What is inverse oflaplace 7/(s^3+2s^2+9s+18) ?

The answer to inverse oflaplace 7/(s^3+2s^2+9s+18) is 7/13 e^{-2t}-7/13 cos(3t)+14/39 sin(3t)

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inverse oflaplace 7/(s^3+2s^2+9s+18)

inverse laplace

\frac{7}{s ^{3} + 2 s ^{2} + 9 s + 18}

\frac{7}{13} e^{- 2 t} - \frac{7}{13} cos (3 t) + \frac{14}{39} sin (3 t)

Solution steps

L^{- 1} {\frac{7}{s ^{3} + 2 s ^{2} + 9 s + 18}}

Take the partial fraction of

\frac{7}{s ^{3} + 2 s ^{2} + 9 s + 18} : \frac{7}{13 ( s + 2 )} + \frac{- 7 s + 14}{13 ( s ^{2} + 9 )}

= L^{- 1} {\frac{7}{13 ( s + 2 )} + \frac{- 7 s + 14}{13 ( s ^{2} + 9 )}}

Expand

= L^{- 1} {- \frac{7 s}{13 ( s ^{2} + 9 )} + \frac{7}{13 ( s + 2 )} + \frac{14}{13 ( s ^{2} + 9 )}}

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)}

= \frac{7}{13} L^{- 1} {\frac{1}{s + 2}} - \frac{7}{13} L^{- 1} {\frac{s}{s ^{2} + 9}} + \frac{14}{13} L^{- 1} {\frac{1}{s ^{2} + 9}}

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

L^{- 1} {\frac{s}{s ^{2} + 9}} : cos (3 t)

L^{- 1} {\frac{1}{s ^{2} + 9}} : \frac{1}{3} sin (3 t)

= \frac{7}{13} e^{- 2 t} - \frac{7}{13} cos (3 t) + \frac{14}{13} \cdot \frac{1}{3} sin (3 t)

Refine

\frac{7}{13} e^{- 2 t} - \frac{7}{13} cos (3 t) + \frac{14}{13} \frac{1}{3} sin (3 t) : \frac{7}{13} e^{- 2 t} - \frac{7}{13} cos (3 t) + \frac{14}{39} sin (3 t)

= \frac{7}{13} e^{- 2 t} - \frac{7}{13} cos (3 t) + \frac{14}{39} sin (3 t)

\int \frac{e ^{x}}{e ^{- 6 x}} d x

\frac{d}{d x} (\frac{5 x}{9 x ^{2} + 1})

y^{^{'}} = y - k

(x^{2} - 4) (\frac{d y}{d x}) + 4 y = (x + 2)^{2}

(- 7, - 7), (- 3, 6)

What is inverse oflaplace 7/(s^3+2s^2+9s+18) ?
The answer to inverse oflaplace 7/(s^3+2s^2+9s+18) is 7/13 e^{-2t}-7/13 cos(3t)+14/39 sin(3t)