laplacetransform f(t)=-4t^2+16t+9

解答

拉普拉斯变换

f (t) = - 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{d}{d x} ((3 x^{2} - 4 x) e^{x})

\int \frac{4 x ^{3} - 7 x - 12}{x ^{3} - x ^{2} - 6 x} d x

\frac{d}{d x} (\frac{7}{2 x})

x \to - 1 lim (4 x^{3} + x^{2} + 2 x + 1)

\frac{d y}{d x} = y x^{9} sin (x^{10})