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

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

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

inverse laplace

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

\frac{1}{2} cos (t) - \frac{1}{2} sin (t)

Solution steps

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

Expand

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

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

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

= \frac{1}{2} cos (t) - \frac{1}{2} sin (t)

2 x^{2} - 5 x + 3

sin (x - x^{2})

\frac{d}{d x} (2^{2^{x}})

\int tan^{2} (3 x) sec^{2} (3 x) d x

\int - \frac{cos ( x )}{sin ^{2} ( x )} d x

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