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

解答

的拉普拉斯逆变换

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

t sin (t)

求解步骤

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

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

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

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

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

在点

0 = 0

处计算

sin (t) - t cos (t)

= t sin (t)

\int sin (\frac{1}{2 x}) d x

t an g e n t y = 16 x

\int \frac{3 x ^{4}}{( 2 x ^{5} - 5 ) ^{4}} d x

\int 6 t^{2} d t

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