GATE 2025 EC – Question 32
The function $y(t)$ satisfies
$$t^2y''(t)-2ty'(t)+2y(t)=0,$$
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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Show answer and explanation
Correct answer: 0.24 to 0.26
Explanation
This is a Cauchy-Euler equation. Trying $y=t^m$ gives $m(m-1)-2m+2=m^2-3m+2=0$, so $m=1,2$ and $y=c_1t+c_2t^2$. Then $y'=c_1+2c_2t$; $y'(0)=1$ gives $c_1=1$ and $y'(1)=1+2c_2=-1$ gives $c_2=-1$. So $y=t-t^2$, which is maximum at $t=\tfrac12$ with value $\tfrac14=0.25$.