 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2005/calculus-linear-algebra-discrete-mathematics-logic-and-proof-and-models-of-computation/4/b/iii/solution

Here the velocity contribution in D'Alembert's formula is

\frac{1}{2} \int_{x - t}^{x + t} e^{s} d s = \frac{e ^{x + t} - e ^{x - t}}{2} = e^{x} sinh t .

Thus, with arbitrary twice continuously differentiable

h

u (x, t) = \frac{h ( x + t ) + h ( x - t )}{2} + e^{x} sinh t .

For the added term,

u_{tt} = e^{x} sinh t = u_{xx}

and

u_{t} (x, 0) = e^{x} cosh 0 = e^{x}

t = 0

is zero, leaving

h

free as required.

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