laplacetransform (5e^{9t}-4)e^{-6t}

解答

拉普拉斯变换

(5 e^{9 t} - 4) e^{- 6 t}

\frac{5}{s - 3} - \frac{4}{s + 6}

求解步骤

L {e^{- 6 t} (5 e^{9 t} - 4)}

应用变换法则: if

L {f (t)} = F (s)

则

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

对于

(5 e^{9 t} - 4) e^{- 6 t} : f (t) = (5 e^{9 t} - 4), a = - 6

= L {(5 e^{9 t} - 4)} (s + 6)

L {(5 e^{9 t} - 4)} : \frac{5}{s - 9} - \frac{4}{s}

= \frac{5}{( s + 6 ) - 9} - \frac{4}{( s + 6 )}

= \frac{5}{s - 3} - \frac{4}{s + 6}

l a pl a ce t r an s f or m (t)^{2}

\int \frac{4}{81 - ( x + 9 ) ^{2}} d x

y^{^{'}} + 7 y = 3

\int \frac{x - 4}{x ( x - 1 ) ^{2}} d x

x \to 2 lim (\frac{x - 2}{x - 2})