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

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

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

inverse laplace

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

cos (t) - sin (t)

Solution steps

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

Expand

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

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

= cos (t) - sin (t)

\frac{\partial}{\partial y} (x e^{y} + 1)

\int \frac{1}{csc ( 2 x )} d x

x \to - 1 lim (x + 2)

\int (- 8 sin (4 x) + 3 cos (4 x)) d x

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

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