GATE 2025 EC Q.32: solution and answer

GATE 2025 · ECQ.32 · 1 mark · NAT

The function $y (t)$ satisfies

t^{2} y^{''} (t) - 2 t y^{'} (t) + 2 y (t) = 0,

where $y^{'} (t)$ and $y^{''} (t)$ denote the first and second derivatives of $y (t)$ , respectively.

Given $y^{'} (0) = 1$ and $y^{'} (1) = - 1$ , the maximum value of $y (t)$ over $[0, 1]$ is ______ (rounded off to two decimal places).

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Answer and solution

ANSWER

$y (t) = t - t^{2}$ , so the maximum over $[0, 1]$ is $y (\frac{1}{2}) = 0.25$ .

Official answer: 0.25Source: Official answer key, GATE 2025 EC (IIT Roorkee)

The equation $t^{2} y^{''} - 2 t y^{'} + 2 y = 0$ is a Cauchy-Euler equation: each derivative $y^{(k)}$ is multiplied by $t^{k}$ . So try $y = t^{m}$ , which gives $y^{'} = m t^{m - 1}$ and $y^{''} = m (m - 1) t^{m - 2}$ .
Substituting and dividing by $t^{m}$ gives the auxiliary equation $m (m - 1) - 2 m + 2 = 0 \Rightarrow m^{2} - 3 m + 2 = 0 \Rightarrow (m - 1) (m - 2) = 0$ , so $m = 1$ or $m = 2$ .
The roots are real and distinct, so the general solution is $y (t) = c_{1} t + c_{2} t^{2}$ , with $y^{'} (t) = c_{1} + 2 c_{2} t$ .
Apply the given conditions, which are on the derivative: $y^{'} (0) = c_{1} = 1$ , and $y^{'} (1) = c_{1} + 2 c_{2} = - 1 \Rightarrow 2 c_{2} = - 2 \Rightarrow c_{2} = - 1$ .
So $y (t) = t - t^{2}$ . It is a polynomial, so it is smooth on all of $[0, 1]$ , including the singular point $t = 0$ , and it satisfies the equation there too: $0 - 0 + 0 = 0$ .
For the maximum, set $y^{'} (t) = 1 - 2 t = 0 \Rightarrow t = \frac{1}{2}$ , which lies inside $[0, 1]$ . Since $y^{''} (t) = - 2 < 0$ , this point is a maximum.
$y (\frac{1}{2}) = \frac{1}{2} - \frac{1}{4} = 0.25$ At the endpoints $y (0) = y (1) = 0$ , so the maximum of $y (t)$ over $[0, 1]$ is $0.25$ .

KEY IDEA

This is a Cauchy-Euler equation, where each derivative $y^{(k)}$ is multiplied by $t^{k}$ . Substituting $y = t^{m}$ turns it into the algebraic equation $m (m - 1) - 2 m + 2 = 0$ . Its roots $m = 1, 2$ give the polynomial solution $y = c_{1} t + c_{2} t^{2}$ , and the two conditions on $y^{'}$ fix $c_{1}$ and $c_{2}$ . What is left is an ordinary maximisation over a closed interval: check the stationary point and both endpoints.

WATCH OUT FOR

The most common slip is reading the conditions as $y (0) = 1$ and $y (1) = - 1$ when they are on $y^{'}$ . This cannot even work, because every solution has $y (0) = 0$ . The second is writing the auxiliary equation as $m^{2} - 2 m + 2 = 0$ , which drops the $- m$ from $m (m - 1)$ and gives false complex roots. The third is forgetting to compare the stationary value with the endpoint values $y (0)$ and $y (1)$ .

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GATE 2025 EC Q.32: solution and answer

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