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

The answer to inverse oflaplace (s+2)/(s(s+1)(s+3)) is 2/3 \H(t)-1/2 e^{-t}-1/6 e^{-3t}

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

inverse laplace

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

\frac{2}{3} H (t) - \frac{1}{2} e^{- t} - \frac{1}{6} e^{- 3 t}

Solution steps

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

Take the partial fraction of

\frac{s + 2}{s ( s + 1 ) ( s + 3 )} : \frac{2}{3 s} - \frac{1}{2 ( s + 1 )} - \frac{1}{6 ( s + 3 )}

= L^{- 1} {\frac{2}{3 s} - \frac{1}{2 ( s + 1 )} - \frac{1}{6 ( s + 3 )}}

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{2}{3 s}} - L^{- 1} {\frac{1}{2 ( s + 1 )}} - L^{- 1} {\frac{1}{6 ( s + 3 )}}

L^{- 1} {\frac{2}{3 s}} : \frac{2}{3} H (t)

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

L^{- 1} {\frac{1}{6 ( s + 3 )}} : \frac{1}{6} e^{- 3 t}

= \frac{2}{3} H (t) - \frac{1}{2} e^{- t} - \frac{1}{6} e^{- 3 t}

\frac{\partial}{\partial x} (sin (2 x) + cos (3 y))

e^{x} (1)

\frac{d y}{d x} = 8 - 2 x

\frac{d}{d x} (e^{x^{5}} + lo g_{2} (π))

\int \frac{cos ( u )}{u} d u

What is inverse oflaplace (s+2)/(s(s+1)(s+3)) ?
The answer to inverse oflaplace (s+2)/(s(s+1)(s+3)) is 2/3 \H(t)-1/2 e^{-t}-1/6 e^{-3t}