 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/3/b/solution

The left-hand value at the junction is

y (μ -) = 1 - e^{- μ} - e^{- μ} sinh μ = \frac{1}{2} - e^{- μ} + \frac{1}{2} e^{- 2 μ} .

The right-hand value is

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

y (μ) = \frac{1}{2} (1 + e^{- 2 μ} - 2 e^{- μ}) = \frac{1}{2} (1 - e^{- μ})^{2} .

Its derivative is continuous as well: both one-sided values equal

- \frac{1}{2} (1 - e^{- μ})^{2}

. Only the second derivative jumps, consistently with the step in the forcing and the absence of a Dirac delta source.

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