inverselaplace 8/((s^2+1)^2)

解答

的拉普拉斯逆变换

\frac{8}{( s ^{2} + 1 ) ^{2}}

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

求解步骤

L^{- 1} {\frac{8}{( s ^{2} + 1 ) ^{2}}}

= L^{- 1} {4 \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)}

= 4 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)

= 4 (sin (t) - t cos (t))

x^{2} \frac{d y}{d x} - y^{2} = 0

x \to 0 lim (x (ln (x))^{2})

\int \frac{4}{2 x + 1} d x

t an g e n t 3 x^{4} + 2 e^{x}, at x = 0

(x^{2} e^{- x})^{^{'}}