ISRO EC 2025 Q.35: solution and answer

ISRO 2025 · ECQ.35 · 1 mark · MCQ

Solution of the below PDE $\frac{\partial u}{\partial t} = 4 \frac{\partial ^{2} u}{\partial x ^{2}}$ , $u (0, t) = u (π, t) = 0$ , $u (x, 0) = sin (3 x)$

(A)
$u (x, t) = e^{- 36 t} sin (3 x)$
(B)
$u (x, t) = - e^{- 36 t} sin (3 x)$
(C)
$u (x, t) = - e^{- 9 t} sin (3 x)$
(D)
$u (x, t) = e^{- 9 t} sin (3 x)$

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

$u (x, t) = e^{- 36 t} sin (3 x)$ , option (A).

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

This is the heat equation $u_{t} = 4 u_{xx}$ on $0 < x < π$ with zero (Dirichlet) ends. Separate variables: $u = X (x) T (t)$ , so $\frac{T ^{'}}{4 T} = \frac{X ^{''}}{X} = - λ$ .
Spatial part: $X^{''} + λ X = 0$ with $X (0) = X (π) = 0$ gives $X_{n} = sin (n x)$ and $λ = n^{2}$ , $n = 1, 2, 3, \dots$
Time part: $T^{'} = - 4 n^{2} T$ , so $T_{n} = e^{- 4 n^{2} t}$ .
General solution: $u (x, t) = \sum_{n = 1}^{\infty} b_{n} e^{- 4 n^{2} t} sin (n x)$ .
Initial condition: $u (x, 0) = sin (3 x)$ is already a single sine mode, so $b_{3} = 1$ and every other $b_{n} = 0$ by orthogonality of $sin (n x)$ on $[0, π]$ .
Decay rate for $n = 3$ : $4 n^{2} = 4 \times 9 = 36$ .
Hence $u (x, t) = e^{- 36 t} sin (3 x)$ . Check: $u_{t} = - 36 u$ and $4 u_{xx} = 4 (- 9) u = - 36 u$ ; $u (0, t) = u (π, t) = 0$ ; $u (x, 0) = sin 3 x$ .
Option (b) has the wrong sign at $t = 0$ . Options (c) and (d) use $e^{- 9 t}$ , which leaves out the factor 4 in the decay rate.

KEY IDEA

For $u_{t} = c u_{xx}$ with $u = 0$ at $x = 0, π$ , the mode $sin (n x)$ decays as $e^{- c n^{2} t}$ . A single-mode initial condition stays a single mode.

WATCH OUT FOR

Forgetting the diffusion constant 4 and writing the decay as $e^{- n^{2} t} = e^{- 9 t}$ , or putting a minus sign in front of the solution, which breaks the initial condition.

From the ISRO EC 2025 paper. Every question of this paper · The full paper (PDF, ISRO) · All 1 ISRO Differential equations question, with notes