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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

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

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

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

f (x) = cos (3 x + 2) - 7 x

\frac{x sec ^{2} ( x ) - tan ( x )}{4 x ^{2}}

\frac{d y}{d x} = 3 x^{2} + 2 x - 7

\frac{\partial}{\partial x} (\frac{x ^{2} y ^{2}}{2})

n = 1 \sum \infty n^{- s i n (2)}

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