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

The answer to inverse oflaplace 1/(2s^2+3s+1) is e^{-t/2}-e^{-t}

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

inverse laplace

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

e^{- \frac{t}{2}} - e^{- t}

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= e^{- \frac{t}{2}} - e^{- t}

x^{6} cos (x)

y (s) = \frac{1}{s + k e ^{s}}

n = 1 \sum \infty arcsin (n)

y^{^{''}} + 4 y^{^{'}} - 5 y = e^{3 x}

\frac{d}{d x} (\frac{x ^{9}}{45})

What is inverse oflaplace 1/(2s^2+3s+1) ?
The answer to inverse oflaplace 1/(2s^2+3s+1) is e^{-t/2}-e^{-t}