What is inverse oflaplace 1/((s-1)(s^2+9)) ?

The answer to inverse oflaplace 1/((s-1)(s^2+9)) is 1/10 e^t-1/10 cos(3t)-1/30 sin(3t)

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inverse oflaplace 1/((s-1)(s^2+9))

inverse laplace

\frac{1}{( s - 1 ) ( s ^{2} + 9 )}

\frac{1}{10} e^{t} - \frac{1}{10} cos (3 t) - \frac{1}{30} sin (3 t)

Solution steps

L^{- 1} {\frac{1}{( s - 1 ) ( s ^{2} + 9 )}}

Take the partial fraction of

\frac{1}{( s - 1 ) ( s ^{2} + 9 )} : \frac{1}{10 ( s - 1 )} + \frac{- s - 1}{10 ( s ^{2} + 9 )}

= L^{- 1} {\frac{1}{10 ( s - 1 )} + \frac{- s - 1}{10 ( s ^{2} + 9 )}}

Expand

= L^{- 1} {- \frac{s}{10 ( s ^{2} + 9 )} + \frac{1}{10 ( s - 1 )} - \frac{1}{10 ( 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{1}{10} L^{- 1} {\frac{1}{s - 1}} - \frac{1}{10} L^{- 1} {\frac{s}{s ^{2} + 9}} - \frac{1}{10} L^{- 1} {\frac{1}{s ^{2} + 9}}

L^{- 1} {\frac{1}{s - 1}} : e^{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{1}{10} e^{t} - \frac{1}{10} cos (3 t) - \frac{1}{10} \cdot \frac{1}{3} sin (3 t)

Refine

\frac{1}{10} e^{t} - \frac{1}{10} cos (3 t) - \frac{1}{10} \frac{1}{3} sin (3 t) : \frac{1}{10} e^{t} - \frac{1}{10} cos (3 t) - \frac{1}{30} sin (3 t)

= \frac{1}{10} e^{t} - \frac{1}{10} cos (3 t) - \frac{1}{30} sin (3 t)

2 x^{2} - 8 x - 55, - x^{2} - 2 x + 50, - 5, 7

\frac{\partial}{\partial y} (e^{x z} y^{3} - ln (x))

\frac{d}{d x} (f (x) (x^{3}))

\int x (x + 3)^{7} d x

x^{2} \frac{d y}{d x} = y (4 x + 3)

What is inverse oflaplace 1/((s-1)(s^2+9)) ?
The answer to inverse oflaplace 1/((s-1)(s^2+9)) is 1/10 e^t-1/10 cos(3t)-1/30 sin(3t)