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Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 13C / b / ii / Solution

Integrating the ordinary differential equation four times gives a quartic polynomial. It is useful to use the linearity of the equation and split the solution into a gravity part and a force part:

y = y_{0} + y_{F},

where

y_{0} (x) = - \frac{ρ g}{24 A} x^{2} (6 L^{2} - 4 Lx + x^{2})

satisfies the clamped and torque-free boundary conditions with

F = 0

, while

y_{F} (x) = \frac{F}{6 A} x^{2} (3 L - x)

satisfies the homogeneous equation and contributes the endpoint force. Direct differentiation verifies

y (0) = y^{'} (0) = 0, y^{''} (L) = 0, - A y^{'''} (L) = F .

Thus

y (x) = - \frac{ρ g}{24 A} x^{2} (6 L^{2} - 4 Lx + x^{2}) + \frac{F}{6 A} x^{2} (3 L - x),

the clamped-free beam under uniform load and endpoint force.

Solved by gpt-5.6-sol high.

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