GATE 2026 IN Q.54: solution and answer

GATE 2026 · INQ.54 · 2 marks · MSQ

Given a differential equation $\frac{d ^{2} x}{d t ^{2}} + x = 0$ with $x (0) \neq = 0$ .

Which of the following statements is/are true?

(A)
$∣ x (t) ∣^{2} + \frac{d x ( t )}{d t}^{2} = 0$ for all $t \geq 0$
(B)
$∣ x (t) ∣^{2} + \frac{d x ( t )}{d t}^{2} = c$ for all $t \geq 0$ , for some real constant $c > 0$
(C)
$∣ x (t) ∣^{2} + \frac{d x ( t )}{d t}^{2} = e^{j t}$ for all $t \geq 0$
(D)
$∣ x (t) ∣^{2} + \frac{d x ( t )}{d t}^{2} = sin t$ for all $t \geq 0$

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Answer and solution

ANSWER

Only (B) is true: $∣ x (t) ∣^{2} + \frac{d x}{d t}^{2}$ is a real constant $c = ∣ x (0) ∣^{2} + ∣ \overset{x}{˙} (0) ∣^{2} > 0$ for all $t$ .

Official answer: BSource: Official answer key, GATE 2026 IN (IIT Guwahati)

Solve the ODE. The auxiliary equation $m^{2} + 1 = 0$ gives $m = \pm j$ , so the general real solution is $x (t) = k_{1} cos t + k_{2} sin t$ , with $k_{1} = x (0) \neq = 0$ and $k_{2} = \overset{x}{˙} (0)$ .
Differentiate: $\frac{d x}{d t} = - k_{1} sin t + k_{2} cos t$ .
Add the squares: $x^{2} + (\frac{d x}{d t})^{2} = (k_{1} cos t + k_{2} sin t)^{2} + (- k_{1} sin t + k_{2} cos t)^{2} = k_{1}^{2} (cos^{2} t + sin^{2} t) + k_{2}^{2} (sin^{2} t + cos^{2} t) = k_{1}^{2} + k_{2}^{2} .$ The cross terms $+ 2 k_{1} k_{2} sin t cos t$ and $- 2 k_{1} k_{2} sin t cos t$ cancel.
Quick check without solving (energy argument): $\frac{d}{d t} [x^{2} + \overset{x}{˙}^{2}] = 2 x \overset{x}{˙} + 2 \overset{x}{˙} \overset{x}{¨} = 2 \overset{x}{˙} (x + \overset{x}{¨}) = 0$ , because $\overset{x}{¨} + x = 0$ . So the sum does not change with time. The modulus signs in the options change nothing: if $x$ were allowed to be complex, $\frac{d}{d t} [∣ x ∣^{2} + ∣ \overset{x}{˙} ∣^{2}] = 2 Re [\overset{ˉ}{\overset{x}{˙}} (x + \overset{x}{¨})] = 0$ as well.
So the sum is a constant, fixed by its value at $t = 0$ : $c = ∣ x (0) ∣^{2} + ∣ \overset{x}{˙} (0) ∣^{2} \geq ∣ x (0) ∣^{2} > 0$ (for the real solution, $c = k_{1}^{2} + k_{2}^{2}$ ). The constant is real and strictly positive, so (B) is true.
(A) is false: the sum equals $c \geq ∣ x (0) ∣^{2} > 0$ at every $t$ , so it is never $0$ . It could only be $0$ for the trivial solution $x \equiv 0$ , which $x (0) \neq = 0$ rules out.
(C) is false: the left side is a real constant, but $e^{j t} = cos t + j sin t$ changes with $t$ and is not even real in general (at $t = π /2$ it equals $j$ ).
(D) is false: $sin t$ changes with $t$ , and at $t = 0$ it gives $0$ , but the left side is $c > 0$ . Only (B) is true.

KEY IDEA

$\overset{x}{¨} + x = 0$ is simple harmonic motion, and $x^{2} + \overset{x}{˙}^{2}$ is its conserved energy. Multiply the equation by $\overset{x}{˙}$ and you get $\frac{d}{d t} [\frac{1}{2} (x^{2} + \overset{x}{˙}^{2})] = 0$ . In the phase plane the trajectory is a circle of radius $c$ , and $x (0) \neq = 0$ means that radius cannot be zero.

WATCH OUT FOR

Picking (C) because the characteristic roots are $\pm j$ and the solutions contain $e^{\pm j t}$ . That mixes up the form of $x (t)$ with the value of $∣ x ∣^{2} + ∣ \overset{x}{˙} ∣^{2}$ , which is a real constant. Another slip is choosing (A) because the equation is homogeneous and has no forcing. The sum is zero only for the trivial solution, and $x (0) \neq = 0$ rules that out. Some students also expand the squares, forget that the $\pm 2 k_{1} k_{2} sin t cos t$ cross terms cancel, and wrongly decide the sum depends on $t$ .

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GATE 2026 IN Q.54: solution and answer

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