 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/15/c/ii/solution

For the general solution in part (i),

y (0) = A + B = 1, y^{'} (0) = - \frac{3}{4} - 2 B = 0.

Thus

B = - 3/8

and

A = 11/8

y (x) = \frac{3}{4} x^{2} - \frac{3}{4} x + \frac{11}{8} - \frac{3}{8} e^{- 2 x} .

Direct substitution checks both the differential equation and the two initial conditions.

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