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

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

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

inverse laplace

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

\frac{1}{2} H (t) + \frac{5}{2} e^{- 2 t}

Solution steps

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

Take the partial fraction of

\frac{3 s + 1}{s ( s + 2 )} : \frac{1}{2 s} + \frac{5}{2 ( s + 2 )}

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

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

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

= \frac{1}{2} H (t) + \frac{5}{2} e^{- 2 t}

\frac{1}{2 x + 2}, at x = 5

(4 - x^{2}) \frac{d y}{d x} = 2 y

y = (\frac{x + 1}{x - 3})^{2}, (7, 4)

\int 2 cos^{2} (\frac{x}{2}) d x

\int \frac{1}{t ln ( t )} d t

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