GATE 2020 IN Q.27: solution and answer

GATE 2020 · INQ.27 · 2 marks · MCQ

Consider the differential equation $\frac{d x}{d t} = sin (x)$ , with the initial condition $x (0) = 0$ .

The solution to this ordinary differential equation is ____

(A)
$x (t) = 0$
(B)
$x (t) = sin (t)$
(C)
$x (t) = cos (t)$
(D)
$x (t) = sin (t) - cos (t)$

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Answer and solution

ANSWER

The solution is $x (t) = 0$ : the initial value $x (0) = 0$ is an equilibrium of $\frac{d x}{d t} = sin x$ , and because the solution is unique, $x$ stays at 0 for all $t$ . Option (A).

Official answer: ASource: Official answer key, GATE 2020 IN (IIT Delhi)

The right-hand side is $sin (x)$ , a function of the dependent variable $x$ , not of $t$ . So this is an autonomous equation $\frac{d x}{d t} = f (x)$ with $f (x) = sin x$ . It is not $\frac{d x}{d t} = sin t$ .
Equilibrium (constant) solutions satisfy $f (x) = 0$ : $sin x = 0 \Rightarrow x = nπ$ . The initial value $x (0) = 0$ is one of these equilibria.
Check $x (t) \equiv 0$ : $\frac{d x}{d t} = 0$ and $sin (0) = 0$ , so the ODE holds. The initial condition $x (0) = 0$ holds too.
Uniqueness (Picard–Lindelöf theorem): $f (x) = sin x$ has $∣ f^{'} (x) ∣ = ∣ cos x ∣ \leq 1$ , so $f$ is Lipschitz. The initial value problem therefore has exactly one solution, and that solution is $x (t) = 0$ .
Rule out the other options directly. (C): $x (0) = cos 0 = 1 \neq = 0$ . (D): $x (0) = sin 0 - cos 0 = - 1 \neq = 0$ . (B) meets the initial condition, but $\frac{d x}{d t} = cos t$ while $sin (x) = sin (sin t)$ . These are not equal: at $t = 0$ , $cos 0 = 1$ but $sin (sin 0) = 0$ .
So the solution is $x (t) = 0$ , option (A).

KEY IDEA

In an autonomous equation $\frac{d x}{d t} = f (x)$ , every root of $f (x) = 0$ gives a constant solution. If the initial value sits on such a root and $f$ is Lipschitz, uniqueness says the solution never leaves it. Here $sin 0 = 0$ , so $x (t) \equiv 0$ .

WATCH OUT FOR

Reading $sin (x)$ as $sin (t)$ and integrating gives $x = 1 - cos t$ . That is not an option, and it tempts students to pick (B) or (C) by pattern. A second trap: separating variables as $\frac{d x}{sin x} = d t$ divides by $sin x$ , which is zero at $x = 0$ . That throws away exactly the equilibrium solution the initial condition picks out, and $ln tan \frac{x}{2} = t + C$ then has no finite $C$ for $x (0) = 0$ .

From the GATE 2020 IN paper. Every question of this paper · Download the full paper · All 19 Differential equations questions, with notes

GATE 2020 IN Q.27: solution and answer

Answer and solution

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