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

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

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

inverse laplace

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

\frac{3}{2} H (t) - \frac{3}{2} e^{- 2 t}

Solution steps

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

Expand

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

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

Take the partial fraction of

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

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

Expand

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

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

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

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

= \frac{3}{2} H (t) - \frac{3}{2} e^{- 2 t}

y = x e^{- k x}

f (x) = x^{2} - 7

41 (6 t^{3} - 7)^{\frac{6}{5}}

\int_{0}^{\frac{π}{2}} t sin (2 t) d t

y = arctan (2 x)

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