What is inverse oflaplace (s^2-1)/(s^4+3s^2+2) ?

The answer to inverse oflaplace (s^2-1)/(s^4+3s^2+2) is -2sin(t)+3/(sqrt(2))sin(sqrt(2)t)

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inverse oflaplace (s^2-1)/(s^4+3s^2+2)

inverse laplace

\frac{s ^{2} - 1}{s ^{4} + 3 s ^{2} + 2}

- 2 sin (t) + \frac{3}{2} sin (2 t)

Solution steps

L^{- 1} {\frac{s ^{2} - 1}{s ^{4} + 3 s ^{2} + 2}}

Take the partial fraction of

\frac{s ^{2} - 1}{s ^{4} + 3 s ^{2} + 2} : - \frac{2}{s ^{2} + 1} + \frac{3}{s ^{2} + 2}

= L^{- 1} {- \frac{2}{s ^{2} + 1} + \frac{3}{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{2}{s ^{2} + 1}} + L^{- 1} {\frac{3}{s ^{2} + 2}}

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

L^{- 1} {\frac{3}{s ^{2} + 2}} : \frac{3}{2} sin (2 t)

= - 2 sin (t) + \frac{3}{2} sin (2 t)

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

- t

\frac{\partial}{\partial x} (- 2 s t)

x \to 0 + lim (\frac{x}{cos ( 4 x ) - 1})

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

What is inverse oflaplace (s^2-1)/(s^4+3s^2+2) ?
The answer to inverse oflaplace (s^2-1)/(s^4+3s^2+2) is -2sin(t)+3/(sqrt(2))sin(sqrt(2)t)