laplacetransform-5(t-2)e^{t-2}

解答

拉普拉斯变换

- 5 (t - 2) e^{t - 2}

- \frac{5 ( 1 - 2 ( s - 1 ) )}{e ^{2} ( s - 1 ) ^{2}}

求解步骤

L {- 5 e^{t - 2} (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)}

= - 5 e^{- 2} L {(t - 2) e^{t}}

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

= - 5 e^{- 2} (\frac{1}{( s - 1 ) ^{2}} - \frac{2}{s - 1})

整理

- 5 e^{- 2} (\frac{1}{( s - 1 ) ^{2}} - \frac{2}{s - 1}) : - \frac{5 ( 1 - 2 ( s - 1 ) )}{e ^{2} ( s - 1 ) ^{2}}

= - \frac{5 ( 1 - 2 ( s - 1 ) )}{e ^{2} ( s - 1 ) ^{2}}

\int x sin (1 - x^{2}) d x

\int \frac{7 arcsin ( x )}{x ^{2}} d x

\frac{d y}{d x} = 5 x (y - 1)^{\frac{1}{5}}

s l o p e - 2 x^{5} + 11 x^{3}

\frac{d}{d x} ((3 - x) (2 - x) (5 - x))