inverselaplace 1/(9s+1)

解答

的拉普拉斯逆变换

\frac{1}{9 s + 1}

\frac{1}{9} e^{- \frac{t}{9}}

求解步骤

L^{- 1} {\frac{1}{9 s + 1}}

= L^{- 1} {\frac{1}{9} \cdot \frac{1}{s + \frac{1}{9}}}

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)}

= \frac{1}{9} L^{- 1} {\frac{1}{s + \frac{1}{9}}}

使用逆拉普拉斯变换表:

L^{- 1} {\frac{1}{s - a}} = e^{a t}

L^{- 1} {\frac{1}{s + \frac{1}{9}}} = e^{- \frac{t}{9}}

= \frac{1}{9} e^{- \frac{t}{9}}

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

y^{^{''}} - 5 y^{^{'}} + 6 y = e^{2 t}

x \to 3 lim (\frac{1}{( x - 3 ) ^{2}})

f^{^{''}} (x) = \frac{1}{x}

y^{^{''}} - 4 y^{^{'}} + 20 y = 0, y (0) = - 1, y^{^{'}} (0) = - 3