inverselaplace 1/((s^2-9))

解答

的拉普拉斯逆变换

\frac{1}{( s ^{2} - 9 )}

- \frac{1}{6} e^{- 3 t} + \frac{1}{6} e^{3 t}

求解步骤

L^{- 1} {\frac{1}{( s ^{2} - 9 )}}

将

\frac{1}{s ^{2} - 9}

用部份分式展开:

- \frac{1}{6 ( s + 3 )} + \frac{1}{6 ( s - 3 )}

= L^{- 1} {- \frac{1}{6 ( s + 3 )} + \frac{1}{6 ( s - 3 )}}

利用逆拉普拉斯变换的线性特性:

对于函数

f (s), g (s)

和常数

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= - L^{- 1} {\frac{1}{6 ( s + 3 )}} + L^{- 1} {\frac{1}{6 ( s - 3 )}}

L^{- 1} {\frac{1}{6 ( s + 3 )}} : \frac{1}{6} e^{- 3 t}

L^{- 1} {\frac{1}{6 ( s - 3 )}} : \frac{1}{6} e^{3 t}

= - \frac{1}{6} e^{- 3 t} + \frac{1}{6} e^{3 t}

\frac{d y}{d x} + 2 x^{2} y = 11 x^{2}

\int \frac{e ^{2 x} + 1}{e ^{2 x} - 1} d x

x \to 0 - lim (1 + \frac{1}{x})

\int_{- \infty}^{6} 10 x e^{2 x} d x

\frac{d}{d x} (ln (x x^{2} - 2))