laplacetransform (t-2)^3

解答

拉普拉斯变换

(t - 2)^{3}

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

求解步骤

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

展开

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

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

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

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

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

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

L {8} : \frac{8}{s}

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

整理

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

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

d er i v a t i v e - \frac{x}{2}

\int \frac{( 3 )}{( u ^{2} - 1 )} d u

a re a y = (x - 2)^{2} - 4, y = x + 1

\int \frac{cos ( \frac{π}{x} )}{x ^{2}} d x

\frac{d}{d x} (x^{2} - 16 x)