GATE 2023 EE Q.34: solution and answer

GATE 2023 · EEQ.34 · 1 mark · NAT

In the following differential equation, the numerically obtained value of $y (t)$ , at $t = 1$ , is ________ (Round off to 2 decimal places).

\frac{d y}{d t} = \frac{e ^{- α t}}{2 + α t}, α = 0.01 and y (0) = 0

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Answer and solution

ANSWER

y(1) = ∫₀¹ e^(−0.01t)/(2 + 0.01t) dt ≈ 0.4963, which rounds to 0.50.

Official answer: 0.48 to 0.52Source: Official answer key, GATE 2023 EE (IIT Kanpur)

The right-hand side depends only on $t$ , not on $y$ , so the ODE integrates directly: $y (1) = y (0) + \int_{0}^{1} \frac{e ^{- 0.01 t}}{2 + 0.01 t} d t = \int_{0}^{1} f (t) d t .$ Substituting $u = 2 + 0.01 t$ turns this into $\frac{e ^{2}}{0.01} \int_{2}^{2.01} \frac{e ^{- u}}{u} d u$ , an exponential integral with no elementary closed form. So we evaluate it numerically.
Evaluate the integrand at the nodes for Simpson's 1/3 rule with $h = 0.5$ : $f (0) = \frac{1}{2} = 0.500000$ , $f (0.5) = \frac{e ^{- 0.005}}{2.005} = \frac{0.995012}{2.005} = 0.496266$ , $f (1) = \frac{e ^{- 0.01}}{2.01} = \frac{0.990050}{2.01} = 0.492562$ .
Apply Simpson's 1/3 rule, $\int_{0}^{1} f d t \approx \frac{h}{3} [f_{0} + 4 f_{1} + f_{2}]$ : $y (1) \approx \frac{0.5}{3} [0.500000 + 4 (0.496266) + 0.492562] = \frac{0.5}{3} (2.977626) = 0.496271.$
Cross-check with the trapezoidal rule on one interval: $\frac{1}{2} (0.500000 + 0.492562) = 0.496281$ . The two results agree to four decimal places ( $0.4963$ ) because $f$ is almost linear on $[0, 1]$ when $α$ is this small. Even a single Euler step, $y (1) \approx y (0) + 1 \cdot f (0) = 0.5$ , gives the same answer to 2 decimal places.
Rounded to 2 decimal places: $y (1) \approx 0.4963 \approx 0.50$ .

KEY IDEA

When dy/dt depends only on t, the initial-value problem is really a definite integral: y(1) = y(0) + ∫₀¹ f(t) dt. Here α = 0.01 is small, so the integrand barely moves from 1/2 across [0, 1] and y(1) ≈ 0.50. Simpson's rule, the trapezoidal rule and a single Euler step all agree on this to 2 decimal places.

WATCH OUT FOR

Reporting the slope at t = 1, f(1) ≈ 0.4926, as the value of y(1). y is the area accumulated under the slope curve from 0 to 1, not the slope at the end point. That answer only lands near the key here because f is almost constant. The other common slip is to assume the right-hand side depends on y and start a multi-step Runge–Kutta calculation. Nothing here needs that: it is a single quadrature.

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GATE 2023 EE Q.34: solution and answer

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