GATE 2026 EE Q.50: solution and answer

GATE 2026 · EEQ.50 · 2 marks · MCQ

Consider the second-order differential equation

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + y = 0

with initial conditions

y (0) = 1, \frac{d y}{d x}_{x = 0} = 1

The solution is given by

(A)
$y (x) = exp (- x /2) (cos (\frac{3 x}{2}) + 3 sin (\frac{3 x}{2}))$
(B)
$y (x) = exp (- x /2) (cos (\frac{3 x}{2}) + \frac{1}{3} sin (\frac{3 x}{2}))$
(C)
$y (x) = exp (- x /2) (cos (\frac{3 x}{2}) - \frac{1}{3} sin (\frac{3 x}{2}))$
(D)
$y (x) = exp (- x /2) (cos (\frac{3 x}{2}) - 3 sin (\frac{3 x}{2}))$

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Answer and solution

ANSWER

(A) $y (x) = exp (- x /2) (cos (\frac{3 x}{2}) + 3 sin (\frac{3 x}{2}))$

Official answer: ASource: Official answer key, GATE 2026 EE (IIT Guwahati)

Auxiliary (characteristic) equation for this constant-coefficient equation: $m^{2} + m + 1 = 0$ , so $m = \frac{- 1 \pm 1 - 4}{2} = - \frac{1}{2} \pm j \frac{3}{2}$ .
The roots are complex, $α \pm j β$ with $α = - \frac{1}{2}$ and $β = \frac{3}{2}$ , so the general solution is $y (x) = e^{- x /2} (A cos \frac{3 x}{2} + B sin \frac{3 x}{2})$
Apply $y (0) = 1$ : $e^{0} (A \cdot 1 + B \cdot 0) = 1$ , so $A = 1$ .
Differentiate using the product rule: $y^{'} (x) = - \frac{1}{2} e^{- x /2} (A cos \frac{3 x}{2} + B sin \frac{3 x}{2}) + e^{- x /2} \frac{3}{2} (- A sin \frac{3 x}{2} + B cos \frac{3 x}{2})$
Apply $y^{'} (0) = 1$ : $- \frac{A}{2} + \frac{3}{2} B = 1$ . Putting in $A = 1$ gives $\frac{3}{2} B = \frac{3}{2}$ , so $B = \frac{3}{3} = 3$ .
So $y (x) = exp (- x /2) (cos \frac{3 x}{2} + 3 sin \frac{3 x}{2})$ , which is option (A).
Quick check that rules out the others: every option already gives $y (0) = 1$ , so only $y^{'} (0) = - \frac{1}{2} + \frac{3}{2} B$ tells them apart. It is $1$ for $B = 3$ (A), $0$ for $B = \frac{1}{3}$ (B), $- 1$ for $B = - \frac{1}{3}$ (C) and $- 2$ for $B = - 3$ (D). Only (A) meets $y^{'} (0) = 1$ .

KEY IDEA

For $y^{''} + a y^{'} + b y = 0$ with complex roots $α \pm j β$ , the solution is $e^{α x} (A cos β x + B sin β x)$ . Then $y (0)$ fixes $A$ , and $y^{'} (0) = α A + β B$ fixes $B$ . All four options share the same envelope, the same frequency and $A = 1$ , so only the $y^{'} (0)$ condition decides between them.

WATCH OUT FOR

Leaving out the $α A$ term that the product rule gives to $y^{'} (0)$ , and writing $β B = 1$ instead of $α A + β B = 1$ . That gives $B = \frac{2}{3}$ , which is not an option. Another slip is dropping the sign of $α$ and writing $+ \frac{A}{2}$ , which gives $\frac{3}{2} B = \frac{1}{2}$ , $B = \frac{1}{3}$ , and so the wrong option (B).

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GATE 2026 EE Q.50: solution and answer

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