Engineering Mathematics III, Tutorial 5: Laplace transforms
Course: MA 301 Engineering Mathematics III · Author: Prof. Arjun Menon · Date: 20 Oct 2026
| Total marks | Due | Submit by | Collaboration |
|---|---|---|---|
| 30 | Monday 27 Oct, 11:00 | Share your copy with the course account | Individual |
Instructions
- Show every step. A correct answer with no working earns at most one mark.
- You may use the table of transforms under Useful results without proof.
- Calculators are allowed for the final numerical value in Problem 5.
Topics covered
- Definition and linearity of the Laplace transform (Lecture 12).
- First shifting theorem (Lecture 13).
- Inverse transforms by partial fractions, and solving linear ODEs (Lecture 14).
Useful results
\mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}}, \qquad \mathcal{L}\{e^{at}\} = \frac{1}{s-a}, \qquad \mathcal{L}\{\sin bt\} = \frac{b}{s^2+b^2}
\mathcal{L}\{e^{at} f(t)\} = F(s-a), \qquad \mathcal{L}\{f'(t)\} = sF(s) - f(0), \qquad \mathcal{L}\{f''(t)\} = s^2F(s) - sf(0) - f'(0)
Problems
Problem 1 [4 marks]
Find \mathcal{L}\{3t^2 + e^{-2t}\}.
Hint
Use linearity, then the first two entries of the table.
Solution
By linearity, \mathcal{L}\{3t^2\} = 3 \cdot \frac{2!}{s^3} = \frac{6}{s^3} and \mathcal{L}\{e^{-2t}\} = \frac{1}{s+2}. So
\mathcal{L}\{3t^2 + e^{-2t}\} = \frac{6}{s^3} + \frac{1}{s+2}
Problem 2 [4 marks]
Find \mathcal{L}\{e^{-3t}\sin 2t\}.
Hint
Transform \sin 2t first, then apply the first shifting theorem.
Solution
\mathcal{L}\{\sin 2t\} = \frac{2}{s^2+4}. Shifting s to s+3 gives
\mathcal{L}\{e^{-3t}\sin 2t\} = \frac{2}{(s+3)^2+4}
Problem 3 [6 marks]
Find the inverse transform of F(s) = \dfrac{s+5}{s^2+3s+2}.
Hint
Factor the denominator, then split into partial fractions with the cover-up rule.
Solution
s^2+3s+2 = (s+1)(s+2), so F(s) = \frac{A}{s+1} + \frac{B}{s+2}.
Cover-up at s=-1: A = \frac{-1+5}{-1+2} = 4. Cover-up at s=-2: B = \frac{-2+5}{-2+1} = -3.
f(t) = 4e^{-t} - 3e^{-2t}
Problem 4 [8 marks]
Solve y'' + 3y' + 2y = 4 with y(0) = 0 and y'(0) = 0.
Hint
Transform both sides, solve for Y(s), then use partial fractions with three terms.
Solution
With zero initial conditions, (s^2 + 3s + 2)Y = \frac{4}{s}, so
Y(s) = \frac{4}{s(s+1)(s+2)} = \frac{2}{s} - \frac{4}{s+1} + \frac{2}{s+2}
Therefore y(t) = 2 - 4e^{-t} + 2e^{-2t}.
Check: y(0) = 2 - 4 + 2 = 0, and y'(t) = 4e^{-t} - 4e^{-2t} gives y'(0) = 0. As t \to \infty, y \to 2, which matches the steady state 2y = 4.
Problem 5 [8 marks]
A series RC circuit has R = 1\,\text{k}\Omega and C = 100\,\mu\text{F}. The capacitor is uncharged when a 5 V step is applied at t = 0. Use the Laplace transform to find the capacitor voltage v_C(t), and evaluate it at t = 0.1 s.
Hint
Kirchhoff's voltage law gives RC\,v_C' + v_C = 5. Work out RC first.
Solution
RC = 1000 \times 100 \times 10^{-6} = 0.1 s. Transforming 0.1\,v_C' + v_C = 5 with v_C(0) = 0:
(0.1s + 1)V_C(s) = \frac{5}{s} \quad\Rightarrow\quad V_C(s) = \frac{50}{s(s+10)} = \frac{5}{s} - \frac{5}{s+10}
So v_C(t) = 5\left(1 - e^{-10t}\right) V. At t = 0.1 s, v_C = 5(1 - e^{-1}) \approx 3.16 V, which is 63.2% of the supply, as expected one time constant after the step.
Progress
Submission notes
- Work in your own copy (File → Make a copy) and rename it "MA301 T5, your roll number".
- Share the copy with the course account as a viewer before the deadline. Edits after 11:00 on Monday are not marked, and version history shows when each change was saved.
- Late submissions lose 3 marks per day, up to two days.
- Questions go in the course discussion doc, not by email, so everyone sees the answer.
Marking scheme
| Problem | Marks for method | Marks for answer | Total |
|---|---|---|---|
| 1 | 2 | 2 | 4 |
| 2 | 2 | 2 | 4 |
| 3 | 4 | 2 | 6 |
| 4 | 6 | 2 | 8 |
| 5 | 6 | 2 | 8 |