Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 15 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

For the homogeneous linear differential equation, the characteristic equation is

r^{2} - 2 r + 2 = 0

, with polynomial roots

1 \pm i

. Its homogeneous solution is therefore

e^{x} (A cos x + B sin x)

.

Use the method of undetermined coefficients for a particular solution

y_{p} = a x^{2} + b x + c

. Substitution gives

2 a x^{2} + (2 b - 4 a) x + (2 a - 2 b + 2 c) = 2 x^{2} .

Equating coefficients yields

a = 1

,

b = 2

and

c = 1

. Hence the general solution is

y = e^{x} (A cos x + B sin x) + x^{2} + 2 x + 1.

ii

Solution

The characteristic polynomial factors as

(r + 1) (r + 2)

, so the homogeneous solution is

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

e^{- x}

already occurs in that solution, so a constant multiple of

e^{- x}

cannot be a particular solution. Use

y_{p} = C x e^{- x}

instead. Its derivatives are

C (1 - x) e^{- x}

and

C (x - 2) e^{- x}

, so substitution gives

C e^{- x}

and fixes

C = 1

. Thus

y = A e^{- x} + B e^{- 2 x} + x e^{- x} .

The extra factor of

x

accounts for the repeated forcing exponent in the method of undetermined coefficients.

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