Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2005 / Paper 2 / 5

Table of contents
- Solution 5

Solution

The one-sided Laplace transform is

\overset{u}{ˉ} (p) = \int_{0}^{\infty} e^{- pt} u (t) d t

for sufficiently large

Re p

. Integration by parts, with the exponential suppressing the upper boundary, gives

\overline{u^{'}} = p \overset{u}{ˉ} - u (0), \overline{u^{''}} = p^{2} \overset{u}{ˉ} - p u (0) - u^{'} (0), \overline{u^{'''}} = p^{3} \overset{u}{ˉ} - p^{2} u (0) - p u^{'} (0) - u^{''} (0) .

Differentiation under the integral gives

d \overset{u}{ˉ} / d p = - \int_{0}^{\infty} t e^{- pt} u (t) d t

, proving

\overline{t u} = - d \overset{u}{ˉ} / d p

. Start with

\overline{e^{- t}} = 1/ (p + 1)

and apply this rule repeatedly:

\overset{v}{ˉ}_{0} = \frac{1}{p + 1}, \overset{v}{ˉ}_{1} = \frac{1}{( p + 1 ) ^{2}}, \overset{v}{ˉ}_{2} = \frac{2}{( p + 1 ) ^{3}}, \overset{v}{ˉ}_{3} = \frac{6}{( p + 1 ) ^{4}} .

With all three initial data zero, transforming the forced equation gives

(p + 1)^{3} \overset{y}{ˉ} = 1/ (p + 1)

. Thus

\overset{y}{ˉ} = (p + 1)^{- 4}

and the listed inverse transform gives

y (t) = \frac{t ^{3} e ^{- t}}{6}, t \geq 0.

The cubic factor is also the resonant exponential particular solution associated with the triple characteristic root

- 1

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