laplacetransform (5e^{2t}-3)^2

解答

拉普拉斯变换

(5 e^{2 t} - 3)^{2}

\frac{25}{s - 4} - \frac{30}{s - 2} + \frac{9}{s}

求解步骤

L {(5 e^{2 t} - 3)^{2}}

展开

(5 e^{2 t} - 3)^{2} : 25 e^{4 t} - 30 e^{2 t} + 9

= L {25 e^{4 t} - 30 e^{2 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)}

= 25 L {e^{4 t}} - 30 L {e^{2 t}} + L {9}

L {e^{4 t}} : \frac{1}{s - 4}

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

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

= 25 \cdot \frac{1}{s - 4} - 30 \cdot \frac{1}{s - 2} + \frac{9}{s}

整理

25 \frac{1}{s - 4} - 30 \frac{1}{s - 2} + \frac{9}{s} : \frac{25}{s - 4} - \frac{30}{s - 2} + \frac{9}{s}

= \frac{25}{s - 4} - \frac{30}{s - 2} + \frac{9}{s}

x \to 6 lim (7 x - 6)

y^{^{''}} + y = 2 t cos (t)

d er i v a t i v e y = (4 x^{2} + 5 x)^{2}

\int 2 t^{2} e^{- s t} d t

t an g e n t f (x) = 2 x^{3} - 4 x + 6, at x = 1