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受欢迎的微积分 >

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

解答

的拉普拉斯逆变换

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

解答

\frac{t cos ( t ) + 2 t - 3 sin ( t )}{2}

求解步骤

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

将

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

用部份分式展开:

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

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

L^{- 1} {\frac{1}{s ^{2}}} : t

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

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

= t - sin (t) - \frac{1}{2} (sin (t) - t cos (t))

化简

t - sin (t) - \frac{1}{2} (sin (t) - t cos (t)) : \frac{t cos ( t ) + 2 t - 3 sin ( t )}{2}

= \frac{t cos ( t ) + 2 t - 3 sin ( t )}{2}

流行的例子

limit as x approaches 0+of (1+4x)^{cot(x)}

x \to 0 + lim ((1 + 4 x)^{c o t (x)})

泰勒 sqrt(x^2+1)

t a y l or x^{2} + 1

y^'=sqrt(2y+39)

y^{^{'}} = 2 y + 39

derivative f(x)= 4/(x-2)

d er i v a t i v e f (x) = \frac{4}{x - 2}

limit as x approaches 4-of e^{3/(4-x)}

x \to 4 - lim (e^{\frac{3}{4 - x}})