Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability / 4 / b

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- Solution b

Solution

The linear ordinary differential equation has differential operator

(D - 3)^{2}

. Put

y = e^{3 x} v

and use the product rule twice:

y^{'} = e^{3 x} (3 v + v^{'}), y^{''} = e^{3 x} (9 v + 6 v^{'} + v^{''}) .

Substitution cancels the

v

and

v^{'}

terms, leaving

v^{''} = 4

. Two integrations give

v = 2 x^{2} + C_{1} x + C_{2}

. Hence the general solution is

y (x) = e^{3 x} (2 x^{2} + C_{1} x + C_{2}), C_{1}, C_{2} \in R .

The two arbitrary constants account for the order of the differential equation; the

x^{2}

factor in the particular term reflects the repeated root

3

of the characteristic polynomial.

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