What is inverse oflaplace 5/(2s^2+3s+1) ?

The answer to inverse oflaplace 5/(2s^2+3s+1) is 5e^{-t/2}-5e^{-t}

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inverse oflaplace 5/(2s^2+3s+1)

inverse laplace

\frac{5}{2 s ^{2} + 3 s + 1}

5 e^{- \frac{t}{2}} - 5 e^{- t}

Solution steps

L^{- 1} {\frac{5}{2 s ^{2} + 3 s + 1}}

Take the partial fraction of

\frac{5}{2 s ^{2} + 3 s + 1} : \frac{10}{2 s + 1} - \frac{5}{s + 1}

= L^{- 1} {\frac{10}{2 s + 1} - \frac{5}{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{10}{2 s + 1}} - L^{- 1} {\frac{5}{s + 1}}

L^{- 1} {\frac{10}{2 s + 1}} : 5 e^{- \frac{t}{2}}

L^{- 1} {\frac{5}{s + 1}} : 5 e^{- t}

= 5 e^{- \frac{t}{2}} - 5 e^{- t}

(2 x + 1)^{3}

\int x^{2} sin (π x) d x

f (t) = e^{- 3 t} + 3

\frac{d}{d x} (x^{2} - ln (x))

sin (x^{2} + π)

What is inverse oflaplace 5/(2s^2+3s+1) ?
The answer to inverse oflaplace 5/(2s^2+3s+1) is 5e^{-t/2}-5e^{-t}