What is inverse oflaplace 2/((s^3-7s^2+2s+40)) ?

The answer to inverse oflaplace 2/((s^3-7s^2+2s+40)) is 1/21 e^{-2t}-1/3 e^{4t}+2/7 e^{5t}

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inverse oflaplace 2/((s^3-7s^2+2s+40))

inverse laplace

\frac{2}{( s ^{3} - 7 s ^{2} + 2 s + 40 )}

\frac{1}{21} e^{- 2 t} - \frac{1}{3} e^{4 t} + \frac{2}{7} e^{5 t}

Solution steps

L^{- 1} {\frac{2}{( s ^{3} - 7 s ^{2} + 2 s + 40 )}}

Take the partial fraction of

\frac{2}{s ^{3} - 7 s ^{2} + 2 s + 40} : \frac{1}{21 ( s + 2 )} - \frac{1}{3 ( s - 4 )} + \frac{2}{7 ( s - 5 )}

= L^{- 1} {\frac{1}{21 ( s + 2 )} - \frac{1}{3 ( s - 4 )} + \frac{2}{7 ( s - 5 )}}

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

L^{- 1} {\frac{1}{21 ( s + 2 )}} : \frac{1}{21} e^{- 2 t}

L^{- 1} {\frac{1}{3 ( s - 4 )}} : \frac{1}{3} e^{4 t}

L^{- 1} {\frac{2}{7 ( s - 5 )}} : \frac{2}{7} e^{5 t}

= \frac{1}{21} e^{- 2 t} - \frac{1}{3} e^{4 t} + \frac{2}{7} e^{5 t}

arccos (∣ x + 1 ∣)

f (x) = 2 x^{2} - 5 x + 3, at a = 2

\frac{\partial}{\partial x} (ln (2 x^{2} + x y))

\int \frac{1}{x 100 x ^{2} - 100} d x

f (x) = x^{9} e^{x}

What is inverse oflaplace 2/((s^3-7s^2+2s+40)) ?
The answer to inverse oflaplace 2/((s^3-7s^2+2s+40)) is 1/21 e^{-2t}-1/3 e^{4t}+2/7 e^{5t}