laplacetransform 2t^2-e^{-1}

解答

拉普拉斯变换

2 t^{2} - e^{- 1}

\frac{4}{s ^{3}} - \frac{1}{es}

求解步骤

L {2 t^{2} - e^{- 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)}

= 2 L {t^{2}} - L {e^{- 1}}

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

L {e^{- 1}} : \frac{1}{es}

= 2 \cdot \frac{2}{s ^{3}} - \frac{1}{es}

整理

2 \frac{2}{s ^{3}} - \frac{1}{es} : \frac{4}{s ^{3}} - \frac{1}{es}

= \frac{4}{s ^{3}} - \frac{1}{es}

n = 1 \sum \infty \frac{1}{4}

t an g e n t f (x) = \frac{x ^{2}}{x + 2}, at x = 2

\int y^{2} e^{x y^{2}} d x

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

d er i v a t i v e \frac{x ^{2}}{x - 3}