 Solution

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

r^{2} + 2 r + 2 = 0

, with roots

- 1 \pm i

y_{1} (x) = e^{- x} cos x, y_{2} (x) = e^{- x} sin x .

Their Wronskian, with the convention

W = y_{1} y_{2}^{'} - y_{1}^{'} y_{2}

, is

W (x) = e^{- 2 x} {cos x (cos x - sin x) + (sin x + cos x) sin x} = e^{- 2 x} \neq = 0 .

Thus the two functions are linearly independent throughout the interval and span its homogeneous solutions.

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