 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-2/13c/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

y + ε η

η

obeys

η (a) = η (b) = η^{'} (a) = η^{'} (b) = 0

because both

y

and its first derivative have fixed endpoint values. The first variation of the functional is

δ L = \int_{a}^{b} (F_{y} η + F_{y^{'}} η^{'} + F_{y^{''}} η^{''}) d x .

Applying integration by parts once to the second term and twice to the third gives

δ L = \int_{a}^{b} [F_{y} - \frac{d}{d x} F_{y^{'}} + \frac{d ^{2}}{d x ^{2}} F_{y^{''}}] η d x + [F_{y^{'}} η + F_{y^{''}} η^{'} - \frac{d}{d x} (F_{y^{''}}) η]_{a}^{b} .

The endpoint conditions on

η

and

η^{'}

make every boundary term zero. Since the remaining integral vanishes for every admissible variation, the fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equation

F_{y} - \frac{d}{d x} F_{y^{'}} + \frac{d ^{2}}{d x ^{2}} F_{y^{''}} = 0 .

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