inverselaplace 1/(s^2-4s+4)

解答

的拉普拉斯逆变换

\frac{1}{s ^{2} - 4 s + 4}

e^{2 t} t

求解步骤

L^{- 1} {\frac{1}{s ^{2} - 4 s + 4}}

\frac{1}{s ^{2} - 4 s + 4} = \frac{1}{( s - 2 ) ^{2}}

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

应用逆变换法则: if

L^{- 1} {F (s)} = f (t)

则

L^{- 1} {F (s - a)} = e^{a t} f (t)

对于

\frac{1}{( s - 2 ) ^{2}} : a = 2, F (s) = \frac{1}{s ^{2}}

= e^{2 t} L^{- 1} {\frac{1}{s ^{2}}}

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

= e^{2 t} t

\frac{d}{d x} (\frac{2}{x y})

f^{^{'}} (x) = x^{2} e^{x^{2}}

t an g e n t \frac{3 x}{x + 1}

d er i v a t i v e 2 x^{2} - \frac{1}{16} ln (x)

y = 8 i n^{x} + c