What is inverse oflaplace 1/(s(s^2+3s+2)) ?

The answer to inverse oflaplace 1/(s(s^2+3s+2)) is 1/2 \H(t)-e^{-t}+1/2 e^{-2t}

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inverse oflaplace 1/(s(s^2+3s+2))

inverse laplace

\frac{1}{s ( s ^{2} + 3 s + 2 )}

\frac{1}{2} H (t) - e^{- t} + \frac{1}{2} e^{- 2 t}

Solution steps

L^{- 1} {\frac{1}{s ( s ^{2} + 3 s + 2 )}}

Take the partial fraction of

\frac{1}{s ( s ^{2} + 3 s + 2 )} : \frac{1}{2 s} - \frac{1}{s + 1} + \frac{1}{2 ( s + 2 )}

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

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

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

L^{- 1} {\frac{1}{2 ( s + 2 )}} : \frac{1}{2} e^{- 2 t}

= \frac{1}{2} H (t) - e^{- t} + \frac{1}{2} e^{- 2 t}

\frac{\partial}{\partial x} (x y e^{4 x - 2 z})

\frac{d y}{d x} + 9 y = e^{8 x}

x^{2}

- x, x^{2} - 12

n = 0 \sum \infty \frac{1}{13 n + 42}

What is inverse oflaplace 1/(s(s^2+3s+2)) ?
The answer to inverse oflaplace 1/(s(s^2+3s+2)) is 1/2 \H(t)-e^{-t}+1/2 e^{-2t}