 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2/15/b/solution

m^{2} + 7 m + 12 = (m + 3) (m + 4) = 0

1/2

, so

y = \frac{1}{2} + A e^{- 3 x} + B e^{- 4 x} .

A + B = - 1/2

. The second gives

A e^{- 1} + B e^{- 4/3} = - e^{- 1} /2

. Eliminating

A

yields

B (e^{- 4/3} - e^{- 1}) = 0

, hence

B = 0

and

A = - 1/2

. Therefore

y (x) = \frac{1 - e ^{- 3 x}}{2}, y (1) = \frac{1 - e ^{- 3}}{2} .

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