laplacetransform-4t^2+16t+9

解答

拉普拉斯变换

- 4 t^{2} + 16 t + 9

- \frac{8}{s ^{3}} + \frac{16}{s ^{2}} + \frac{9}{s}

求解步骤

L {- 4 t^{2} + 16 t + 9}

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)}

= - 4 L {t^{2}} + 16 L {t} + L {9}

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

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

L {9} : \frac{9}{s}

= - 4 \cdot \frac{2}{s ^{3}} + 16 \cdot \frac{1}{s ^{2}} + \frac{9}{s}

整理

- 4 \frac{2}{s ^{3}} + 16 \frac{1}{s ^{2}} + \frac{9}{s} : - \frac{8}{s ^{3}} + \frac{16}{s ^{2}} + \frac{9}{s}

= - \frac{8}{s ^{3}} + \frac{16}{s ^{2}} + \frac{9}{s}

\frac{\partial}{\partial x} (4 x (2 + y)^{- 1})

t an g e n t f (x) = \frac{x}{1 + x ^{2}}, at x = 2

\int \frac{x}{x - 10} d x

a re a 9.86 x^{2}, 4.92 x - x^{2}

\frac{\partial}{\partial t} (t^{3})