inverselaplace (3s)/((s^2+1))

解答

的拉普拉斯逆变换

\frac{3 s}{( s ^{2} + 1 )}

3 cos (t)

求解步骤

L^{- 1} {\frac{3 s}{( s ^{2} + 1 )}}

应用逆变换法则: if

L^{- 1} {F (s)} = f (t)

则

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

对于

\frac{3 s}{s ^{2} + 1} : F (s) = \frac{3}{s ^{2} + 1}

= \frac{d}{d t} (L^{- 1} {\frac{3}{s ^{2} + 1}}) + L^{- 1} {\frac{3}{s ^{2} + 1}} (0)

L^{- 1} {\frac{3}{s ^{2} + 1}} : 3 sin (t)

\frac{d}{d t} (3 sin (t)) = 3 cos (t)

在点

0 = 0

处计算

3 sin (t)

= 3 cos (t)

e x t re m e f (x) = r x + n

\frac{\partial}{\partial t} (t - s)

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

d er i v a t i v e f (x) = e^{4 t s i n (2 t)}

y^{^{'}} = e^{(t + y)}, y (0) = 0