laplacetransform (t+1)^3

解答

拉普拉斯变换

(t + 1)^{3}

\frac{6}{s ^{4}} + \frac{6}{s ^{3}} + \frac{3}{s ^{2}} + \frac{1}{s}

求解步骤

L {(t + 1)^{3}}

展开

(t + 1)^{3} : t^{3} + 3 t^{2} + 3 t + 1

= L {t^{3} + 3 t^{2} + 3 t + 1}

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^{3}} + 3 L {t^{2}} + 3 L {t} + L {1}

L {t^{3}} : \frac{6}{s ^{4}}

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

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

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

= \frac{6}{s ^{4}} + 3 \cdot \frac{2}{s ^{3}} + 3 \cdot \frac{1}{s ^{2}} + \frac{1}{s}

整理

\frac{6}{s ^{4}} + 3 \frac{2}{s ^{3}} + 3 \frac{1}{s ^{2}} + \frac{1}{s} : \frac{6}{s ^{4}} + \frac{6}{s ^{3}} + \frac{3}{s ^{2}} + \frac{1}{s}

= \frac{6}{s ^{4}} + \frac{6}{s ^{3}} + \frac{3}{s ^{2}} + \frac{1}{s}

d er i v a t i v e 3 - x

\frac{\partial}{\partial x} (- 4 + 2 e^{x} - 3)

\int (tan (x))^{2} d x

d er i v a t i v e y = 2 x^{2} + 4 x - 10

y^{^{''}} + 3 y^{^{'}} + 2 y = 0