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

解答

的拉普拉斯逆变换

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

t cos (t)

求解步骤

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

将

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

用部份分式展开:

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

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

利用逆拉普拉斯变换的线性特性:

对于函数

f (s), g (s)

和常数

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

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

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

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

化简

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

= t cos (t)

d er i v a t i v e 3 x^{2} e^{x} + e^{x} (x^{3} + 3)

d er i v a t i v e arctan (3 x)

\int cos^{4} (2 θ) d θ

\frac{d}{d x} (e^{a x} cos (- b x))

\int_{1}^{3} x 2^{(x^{2})} d x