 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2006/differential-equations/9/iv/solution

p

replaced by

p + c

gives

\overline{t e^{- c t}} (p) = \frac{1}{( p + c ) ^{2}} (Re (p + c) > 0) .

For the remaining unnumbered requests, the derivative and convolution formulas, under the usual convergence hypotheses, are

\overline{f^{'}} (p) = p \overset{ˉ}{f} (p) - f (0), \overline{f * g} (p) = \overset{ˉ}{f} (p) \overset{g}{ˉ} (p), (f * g) (t) = \int_{0}^{t} f (t - s) g (s) d s .

Applying them to the general equation yields

p \overset{x}{ˉ} - a + \overset{ˉ}{k} \overset{x}{ˉ} = \overset{y}{ˉ}

and therefore

\overset{x}{ˉ} (p) = \frac{y ˉ ( p ) + a}{p + k ˉ ( p )} .

For the particular data,

\overset{y}{ˉ} = 1/ p

\overset{ˉ}{k} = 1/ (p + 2)

and

a = 0

, so

\overset{x}{ˉ} (p) = \frac{p + 2}{p ( p + 1 ) ^{2}} = \frac{2}{p} - \frac{2}{p + 1} - \frac{1}{( p + 1 ) ^{2}} .

The four transforms above give

x (t) = 2 - (2 + t) e^{- t}, t \to \infty lim x (t) = 2.

As a direct check,

x (0) = 0

x^{'} = (1 + t) e^{- t}

, and

\int_{0}^{t} e^{- 2 (t - s)} x (s) d s = 1 - (1 + t) e^{- t} .

Adding the last two expressions gives one, verifying the original equation as well as the required limit.

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