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

解答

的拉普拉斯逆变换

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

- (sin (t) - t cos (t))

求解步骤

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

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

利用逆拉普拉斯变换的常数乘法特性:

对于函数

f (t)

和常数

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

使用逆拉普拉斯变换表:

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

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

= - 1 \cdot (sin (t) - t cos (t))

= - (sin (t) - t cos (t))

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