laplacetransform 3t^2+3t+1

解答

拉普拉斯变换

3 t^{2} + 3 t + 1

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

求解步骤

L {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)}

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

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

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

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

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

整理

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

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

d er i v a t i v e f (x) = x^{6} \cdot e^{x}

d er i v a t i v e f (x) = \frac{3}{4 x}

d er i v a t i v e (4 x - 5)^{3} (1 - x^{5})^{2}

y^{^{'}} = (x + y)^{2}, y (0) = 1

l a pl a ce t r an s f or m sin (x)