laplacetransform (t-1)^4

解答

拉普拉斯变换

(t - 1)^{4}

\frac{24}{s ^{5}} - \frac{24}{s ^{4}} + \frac{12}{s ^{3}} - \frac{4}{s ^{2}} + \frac{1}{s}

求解步骤

L {(t - 1)^{4}}

展开

(t - 1)^{4} : t^{4} - 4 t^{3} + 6 t^{2} - 4 t + 1

= L {t^{4} - 4 t^{3} + 6 t^{2} - 4 t + 1}

利用拉普拉斯变换的线性特性:

对于函数

f (t), g (t)

和常数

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= L {t^{4}} - 4 L {t^{3}} + 6 L {t^{2}} - 4 L {t} + L {1}

L {t^{4}} : \frac{24}{s ^{5}}

L {t^{3}} : \frac{6}{s ^{4}}

L {t^{2}} : \frac{2}{s ^{3}}

L {t} : \frac{1}{s ^{2}}

L {1} : \frac{1}{s}

= \frac{24}{s ^{5}} - 4 \cdot \frac{6}{s ^{4}} + 6 \cdot \frac{2}{s ^{3}} - 4 \cdot \frac{1}{s ^{2}} + \frac{1}{s}

整理

\frac{24}{s ^{5}} - 4 \frac{6}{s ^{4}} + 6 \frac{2}{s ^{3}} - 4 \frac{1}{s ^{2}} + \frac{1}{s} : \frac{24}{s ^{5}} - \frac{24}{s ^{4}} + \frac{12}{s ^{3}} - \frac{4}{s ^{2}} + \frac{1}{s}

= \frac{24}{s ^{5}} - \frac{24}{s ^{4}} + \frac{12}{s ^{3}} - \frac{4}{s ^{2}} + \frac{1}{s}

\frac{\partial}{\partial x} (25 - x^{2})

\frac{d y}{d x} + \frac{1}{x + 8} y = x^{- 2}

e^{x} y d y = (e^{- y} + e^{- 2 x - y}) d x

x \to \infty lim (2^{2})

n \to 1 lim (\frac{( - \frac{1}{5} ) ^{n}}{n})