 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2/14/c/solution

Differentiate

y^{'} = - y + 3 x

and use the other equation:

y^{''} = - y^{'} + 3 x^{'} = - y^{'} + 12 x - 6 y .

Since

3 x = y^{'} + y

, this becomes

y^{''} - 3 y^{'} + 2 y = 0

. Its roots are one and two, so

y = A e^{t} + B e^{2 t}

. The given data imply

y^{'} (0) = - 2 + 3 = 1

, hence

A + B = 2

and

A + 2 B = 1

. Thus

A = 3

B = - 1

y (t) = 3 e^{t} - e^{2 t} .

Recover the other dependent variable from

x = (y^{'} + y) /3

x (t) = 2 e^{t} - e^{2 t} .

These expressions satisfy both original equations and the two initial values.

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