Rozwiązać poniższe równanie różniczkowe, stosując rachunek operatorowy Laplace’a:

dla następujących danych:

a= 4 f(t)= t

b= 8 x(0)=x’(0)

$$S^{2}x\left( S \right) + 4\text{Sx}\left( S \right) + 8x\left( S \right) = \ \frac{1}{S^{2}}$$

$x\left( S \right) = \frac{1}{S^{2}}*\frac{1}{S^{2} + 4S + 8}$

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$$x\left( S \right) = \frac{A_{a}}{S^{2}} + \frac{BS + C}{S^{2} + 4S + 8}$$

S² + 4 + 8

(S + α)² + ω²

α = 2

ω = 2

$$\frac{BS + C}{S^{2} + 4S + 8} = B\frac{S + 2}{\left( S + 2 \right)^{2} + 2^{2}} + \frac{C - 2B}{2}*\frac{2}{\left( S + 2 \right)^{2} + 2^{2}}$$

Zaś $\frac{A_{a}}{S^{2}} = \frac{A_{1}}{s} + \frac{A_{2}}{s} = \frac{A_{1} + A_{2}}{s} = \ \frac{A}{s}$

Należy również zapamiętać aby $\frac{1}{S^{2}}\ zamienic\ na\ \frac{1}{S}$

$$x\left( S \right) = \frac{A}{S^{2}} + B\frac{S + 2}{{(S + 2)}^{2} + 2^{2}} + \frac{C - 2B}{2}*\frac{2}{\left( S + 2 \right)^{2} + 2^{2}}$$

$$x\left( t \right) = A + Be^{- 2t}\cos{2t} + \frac{C - 2B}{2}e^{- 2t}\sin{2t}$$

1 = A(S²+4S+8) + (BS+C) * S

Dla S = 0

1 = A(0−0+8) + 0

1 = A * 8     

$$A = \frac{1}{8}$$

Dla $S = \pm 1\ \ \ \ A = \frac{1}{8}$

$$\left\{ \begin{matrix} 1 = \frac{1}{8}\left( 1 + 4 + 8 \right) + \left( B + C \right) \\ 1 = \frac{1}{8}\left( 1 - 4 + 8 \right) + \left( B - C \right) \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} 1 = \frac{13}{8} + B + C \\ 1 = \frac{5}{8} + B - C \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} - \frac{5}{8} = B + C \\ \frac{3}{8} = B - C \\ \end{matrix} \right.\ $$

$$- \frac{2}{8} = 2B$$

$$B = \ - \frac{1}{8}$$

$$c = - \frac{1}{2}$$

$$A = \frac{1}{8}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B = - \frac{1}{8}\ \ \ \ \ \ \ \ \ \ \ \ \ \ C = \ - \frac{1}{2}$$

$$x\left( S \right) = \frac{\frac{1}{8}}{S} + \frac{- \frac{1}{8}*S - \frac{15}{2}}{S^{2} + 4S + 8}$$

$$x\left( t \right) = \frac{1}{8} - \frac{1}{8}e^{- 2t}\cos{2t} - \frac{1}{8}e^{- 2t}\sin{2t}$$