12A (Complex Methods)

L {t^{n}} (s) = \frac{n}{s} L {t^{n - 1}} (s), L {1} = \frac{1}{s},

L {t^{n}} = \frac{n !}{s ^{n + 1}} .

e^{a t}

shifts the transform:

L {e^{a t} t^{n}} (s) = \frac{n !}{( s - a ) ^{n + 1}} .

b

The Bromwich inversion integral closes to the left for

t > 0

. Equivalently, partial fractions at the two double poles give

\frac{1}{s ^{2} ( s + 2 ) ^{2}} = - \frac{1}{4 s} + \frac{1}{4 s ^{2}} + \frac{1}{4 ( s + 2 )} + \frac{1}{4 ( s + 2 ) ^{2}} .

Taking residues therefore yields

f (t) = \frac{1}{4} (t - 1 + (t + 1) e^{- 2 t}) .

The transform factors as

s^{- 2} (s + 2)^{- 2}

, whose inverse factors are

t

and

t e^{- 2 t}

f (t) = \int_{0}^{t} τ (t - τ) e^{- 2 (t - τ)}, d τ,

and direct integration produces

f (t) = \frac{1}{4} (t - 1 + (t + 1) e^{- 2 t}),

confirming part (b).

The first three initial values vanish and the third derivative at zero equals one. Transforming the differential equation gives

(s^{4} + 4 s^{3} + 4 s^{2}) F (s) - 1 = 0,

F (s) = 1/ [s^{2} (s + 2)^{2}]

. Hence

f (t) = \frac{1}{4} (t - 1 + (t + 1) e^{- 2 t}) .

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