 Solution

ID: past-exam-of-the-differential-equations-2-course-of-the-university-of-oxford/2023/1/b/ii/solution

h (x) = e^{- x} cos x

already has the prescribed endpoint values. Thus

z = y - h

has zero boundary conditions and satisfies

L z = f

y (x) = e^{- x} cos x + \int_{0}^{π /2} g (x, ξ) f (ξ) d ξ .

Equivalently, splitting the integral at

ξ = x

gives the explicit formula

y (x) = e^{- x} [cos x (1 - \int_{0}^{x} e^{ξ} sin ξ f (ξ) d ξ) - sin x \int_{x}^{π /2} e^{ξ} cos ξ f (ξ) d ξ] .

Any difference between two solutions would be

e^{- x} (C cos x + D sin x)

with zero endpoint values, forcing

C = D = 0

. The displayed solution is therefore unique.

 New to topics? Read the docs here!