Past exam of the mathematics course of the University of Cambridge / 2013 / ia / Paper 2 / 1A (Differential Equations)

Table of contents
- Solution 1A

Solution

The characteristic polynomial factors as

(r - 2) (r + 1)

, so the complementary solution is

A e^{2 t} + B e^{- t}

. Each exponential forcing resonates with a homogeneous mode; the method of undetermined coefficients therefore uses

t e^{2 t}

and

t e^{- t}

. Substitution gives the particular solution

t e^{2 t} - t e^{- t} - 3 t

. Thus

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

The initial conditions give

A + B = 0

and

2 A - B - 3 = 0

. Hence

y (t) = (t + 1) e^{2 t} - (t + 1) e^{- t} - 3 t .

The factors of

t

in the particular solution are essential: an unmodified exponential trial lies in the kernel of the differential operator.

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