What is inverse oflaplace 1/((s+2)(s^2-2s+5)) ?

The answer to inverse oflaplace 1/((s+2)(s^2-2s+5)) is 1/13 e^{-2t}-1/13 e^tcos(2t)+3/26 e^tsin(2t)

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inverse oflaplace 1/((s+2)(s^2-2s+5))

inverse laplace

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

\frac{1}{13} e^{- 2 t} - \frac{1}{13} e^{t} cos (2 t) + \frac{3}{26} e^{t} sin (2 t)

Solution steps

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

Take the partial fraction of

\frac{1}{( s + 2 ) ( s ^{2} - 2 s + 5 )} : \frac{1}{13 ( s + 2 )} + \frac{- s + 4}{13 ( s ^{2} - 2 s + 5 )}

= L^{- 1} {\frac{1}{13 ( s + 2 )} + \frac{- s + 4}{13 ( s ^{2} - 2 s + 5 )}}

Expand

\frac{- s + 4}{13 ( s ^{2} - 2 s + 5 )} : - \frac{1}{13} \cdot \frac{s - 1}{( s - 1 ) ^{2} + 4} + \frac{3}{13} \cdot \frac{1}{( s - 1 ) ^{2} + 4}

= L^{- 1} {\frac{1}{13 ( s + 2 )} - \frac{1}{13} \cdot \frac{s - 1}{( s - 1 ) ^{2} + 4} + \frac{3}{13} \cdot \frac{1}{( s - 1 ) ^{2} + 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}{13 ( s + 2 )}} - \frac{1}{13} L^{- 1} {\frac{s - 1}{( s - 1 ) ^{2} + 4}} + \frac{3}{13} L^{- 1} {\frac{1}{( s - 1 ) ^{2} + 4}}

L^{- 1} {\frac{1}{13 ( s + 2 )}} : \frac{1}{13} e^{- 2 t}

L^{- 1} {\frac{s - 1}{( s - 1 ) ^{2} + 4}} : e^{t} cos (2 t)

L^{- 1} {\frac{1}{( s - 1 ) ^{2} + 4}} : e^{t} \frac{1}{2} sin (2 t)

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

Refine

\frac{1}{13} e^{- 2 t} - \frac{1}{13} e^{t} cos (2 t) + \frac{3}{13} e^{t} \frac{1}{2} sin (2 t) : \frac{1}{13} e^{- 2 t} - \frac{1}{13} e^{t} cos (2 t) + \frac{3}{26} e^{t} sin (2 t)

= \frac{1}{13} e^{- 2 t} - \frac{1}{13} e^{t} cos (2 t) + \frac{3}{26} e^{t} sin (2 t)

\int_{- 3}^{- 1} x^{2} e^{x^{3} + 27} d x

\int tan^{2} (a x) cos (a x) d x

\int e^{x} arctan (e^{x}) d x

x^{2} + 8 x

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

What is inverse oflaplace 1/((s+2)(s^2-2s+5)) ?
The answer to inverse oflaplace 1/((s+2)(s^2-2s+5)) is 1/13 e^{-2t}-1/13 e^tcos(2t)+3/26 e^tsin(2t)