Part BCSIR NET June 2025the-particular-solution-is-x-over-4-and-it-sets-the-boundary-constant-too

The particular solution is x over 4 and it sets the boundary constant too

Let y(x) be the extremal of the functional $J [y] = \int_{0}^{π /4}$ $((y^{'})^{2} - 4 y^{2} + 2$ xy) dx subject to $y (0) = 0, y (π /4) = 1$ . Then y(x) is equal to

A. $(1 - π /4) sin (2 x) + x$
B. $(1 - π /16) sin (2 x) + x /4$ ✓
C. $(1 + π /4) sin (2 x) - x$
D. $(1 + π /16) sin (2 x) - x /4$

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: A weak minimum of a functional is a strong minimum

The particular solution is x over 4 and it sets the boundary constant too

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