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

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

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

inverse laplace

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

\frac{1}{3} H (t) - \frac{1}{3} e^{- 3 t}

Solution steps

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

Take the partial fraction of

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

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

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}{3 s}} - L^{- 1} {\frac{1}{3 ( s + 3 )}}

L^{- 1} {\frac{1}{3 s}} : \frac{1}{3} H (t)

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

= \frac{1}{3} H (t) - \frac{1}{3} e^{- 3 t}

\int \frac{1}{7 x} d x

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\int 3 x^{2} e^{(x^{3})} d x

\frac{d}{d x} (\frac{x ^{2} - 6 x + 12}{x - 4})

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