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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]=∫0π/4J[y] = \int_{0}^{\pi/4} ((y′)2−4y2+2((y')^{2} - 4y^{2} + 2xy) dx subject to y(0)=0,y(π/4)=1y(0) = 0, y(\pi/4) = 1. Then y(x) is equal to

  1. A.(1−π/4)sin⁡(2x)+x(1 - \pi/4)\sin(2x) + x
  2. B.(1−π/16)sin⁡(2x)+x/4(1 - \pi/16)\sin(2x) + x/4✓ Correct
  3. C.(1+π/4)sin⁡(2x)−x(1 + \pi/4)\sin(2x) - x
  4. D.(1+π/16)sin⁡(2x)−x/4(1 + \pi/16)\sin(2x) - 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

More on this topic

The chapter behind this: Euler–Lagrange, first integrals and null Lagrangians — free to read

From Calculus of Variations › Euler–Lagrange equation and standard functionals

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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