GATE 2024 IN Q.34: solution and answer

GATE 2024 · INQ.34 · 1 mark · NAT

Consider a system given by the following first order differential equation:

\frac{d y}{d t} = y + 2 t - t^{2}

where, $y (0) = 1$ and $0 \leq t < \infty$ . Using a step size $h = 0.1$ for the improved Euler method, the value of $y (t)$ at $t = 0.1$ is ________ (rounded off to two decimal places).

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Answer and solution

ANSWER

Using improved Euler (Heun's method) with $h = 0.1$ , $y (0.1) = 1.1145 \approx 1.11$ .

Official answer: 1.10 to 1.12Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

Write the right-hand side as $f (t, y) = y + 2 t - t^{2}$ , with $t_{0} = 0$ , $y_{0} = 1$ , $h = 0.1$ , so $t_{1} = 0.1$ .
Improved Euler (Heun's method) is a predictor-corrector: predict $y_{1}^{*} = y_{0} + h f (t_{0}, y_{0})$ , then correct with $y_{1} = y_{0} + \frac{h}{2} [f (t_{0}, y_{0}) + f (t_{1}, y_{1}^{*})]$ .
Slope at the start: $f (0, 1) = 1 + 2 (0) - 0^{2} = 1$ .
Predictor (a plain Euler step): $y_{1}^{*} = 1 + 0.1 (1) = 1.1$ .
Slope at the predicted end point: $f (0.1, 1.1) = 1.1 + 2 (0.1) - (0.1)^{2} = 1.1 + 0.2 - 0.01 = 1.29$ .
Corrector (average of the two slopes): $y_{1} = 1 + \frac{0.1}{2} (1 + 1.29) = 1 + 0.05 (2.29) = 1.1145$
Rounded to two decimal places, $y (0.1) \approx 1.11$ (the third decimal is 4, so the value rounds down).
Check: the exact solution is $y = t^{2} + e^{t}$ . It satisfies $y (0) = 0 + 1 = 1$ , and $y^{'} = 2 t + e^{t} = (t^{2} + e^{t}) + 2 t - t^{2}$ . So $y (0.1) = 0.01 + e^{0.1} \approx 1.1152$ . The improved Euler value $1.1145$ is close to this, as expected from a second-order method with $h = 0.1$ .

KEY IDEA

Improved Euler predicts with a plain Euler step, then corrects by averaging the slope at the start, $f (t_{0}, y_{0})$ , with the slope at the predicted end point, $f (t_{1}, y_{1}^{*})$ : $y_{1} = y_{0} + \frac{h}{2} [f (t_{0}, y_{0}) + f (t_{1}, y_{1}^{*})]$ . The predicted $y_{1}^{*}$ must go into the second slope evaluation, not $y_{0}$ .

WATCH OUT FOR

Stopping at the plain Euler step gives $1.10$ , which only just touches the edge of the key range and is the wrong method. Another slip is to evaluate the second slope at $(t_{1}, y_{0})$ instead of $(t_{1}, y_{1}^{*})$ , which gives $f (0.1, 1) = 1.19$ and $y_{1} = 1.1095 \approx 1.11$ : the right number by luck, from the wrong working. Also, some textbooks use 'modified Euler' for the midpoint rule. Here that also gives $1.11475 \approx 1.11$ , but Heun's form is the standard reading of 'improved Euler'.

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GATE 2024 IN Q.34: solution and answer

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