s}.\] Both were confirmed numerically. The first of these is exactly what the next question needs.

Solution.

(a)

Using \(\mathcal {L}\{t^a\} = \dfrac {\Gamma (a+1)}{s^{a+1}}\) with \(a = \dfrac 32\), and \(\Gamma \big (\tfrac 52\big ) = \dfrac {3\sqrt \pi }{4}\), \[\mathcal {L}^{-1}\Big \{\frac {3}{s^{5/2}}\Big \} = \frac {3t^{3/2}}{\Gamma (5/2)} = \frac {4t^{3/2}}{\sqrt \pi }.\]

(b)

Each term is read with the second shifting theorem: \[t - (t-1)u(t-1) + u(t-2).\]

(c)

The denominator factors as \(s^4+20s^2+64 = \big (s^2+4\big )\big (s^2+16\big )\), and \[\frac {24}{\big (s^2+4\big )\big (s^2+16\big )} = 2\Big (\frac {1}{s^2+4} - \frac {1}{s^2+16}\Big ),\] so the inverse is \[\sin 2t - \frac {\sin 4t}{2}.\]

(d)

From the previous question, \(\mathcal {L}\Big \{\dfrac {\sinh ^2t}{t}\Big \} = \dfrac 14\ln \dfrac {s^2}{s^2-4}\), so \[\mathcal {L}^{-1}\Big \{\ln \frac {s^2}{s^2-4}\Big \} = \frac {4\sinh ^2t}{t} = \frac {2\big (\cosh 2t-1\big )}{t},\] confirmed numerically at \(s=5\).

(f)

Complete the square: \(s^2+6s+13 = (s+3)^2+4\), and write the numerator to match: \[\frac {2s+3}{(s+3)^2+4} = \frac {2(s+3)-3}{(s+3)^2+4},\] so the inverse is \[e^{-3t}\Big (2\cos 2t - \frac 32\sin 2t\Big ).\]

Solution.

(a)

Transforming and using the initial values, \[\big (s^2Y-s-1\big ) + 4\big (sY-1\big ) + 4Y = \frac {1}{s+4},\] so \[(s+2)^2Y = \frac {1}{s+4} + s + 5 .\] Hence \[Y = \frac {1}{(s+4)(s+2)^2} + \frac {s+5}{(s+2)^2}.\] Partial fractions give \[Y = \frac {1}{4(s+4)} + \frac {3}{4(s+2)} + \frac {7}{2(s+2)^2},\] and inverting, \[y = \frac {e^{-4t}}{4} + \Big (\frac 34 + \frac {7t}{2}\Big )e^{-2t}.\]

(b)

With all initial values zero, \[\big (s^3+2s^2+s\big )Y = \frac {1}{s-3} \implies Y = \frac {1}{s(s+1)^2(s-3)}.\] Partial fractions give \[Y = -\frac {1}{3s} + \frac {1}{48(s-3)} + \frac {5}{16(s+1)} + \frac {1}{4(s+1)^2},\] so \[y = -\frac 13 + \frac {e^{3t}}{48} + \Big (\frac {5}{16} + \frac {t}{4}\Big )e^{-t}.\]

Solution. This is an Euler–Cauchy equation, and the Laplace transform is the wrong tool for it — transforming would produce a differential equation in \(s\) rather than an algebraic one, since the coefficients are not constant.

Trying \(y = t^m\) gives \[m(m-1) + m - 1 = m^2-1 = 0 ,\] so \(m = \pm 1\) and \[y = At + \frac {B}{t}.\]

Solution. The integral is a convolution, \(\big (t^4 * y\big )(t)\), so transforming turns it into a product: \[Y = \frac {2}{s^3} + 6\cdot \frac {4!}{s^5}\,Y = \frac {2}{s^3} + \frac {144}{s^5}Y .\] Solving, \[Y\Big (1-\frac {144}{s^5}\Big ) = \frac {2}{s^3} \implies Y = \frac {2s^2}{s^5-144}.\] That the convolution becomes multiplication is exactly why integral equations of this kind are approached by transform rather than directly.