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)

\int_{1}^{3} ln (x - 1) d x

n = 1 \sum \infty cos (2 n)

y = x^{2} - 1, y = x + 2, x = 0, x + 1

x \to - 1 lim (5 - 3 x - x^{2})

(3.1) (0)

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)