ISRO EC 2025 Q.13: solution and answer

ISRO 2025 · ECQ.13 · 1 mark · MCQ

Inverse of Laplace transform of $\frac{10}{s ^{2} + 5 s + 4}$ is

(A)
$10 [e^{- t} + e^{- 4 t}]$
(B)
$10 [e^{- t} - e^{- 4 t}]$
(C)
$\frac{10}{3} [e^{- t} + e^{- 4 t}]$
(D)
$\frac{10}{3} [e^{- t} - e^{- 4 t}]$

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Answer and solution

ANSWER

The inverse Laplace transform is $\frac{10}{3} [e^{- t} - e^{- 4 t}] u (t)$ , which is option (D).

Official answer: DSource: ISRO's final answer key of 15 May 2026

Factorise the denominator: $s^{2} + 5 s + 4 = (s + 1) (s + 4)$ , so the poles are at $s = - 1$ and $s = - 4$ .
Write the partial fractions: $\frac{10}{( s + 1 ) ( s + 4 )} = \frac{A}{s + 1} + \frac{B}{s + 4}$ .
Residue at $s = - 1$ : $A = \frac{10}{s + 4}_{s = - 1} = \frac{10}{3}$ .
Residue at $s = - 4$ : $B = \frac{10}{s + 1}_{s = - 4} = \frac{10}{- 3} = - \frac{10}{3}$ .
So $X (s) = \frac{10}{3} [\frac{1}{s + 1} - \frac{1}{s + 4}]$ .
Use $\frac{1}{s + a} ⟷ e^{- a t} u (t)$ to get $x (t) = \frac{10}{3} [e^{- t} - e^{- 4 t}] u (t)$ .
Check by transforming back: $\frac{10}{3} [\frac{1}{s + 1} - \frac{1}{s + 4}] = \frac{10}{3} \cdot \frac{3}{( s + 1 ) ( s + 4 )} = \frac{10}{s ^{2} + 5 s + 4}$ . Also, $x (0^{+}) = \frac{10}{3} (1 - 1) = 0$ , which agrees with $lim_{s \to \infty} s X (s) = 0$ . Options (a) and (c) start at $20$ and $\frac{20}{3}$ , so they are wrong. Option (b) transforms to $\frac{30}{( s + 1 ) ( s + 4 )}$ , three times too large. Only (d) is correct.

KEY IDEA

Factorise the denominator into real poles, expand in partial fractions using the residue at each pole, then use $\frac{1}{s + a} \leftrightarrow e^{- a t} u (t)$ .

WATCH OUT FOR

Forgetting that the residue is $\frac{10}{3}$ , not $10$ : the factor $\frac{1}{( - 1 ) - ( - 4 )} = \frac{1}{3}$ comes from the other pole. Another slip is taking the second residue with a plus sign, which gives $e^{- t} + e^{- 4 t}$ instead of the difference.

From the ISRO EC 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 4 ISRO Laplace questions, with notes

ISRO EC 2025 Q.13: solution and answer

Answer and solution

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