 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability-resit/5/a/iii/solution

Put

a = π /2

. Substituting

x = - a

and

x = a

into part (ii), using

cos (π /4) = sin (π /4) = 1/ 2

, gives

- 1 + \frac{e ^{a}}{2} (A - B) = e^{a}, 1 + \frac{e ^{- a}}{2} (A + B) = e^{- a} .

Thus

A - B = 2 (1 + e^{- a})

and

A + B = 2 (1 - e^{a})

. Solving these two linear equations gives

A = 2 (1 - sinh a), B = - 2 cosh a .

Consequently the solution of the boundary value problem is

y (x) = sin x + 2 e^{- x} [(1 - sinh \frac{π}{2}) cos \frac{x}{2} - cosh \frac{π}{2} sin \frac{x}{2}] .

The two independent boundary conditions determine

A, B

uniquely.

 New to topics? Read the docs here!