laplacetransform (t-5)/5

解答

拉普拉斯变换

\frac{t - 5}{5}

\frac{1}{5 s ^{2}} - \frac{1}{s}

求解步骤

L {\frac{t - 5}{5}}

乘开

= L {\frac{t}{5} - \frac{5}{5}}

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)}

= \frac{1}{5} L {t} - L {\frac{5}{5}}

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

L {\frac{5}{5}} : \frac{1}{s}

= \frac{1}{5} \cdot \frac{1}{s ^{2}} - \frac{1}{s}

整理

\frac{1}{5} \frac{1}{s ^{2}} - \frac{1}{s} : \frac{1}{5 s ^{2}} - \frac{1}{s}

= \frac{1}{5 s ^{2}} - \frac{1}{s}

\int sin^{3} (11 x) d x

\int_{- 3}^{3} (9 - x^{2}) d x

d er i v a t i v e \frac{π}{2}

x \to - 2 lim (\frac{x + 2}{x ^{2} - 4})

x \to 3 lim (- (3 - x))