What is inverse oflaplace (s+5)/(s(s^2+5s+6)) ?

The answer to inverse oflaplace (s+5)/(s(s^2+5s+6)) is 5/6 \H(t)-3/2 e^{-2t}+2/3 e^{-3t}

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inverse oflaplace (s+5)/(s(s^2+5s+6))

inverse laplace

\frac{s + 5}{s ( s ^{2} + 5 s + 6 )}

\frac{5}{6} H (t) - \frac{3}{2} e^{- 2 t} + \frac{2}{3} e^{- 3 t}

Solution steps

L^{- 1} {\frac{s + 5}{s ( s ^{2} + 5 s + 6 )}}

Take the partial fraction of

\frac{s + 5}{s ( s ^{2} + 5 s + 6 )} : \frac{5}{6 s} - \frac{3}{2 ( s + 2 )} + \frac{2}{3 ( s + 3 )}

= L^{- 1} {\frac{5}{6 s} - \frac{3}{2 ( s + 2 )} + \frac{2}{3 ( 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{5}{6 s}} - L^{- 1} {\frac{3}{2 ( s + 2 )}} + L^{- 1} {\frac{2}{3 ( s + 3 )}}

L^{- 1} {\frac{5}{6 s}} : \frac{5}{6} H (t)

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

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

= \frac{5}{6} H (t) - \frac{3}{2} e^{- 2 t} + \frac{2}{3} e^{- 3 t}

y^{^{'''}} - y^{^{''}} = x e^{x}

(x^{2} + 4) y = 8, (- 2, 1)

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

\frac{d}{d x} (\frac{d y}{d x}) + 9 y = 0

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

What is inverse oflaplace (s+5)/(s(s^2+5s+6)) ?
The answer to inverse oflaplace (s+5)/(s(s^2+5s+6)) is 5/6 \H(t)-3/2 e^{-2t}+2/3 e^{-3t}