Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2013/12/a/solution

The second equation gives . Differentiating it gives . Substitution into the first equation cancels the terms and the constant forcing, leaving . Hence
This 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 is
Checking directly, , while . This also verifies the signs introduced during elimination.

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