inverselaplace 1/((s^2+0.5s+1))

解答

的拉普拉斯逆变换

\frac{1}{( s ^{2} + 0.5 s + 1 )}

1.03279 \dots e^{- 0.25 t} sin (0.96824 \dots t)

求解步骤

L^{- 1} {\frac{1}{( s ^{2} + 0.5 s + 1 )}}

\frac{1}{s ^{2} + 0.5 s + 1} = \frac{1}{( s + 0.25 ) ^{2} + 0.9375}

= L^{- 1} {\frac{1}{( s + 0.25 ) ^{2} + 0.9375}}

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 + 0.25 ) ^{2} + 0.9375} : a = - 0.25, F (s) = \frac{1}{s ^{2} + 0.9375}

= e^{- 0.25 t} L^{- 1} {\frac{1}{s ^{2} + 0.9375}}

L^{- 1} {\frac{1}{s ^{2} + 0.9375}} : \frac{1}{0.96824 \dots} sin (0.96824 \dots t)

化简

e^{- 0.25 t} \frac{1}{0.96824 \dots} sin (0.96824 \dots t) : 1.03279 \dots e^{- 0.25 t} sin (0.96824 \dots t)

= 1.03279 \dots e^{- 0.25 t} sin (0.96824 \dots t)

y^{^{''}} - 6 y^{^{'}} + 9 y = \frac{e ^{3 x}}{x ^{2}}

a re a x = 15 - y^{2}, x = y^{2} - 3

\int_{0}^{\frac{π}{8}} arctan (2 x) d x

\frac{\partial}{\partial w} (w arctan (uv))

y^{^{''}} + 2 y - 1 (y^{^{'}}) 2 = y^{^{'}}