GATE 2025 EE Q.54: solution and answer

GATE 2025 · EEQ.54 · 2 marks · NAT

Consider ordinary differential equations given by

$\overset{x}{˙}_{1} (t) = 2 x_{2} (t)$

$\overset{x}{˙}_{2} (t) = r (t)$

with initial conditions $x_{1} (0) = 1$ and $x_{2} (0) = 0$ .

If $r (t) = {1, 0, t \geq 0 t < 0$ , then at $t = 1$ , $x_{1} (t) =$ ________ (round off to the nearest integer).

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Answer and solution

ANSWER

$x_{1} (1) = 2$

Official answer: 2Source: Official answer key, GATE 2025 EE (IIT Roorkee)

The coupling runs one way only: $x_{2}$ depends on nothing but $r (t)$ . So solve for $x_{2}$ first, then substitute it into the equation for $x_{1}$ .
For $t \geq 0$ the input is $r (t) = 1$ , so $\overset{x}{˙}_{2} = 1$ . Integrate from $0$ to $t$ using $x_{2} (0) = 0$ : $x_{2} (t) = x_{2} (0) + \int_{0}^{t} 1 d τ = t .$
Substitute into the first equation: $\overset{x}{˙}_{1} (t) = 2 x_{2} (t) = 2 t$ .
Integrate from $0$ to $t$ using $x_{1} (0) = 1$ : $x_{1} (t) = x_{1} (0) + \int_{0}^{t} 2 τ d τ = 1 + t^{2} .$
Check with the Laplace transform, keeping the initial conditions: from $s X_{2} (s) - x_{2} (0) = \frac{1}{s}$ we get $X_{2} (s) = \frac{1}{s ^{2}}$ . Then $s X_{1} (s) - x_{1} (0) = 2 X_{2} (s)$ gives $s X_{1} (s) - 1 = \frac{2}{s ^{2}}$ , so $X_{1} (s) = \frac{1}{s} + \frac{2}{s ^{3}}$ . Inverting gives $x_{1} (t) = 1 + t^{2}$ , the same result.
At $t = 1$ : $x_{1} (1) = 1 + 1^{2} = 2$ .

KEY IDEA

This is a chain of two integrators, $x_{1} \leftarrow 2 \int x_{2}$ and $x_{2} \leftarrow \int r$ . A unit step fed through it gives a ramp for $x_{2}$ and a parabola for $x_{1}$ . Integrate twice and add the initial condition at each stage: $x_{2} = t$ , and $x_{1} = 1 + t^{2}$ .

WATCH OUT FOR

Dropping the initial condition $x_{1} (0) = 1$ gives $x_{1} (1) = 1$ . This is the zero-state response only, and it misses the zero-input part $1/ s$ . Another common slip is forgetting the factor $2$ in $\overset{x}{˙}_{1} = 2 x_{2}$ , which gives $x_{1} = 1 + t^{2} /2$ , that is $1.5$ at $t = 1$ .

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GATE 2025 EE Q.54: solution and answer

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