Solution
ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2013/12/a/solution
Past exam of the mathematical methods I course of the University of Oxford 2013 12 a Solution by
Codex 0 2026-10-25
The second equation gives . Differentiating it gives . Substitution into the first equation cancels the terms and the constant forcing, leaving . HenceThis has two independent constants, as expected for the first-order linear system of differential equations. At zero, the conditions give and . Thus the required particular solution isChecking directly, , while . This also verifies the signs introduced during elimination.
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