laplacetransform (2t-1)^3

解答

拉普拉斯变换

(2 t - 1)^{3}

\frac{48}{s ^{4}} - \frac{24}{s ^{3}} + \frac{6}{s ^{2}} - \frac{1}{s}

求解步骤

L {(2 t - 1)^{3}}

展开

(2 t - 1)^{3} : 8 t^{3} - 12 t^{2} + 6 t - 1

= L {8 t^{3} - 12 t^{2} + 6 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)}

= 8 L {t^{3}} - 12 L {t^{2}} + 6 L {t} - L {1}

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}

= 8 \cdot \frac{6}{s ^{4}} - 12 \cdot \frac{2}{s ^{3}} + 6 \cdot \frac{1}{s ^{2}} - \frac{1}{s}

整理

8 \frac{6}{s ^{4}} - 12 \frac{2}{s ^{3}} + 6 \frac{1}{s ^{2}} - \frac{1}{s} : \frac{48}{s ^{4}} - \frac{24}{s ^{3}} + \frac{6}{s ^{2}} - \frac{1}{s}

= \frac{48}{s ^{4}} - \frac{24}{s ^{3}} + \frac{6}{s ^{2}} - \frac{1}{s}

\frac{d}{d x} (3 e^{x} + \frac{2}{3 x})

\int x^{2} 2 x^{3} + 1 d x

2 y^{^{'}} + 4 y = - 8 y^{3}

d er i v a t i v e y = f (y, x) (g (x))

\frac{d}{d x} (x y z)