 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2003/calculus-probability-linear-algebra-discrete-mathematics-logic-and-proof-and-models-of-computation/1/a/i/solution

The repeated root of the characteristic equation suggests factoring out

e^{x}

y = e^{x} u

y^{'} = e^{x} (u + u^{'}), y^{''} = e^{x} (u + 2 u^{'} + u^{''}), y^{''} - 2 y^{'} + y = e^{x} u^{''} .

u^{''} = 1

u = x^{2} /2 + A x + B

B = 1

. Since

y^{'} (0) = A + B = 2

, the other initial condition gives

A = 1

. Therefore

y (x) = e^{x} (1 + x + \frac{x ^{2}}{2}), x \in R .

Substitution into the reduced ordinary differential equation and the two initial conditions verifies the result.

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