laplacetransform e^{t-2}

解答

拉普拉斯变换

e^{t - 2}

\frac{1}{e ^{2} ( s - 1 )}

求解步骤

L {e^{t - 2}}

利用拉普拉斯变换的常数乘法特性:

对于函数

f (t)

和常数

a : L {a \cdot f (t)} = a \cdot L {f (t)}

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

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

= e^{- 2} \frac{1}{s - 1}

整理

e^{- 2} \frac{1}{s - 1} : \frac{1}{e ^{2} ( s - 1 )}

= \frac{1}{e ^{2} ( s - 1 )}

\frac{d}{d x} (ln (1 - \frac{1}{x ^{2}}))

\frac{\partial}{\partial x} (a cos^{2} (2 π) (t - x))

t an g e n t \frac{x ^{2}}{x - 5}

d er i v a t i v e f (x) = x^{2} + 7

\frac{d}{d x} (x^{0.1})