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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/17/b/iv/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 17 / b / iv / Solution by

Codex  0 2026-10-18

Impose the five initial conditions at zero on the preceding general solution. They give

C_{0} + C_{1} + C_{2} + C_{3} C_{1} - C_{2} + C_{4} C_{1} + C_{2} - C_{3} - 1 C_{1} - C_{2} - C_{4} C_{1} + C_{2} + C_{3} = 1, = 0, = 0, = 0, = 0.

The second and fourth equations give

C_{1} = C_{2}

and

C_{4} = 0

. The third and fifth then yield

C_{1} = C_{2} = 1/4

and

C_{3} = - 1/2

, while the first gives

C_{0} = 1

. Thus

y (x) = 1 + \frac{1}{4} (e^{x} + e^{- x}) - \frac{1}{2} cos x - \frac{x ^{2}}{2} = 1 + \frac{1}{2} (cosh x - cos x) - \frac{x ^{2}}{2} .

For a direct check, the difference of its fifth and first derivatives is

x

. At zero its value is one; its first, second, third and fourth derivatives vanish. In particular the quadratic term cancels the otherwise nonzero second derivative of the hyperbolic-cosine and cosine combination.

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