# 使用拉普拉斯变换求解微分方程

$$\mathscr{L}\{ -2 y' + y\} = \mathscr{L}\{0 \}$$

$$- 2 \mathscr{L}\{ y'\} + \mathscr{L}\{ y\} = \mathscr{L}\{0 \}$$

$$- 2 ( s Y(s) - y(0)) + Y(s) = 0$$

$$- 2 s Y(s) + 2 y(0) + Y(s) = 0$$

$$- 2 s Y(s) + 2 + Y(s) = 0$$

$$Y(s) (1 - 2 s) = -2$$
$$Y(s) = \dfrac{2}{2 s - 1}$$
$$Y(s) = \dfrac{1}{ s - 1/2}$$

$$\displaystyle y(t) = e^{\frac{1}{2} t }$$

$$- 2 y' + y = - 2 ( (1/2) e^{\frac{1}{2} t } ) + e^{\frac{1}{2} t }$$

$$- e^{\frac{1}{2} t } + e^{\frac{1}{2} t } = 0$$ ；微分方程满足。
$$y(0) = e^{\frac{1}{2} 0 } = e^0 = 1$$ ；初始值也满足。

$$\mathscr{L}\{ y'' - 2 y' -3 y \} = \mathscr{L}\{0 \}$$

$$\mathscr{L}\{ y"\} - 2 \mathscr{L}\{ y'\} - 3 \mathscr{L}\{ y \} = \mathscr{L}\{0 \}$$

$$s^2 Y(s) - s y(0) - y'(0) - 2 (sY(s) - y(0)) - 3 Y(s) = 0$$

$$s^2 Y(s) - 2 s + 1 - 2 s Y(s) + 4 - 3Y(s) = 0$$

$$s^2 Y(s) - 2 s Y(s) - 3 Y(s) = 2 s - 5$$

$$Y(s) (s^2 - 2 s - 3 ) = 2 s - 5$$

$$Y(s) = \dfrac{2s - 5}{s^2 - 2 s - 3}$$

$$Y(s) = \dfrac{7}{4\left(s+1\right)}+\dfrac{1}{4\left(s-3\right)}$$

$$\displaystyle y(t) = \dfrac{7}{4} e^{- t } + \dfrac{1}{4} e^{3 t }$$

$$\mathscr{L}\{ y'' + 2 y' + 2 y \} = \mathscr{L}\{0 \}$$

$$\mathscr{L}\{ y"\} + 2 \mathscr{L}\{ y'\} + 2 \mathscr{L}\{ y \} = \mathscr{L}\{0 \}$$

$$s^2 Y(s) - s y(0) - y'(0) + 2 (sY(s) - y(0)) + 2 Y(s) = 0$$

$$s^2 Y(s) + s - 2 + 2 s Y(s) + 2 + 2 Y(s) = 0$$

$$s^2 Y(s) + 2 s Y(s) + 2 Y(s) = - s$$

$$Y(s) (s^2 + 2 s + 2 ) = - s$$

$$Y(s) = \dfrac{-s}{s^2 + 2 s + 2}$$

$$s^2 + 2 s + 2 = 0$$

$$S_1 = -1 + j$$ 和 $$s_2 = -1 - j$$

$$Y(s) = \dfrac{-s}{(s - s_1)(s - s_2)}$$

$$\dfrac{-s}{(s - s_1)(s - s_2)} = \dfrac{A}{s-s_1} + \dfrac{B}{s-s_2}$$

$$A = \dfrac{-s_1}{s_1-s_2} = \dfrac{-(-1 + j)}{2 j} = -\dfrac{1}{2} - \dfrac{1}{2} j$$

$$B = \dfrac{-s_2}{s_2-s_1} = \dfrac{-(-1 - j)}{-2 j} = - \dfrac{1}{2} + \dfrac{1}{2} j$$

$$y(t) = A e^{s_1 t} + B e^{s_2 t}$$

$$A = -\dfrac{1}{2} - \dfrac{1}{2} j = \frac{\sqrt 2}{2} e^{ \frac{-3\pi}{4} j}$$
$$B = -\dfrac{1}{2} + \dfrac{1}{2} j = \frac{\sqrt 2}{2} e^{ \frac{3\pi}{4} j}$$

$$y(t) = (\frac{\sqrt 2}{2} e^{ \frac{-3\pi}{4} j}) e^{(-1 + j) t} + (\frac{\sqrt 2}{2} e^{ \frac{3\pi}{4} j}) e^{(-1 - j) t}$$

$$y(t) = \dfrac{\sqrt 2}{2} e^{-t} \left[ e^{j t - \frac{3\pi}{4} j } + e^{-j t + \frac{3\pi}{4} j } \right]$$

$$y(t) = \dfrac{\sqrt 2}{2} e^{-t} \left[ \cos(t - \frac{3\pi}{4}) + j\sin(t - \frac{3\pi}{4}) + \cos(-t + \frac{3\pi}{4}) + j\sin(- t + \frac{3\pi}{4}) \right]$$

$$y(t) = \sqrt 2 e^{-t} \cos(t - \frac{3\pi}{4})$$

$$\mathscr{L}\{ y'' - y' - 2 y \} = \mathscr{L}\{ \sin(3t) \}$$

$$\mathscr{L}\{ y"\} - \mathscr{L}\{ y'\} - 2 \mathscr{L}\{ y \} = \dfrac{3}{s^2+3^2}$$

