Laplace & Inverse Laplace — Solutions Part-1

Find the Laplace transforms

a) \(f(t)=3e^{-4t}\)
Use \( \mathcal{L}\{e^{at}\} = \dfrac{1}{s-a}\) (for region of convergence \( \Re(s)> \Re(a)\)). Here \(a=-4\).
\[ \mathcal{L}\{3e^{-4t}\}(s)=3\cdot \frac{1}{s-(-4)}=\frac{3}{s+4}. \]
b) \(f(t)=2t^{2}\)
Use \( \mathcal{L}\{t^{n}\} = \dfrac{n!}{s^{n+1}}\). For \(t^{2}\) we have \(2! = 2\).
\[ \mathcal{L}\{2t^{2}\}(s)=2\cdot\frac{2}{s^{3}}=\frac{4}{s^{3}}. \]
c) \(f(t)=4\cos(5t)\)
Use \( \mathcal{L}\{\cos(at)\} = \dfrac{s}{s^{2}+a^{2}}\).
\[ \mathcal{L}\{4\cos(5t)\}(s)=4\cdot\frac{s}{s^{2}+25}=\frac{4s}{s^{2}+25}. \]
d) \(f(t)=\sin(\pi t)\)
Use \( \mathcal{L}\{\sin(at)\} = \dfrac{a}{s^{2}+a^{2}}\) with \(a=\pi\).
\[ \mathcal{L}\{\sin(\pi t)\}(s)=\frac{\pi}{s^{2}+\pi^{2}}. \]
e) \(f(t)=\begin{cases} 3 & 0<t<2,\\[4pt] -1 & 2\le t<4,\\[4pt] 0 & 4\le t, \end{cases}\)
Compute the Laplace integral piecewise: \[ F(s)=\int_{0}^{2}3e^{-st}\,dt+\int_{2}^{4}(-1)e^{-st}\,dt. \]
Evaluate each integral: \[ \int_{0}^{2}3e^{-st}dt=3\frac{1-e^{-2s}}{s},\qquad \int_{2}^{4}-e^{-st}dt=-e^{-2s}\frac{1-e^{-2s}}{s}. \]
\[ F(s)=\frac{3(1-e^{-2s})-e^{-2s}(1-e^{-2s})}{s} =\frac{3-4e^{-2s}+e^{-4s}}{s}. \]
f) \(f(t)=\sin t\cos t\)
Use the identity \(\sin t\cos t = \tfrac{1}{2}\sin(2t)\) and then Laplace of sine:
\[ \mathcal{L}\{\sin t\cos t\}(s)=\frac12\mathcal{L}\{\sin(2t)\} =\frac12\cdot\frac{2}{s^{2}+4}=\frac{1}{s^{2}+4}. \]

Find the inverse Laplace transforms

a) \(F(s)=\dfrac{2s^{2}-4}{(s-2)(s+1)(s-3)}\)
Perform partial fraction decomposition: \[ \frac{2s^{2}-4}{(s-2)(s+1)(s-3)} = -\frac{1}{6}\cdot\frac{1}{s+1}-\frac{4}{3}\cdot\frac{1}{s-2}+\frac{7}{2}\cdot\frac{1}{s-3}. \]
Take inverse Laplace term-by-term (\(\mathcal{L}^{-1}\{\tfrac{1}{s-a}\}=e^{at}\)):
\[ \mathcal{L}^{-1}\{F(s)\}(t) = -\frac{1}{6}e^{-t}-\frac{4}{3}e^{2t}+\frac{7}{2}e^{3t}. \]
b) \(F(s)=\dfrac{3s+1}{(s-1)(s^{2}+1)}\)
Decompose: \[ \frac{3s+1}{(s-1)(s^{2}+1)} = \frac{2}{s-1} - \frac{2s-1}{s^{2}+1}. \] Then split the second term: \(\dfrac{2s-1}{s^{2}+1}= \dfrac{2s}{s^{2}+1} - \dfrac{1}{s^{2}+1}.\)
Inverse Laplace using \(\mathcal{L}^{-1}\{\tfrac{1}{s-1}\}=e^{t}\), \(\mathcal{L}^{-1}\{\tfrac{s}{s^{2}+1}\}=\cos t\), \(\mathcal{L}^{-1}\{\tfrac{1}{s^{2}+1}\}=\sin t\):
\[ \mathcal{L}^{-1}\{F(s)\}(t)=2e^{t}-2\cos t+\sin t. \]

Summary Table of Results

ProblemTime Domain / Laplace Domain
a) Laplace\(\mathcal{L}\{3e^{-4t}\}=\dfrac{3}{s+4}\)
b) Laplace\(\mathcal{L}\{2t^{2}\}=\dfrac{4}{s^{3}}\)
c) Laplace\(\mathcal{L}\{4\cos(5t)\}=\dfrac{4s}{s^{2}+25}\)
d) Laplace\(\mathcal{L}\{\sin(\pi t)\}=\dfrac{\pi}{s^{2}+\pi^{2}}\)
e) Laplace (piecewise)\(F(s)=\dfrac{3-4e^{-2s}+e^{-4s}}{s}\)
f) Laplace\(\mathcal{L}\{\sin t\cos t\}=\dfrac{1}{s^{2}+4}\)
a) Inverse Laplace\(\mathcal{L}^{-1}\{ \dfrac{2s^{2}-4}{(s-2)(s+1)(s-3)}\}=-\tfrac{1}{6}e^{-t}-\tfrac{4}{3}e^{2t}+\tfrac{7}{2}e^{3t}\)
b) Inverse Laplace\(\mathcal{L}^{-1}\{ \dfrac{3s+1}{(s-1)(s^{2}+1)}\}=2e^{t}-2\cos t+\sin t\)


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