 Solution

ID: past-exam-of-the-differential-equations-and-applications-course-of-the-university-of-oxford/2008/a/3/d/solution

The nonzero kernel of

L

rules out a unique solution for every value of

c

. By the Fredholm alternative, solvability requires orthogonality of the forcing to

ker L^{*} = span {3 a - 5 b}

. Its pairing with

6 a + c b

⟨ 3 a - 5 b, 6 a + c b ⟩ = 9 + c - 10 - \frac{5 c}{4} = - 1 - \frac{c}{4} .

Thus only

c = - 4

is compatible. In that case

6 a - 4 b = 2 y_{- 4}

- y_{- 4} /2

. All solutions are

y (x) = - \frac{3}{2} e^{- x} + e^{- 2 x} + C (5 e^{- x} - 6 e^{- 2 x}), c = - 4.

Therefore infinitely many solutions exist for $c = - 4$ , no solution exists for $c \neq = - 4$ , and no parameter value gives a unique solution.

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