Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 4 / 13D / c

Table of contents
- Solution c

Solution

The Euler-Lagrange equation is

y^{(4)} - 2 y^{''} - y = 0.

Put

A = 1 + 2

,

B = 2 - 1

, and

L = π /2

. The conditions at zero reduce the general solution to

y = C (cosh A x - cos B x) + D (sinh A x - \frac{A}{B} sin B x) .

U = cosh (A L) - cos (B L), V = sinh (A L) - \frac{A}{B} sin (B L),

U^{'} = A sinh (A L) + B sin (B L), V^{'} = A (cosh (A L) - cos (B L)) .

The remaining boundary conditions give

C = \frac{L V ^{'} - V}{U V ^{'} - U ^{'} V}, D = \frac{U - L U ^{'}}{U V ^{'} - U ^{'} V},

which determines the unique extremal explicitly.

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