laplacetransform t^2-4t+4

解答

拉普拉斯变换

t^{2} - 4 t + 4

\frac{2}{s ^{3}} - \frac{4}{s ^{2}} + \frac{4}{s}

求解步骤

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

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)}

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

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

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

L {4} : \frac{4}{s}

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

整理

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

= \frac{2}{s ^{3}} - \frac{4}{s ^{2}} + \frac{4}{s}

x d x + y d y = x y (x d y - y d x)

\frac{d}{d x} (- 6 x^{\frac{1}{4}})

\int \frac{1}{x ( x - 1 )} d x

x \to 1 lim (\frac{1}{2 x - 1})

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + y ^{3}})