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)

x^{2} - 9 x

\frac{d}{d x} (- x sin (x) - cos (x))

(y e^{4 x y} + x) d x + (x e^{4 x y}) d y = 0

\int sin (x) cos (5 x) d x

x \to 0 lim (x^{a})

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)