inverselaplace ((s+3))/(((s+3)^2+25))

解答

的拉普拉斯逆变换

\frac{( s + 3 )}{( ( s + 3 ) ^{2} + 25 )}

e^{- 3 t} cos (5 t)

求解步骤

L^{- 1} ⎩ ⎨ ⎧ \frac{( s + 3 )}{( ( s + 3 ) ^{2} + 25 )} ⎭ ⎬ ⎫

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{s + 3}{( s + 3 ) ^{2} + 25} : a = - 3, F (s) = \frac{s}{s ^{2} + 25}

= e^{- 3 t} L^{- 1} {\frac{s}{s ^{2} + 25}}

L^{- 1} {\frac{s}{s ^{2} + 25}} : cos (5 t)

= e^{- 3 t} cos (5 t)

\int \frac{3 x ^{5} - 7 x ^{3}}{x ^{2}} d x

\int (x + 1) cos (x^{2} + 2 x + 1) d x

\frac{d}{d x} (6 x^{2} - 18 x + 12)

a re a x^{2}, - 4, 0

\int x^{5} x^{3} + 4 d x