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

The answer to inverse oflaplace 1/(s^2(s+sqrt(2))) is (-\H(t)+sqrt(2)t+e^{-sqrt(2)t})/2

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

inverse laplace

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

\frac{- H ( t ) + 2 t + e ^{- 2 t}}{2}

Solution steps

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

Take the partial fraction of

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

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

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

L^{- 1} {\frac{1}{2 s ^{2}}} : \frac{t}{2}

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

= - \frac{1}{2} H (t) + \frac{t}{2} + \frac{1}{2} e^{- 2 t}

Simplify

- \frac{1}{2} H (t) + \frac{t}{2} + \frac{1}{2} e^{- 2 t} : \frac{- H ( t ) + 2 t + e ^{- 2 t}}{2}

= \frac{- H ( t ) + 2 t + e ^{- 2 t}}{2}

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y = \frac{1}{( 1 + x ^{2} ) , ( - 1 , \frac{1}{2} )}

\int (x e^{2 x}) d x

\frac{d ^{2} y}{d x ^{2}} + 6 \frac{d y}{d x} + 5 y = 0

\int \frac{x ^{2} + 18 x - 9}{x ^{3} - 9 x} d x

What is inverse oflaplace 1/(s^2(s+sqrt(2))) ?
The answer to inverse oflaplace 1/(s^2(s+sqrt(2))) is (-\H(t)+sqrt(2)t+e^{-sqrt(2)t})/2