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

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

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inverse oflaplace (3s)/((s-1)(s^2+1))

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

\frac{3 s}{( s - 1 ) ( s ^{2} + 1 )} : \frac{3}{2 ( s - 1 )} + \frac{- 3 s + 3}{2 ( s ^{2} + 1 )}

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

Expand

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

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

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

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

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

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

x \to \infty lim (2 x - 2)

\frac{1}{4 s + 1}

y^{^{'}} + 2 y = 6, y (0) = 0

(2.5) (1.1)

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

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