What is inverse oflaplace (2s+1)/(s^3+7s^2+12s) ?

The answer to inverse oflaplace (2s+1)/(s^3+7s^2+12s) is 1/12 \H(t)+5/3 e^{-3t}-7/4 e^{-4t}

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inverse oflaplace (2s+1)/(s^3+7s^2+12s)

inverse laplace

\frac{2 s + 1}{s ^{3} + 7 s ^{2} + 12 s}

\frac{1}{12} H (t) + \frac{5}{3} e^{- 3 t} - \frac{7}{4} e^{- 4 t}

Solution steps

L^{- 1} {\frac{2 s + 1}{s ^{3} + 7 s ^{2} + 12 s}}

Take the partial fraction of

\frac{2 s + 1}{s ^{3} + 7 s ^{2} + 12 s} : \frac{1}{12 s} + \frac{5}{3 ( s + 3 )} - \frac{7}{4 ( s + 4 )}

= L^{- 1} {\frac{1}{12 s} + \frac{5}{3 ( s + 3 )} - \frac{7}{4 ( s + 4 )}}

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}{12 s}} + L^{- 1} {\frac{5}{3 ( s + 3 )}} - L^{- 1} {\frac{7}{4 ( s + 4 )}}

L^{- 1} {\frac{1}{12 s}} : \frac{1}{12} H (t)

L^{- 1} {\frac{5}{3 ( s + 3 )}} : \frac{5}{3} e^{- 3 t}

L^{- 1} {\frac{7}{4 ( s + 4 )}} : \frac{7}{4} e^{- 4 t}

= \frac{1}{12} H (t) + \frac{5}{3} e^{- 3 t} - \frac{7}{4} e^{- 4 t}

\frac{d}{d x} (cosh (0))

e^{x}, x^{2} - 1, x = - 1, x = 1

\int 29 sec^{- 2} (x t a) n^{3} x

x \to 0 lim (x^{2} + y^{2})

\int e^{7 x + 5} x d x

What is inverse oflaplace (2s+1)/(s^3+7s^2+12s) ?
The answer to inverse oflaplace (2s+1)/(s^3+7s^2+12s) is 1/12 \H(t)+5/3 e^{-3t}-7/4 e^{-4t}