$$s^2 Y(s) - s y(0) - y'(0) - (sY(s) - y(0)) + 2 Y(s) = \dfrac{3}{s^2+3^2}$$

$$s^2 Y(s) - s + 1 - s Y(s) + 1 - 2 Y(s) = \dfrac{3}{s^2+3^2}$$

$$s^2 Y(s) - s Y(s) - 2 Y(s) = \dfrac{3}{s^2+3^2} + s - 2$$

$$Y(s) (s^2 - s - 2 ) = \dfrac{3}{s^2+3^2} + s - 2$$

$$Y(s) = \dfrac{3}{(s^2+3^2)(s^2 - s - 2)} + \dfrac{s-2}{s^2 - s - 2}$$

$$Y(s) = \dfrac{3}{(s^2+3^2)(s-2)(s+1)} + \dfrac{s-2}{(s-2)(s+1)}$$

$$Y(s) = \dfrac{3s}{130(s^2+3^2)} - \dfrac{33}{130(s^2+3^2)} + \dfrac{9}{10(s+1)} + \dfrac{1}{13(s-2)}$$

$$y(t) = \dfrac{3}{130} \cos(3x) - \dfrac{11}{130} \sin(3x) + \dfrac{9}{10} e^{-x} +\dfrac{1}{13} e^{2x}$$

## 附录 A

$$\dfrac{2s - 5}{s^2 - 2 s - 3} = \dfrac{2s - 5}{(s-3)(s+1)}$$

$$\dfrac{2s - 5}{s^2 - 2 s - 3} = \dfrac{A}{s+1} + \dfrac{B}{s-3}$$

$$2s - 5 = A(s-3) + B(s+1)$$      (1)

2(3) - 5 = A(3 -3) + B(3+1)

$$B = 1/4$$

$$A = \dfrac{7}{4}$$

## 附录 B

$$\dfrac{-s}{(s - s_1)(s - s_2)} = \dfrac{A}{s-s_1} + \dfrac{B}{s-s_2}$$

$$- s = A (s-s_2) + B(s - s_1)$$       (1)

$$- s_1 = A (s_1-s_2) + B(s_1 - s_1)$$

$$-s_1 = A (s_1-s_2)$$

$$A = \dfrac{-s_1}{s_1-s_2} = \dfrac{-(-1 + j)}{2 j} = -\dfrac{1}{2} - \dfrac{1}{2} j$$

$$B = \dfrac{-s_2}{s_2-s_1} = \dfrac{-(-1 - j)}{-2 j} = - \dfrac{1}{2} + \dfrac{1}{2} j$$

## 附录 C

$$\dfrac{3}{(s^2+3^2)(s^2 - s - 2)} + \dfrac{s-2}{s^2 - s - 2}$$

$$\dfrac{3}{(s^2+3^2)(s-2)(s+1)} + \dfrac{s-2}{(s-2)(s+1)}$$

$$\dfrac{3}{(s^2+3^2)(s-2)(s+1)} + \dfrac{1}{s+1}$$

$$\dfrac{3}{(s^2+3^2)(s-2)(s+1)} + \dfrac{1}{s+1} = \dfrac{As + B}{s^2+3^2} + \dfrac{C}{s+1} + \dfrac{D}{s-2}$$

$$3 + (s^2+3^2)(s-2) = (As + B)(s-2)(s+1) + C (s^2+3^2)(s-2) + D (s^2+3^2)(s+1)$$     (1)

$$3 + (2^2+3^2)(2-2) = (2 A + B)(2-2)(s+1) + C (2^2+3^2)(2-2) + D (2^2+3^2)(2+1)$$

$$3 = 39 D$$

$$D = \dfrac{1}{13}$$

$$3 + ((-1)^2+3^2)(-1-2) = (-A + B)(-1-2)(-1+1) + C ((-1)^2+3^2)(-1-2) + D ((-1)^2+3^2)(-1+1)$$

$$3 - 30 = - 30 C$$

$$C = \dfrac{9}{10}$$

$$3 +(0^2+3^2)(0-2) = (0 + B)(0-2)(0+1) + C (0^2+3^2)(0-2) + D (0^2+3^2)(0+1)$$

$$3 - 18 = -2 B - 19 C + 9D$$

$$B = -\dfrac{33}{130}$$

$$3 + (1^2+3^2)(1-2) = (A + B)(1-2)(1+1) + C (1^2+3^2)(1-2) + D (1^2+3^2)(1+1)$$

$$A = \dfrac{3}{130}$$

$$\dfrac{3}{(s^2+3^2)(s^2 - s - 2)} + \dfrac{s-2}{s^2 - s - 2}$$

$$\quad \quad = \dfrac{As}{s^2+3^2} + \dfrac{B}{s^2+3^2} + \dfrac{C}{s+1} + \dfrac{D}{s-2}$$

$$\quad \quad = \dfrac{ 3s}{130(s^2+3^2)} - \dfrac{33}{130(s^2+3^2)} + \dfrac{9}{10(s+1)} + \dfrac{1}{13(s-2)}$$