laplacetransform 6e^{-3t}-t^2+2t-8

解答

拉普拉斯变换

6 e^{- 3 t} - t^{2} + 2 t - 8

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

求解步骤

L {6 e^{- 3 t} - t^{2} + 2 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)}

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

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

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

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

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

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

整理

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

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

x \to 0 + lim (x ln (tan (6 x)))

\int - 3 sin (2 x) d x

\int_{- 5}^{0} 25 - x^{2} d x

\frac{d}{d x} (\frac{4 x ^{2} + 8 x + 8}{x})

\frac{d}{d x} (arctan (\frac{x}{b}))