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

The answer to inverse oflaplace 3/((s-3)(s+1)) is 3/4 e^{3t}-3/4 e^{-t}

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

inverse laplace

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

\frac{3}{4} e^{3 t} - \frac{3}{4} e^{- t}

Solution steps

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

Take the partial fraction of

\frac{3}{( s - 3 ) ( s + 1 )} : \frac{3}{4 ( s - 3 )} - \frac{3}{4 ( s + 1 )}

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

L^{- 1} {\frac{3}{4 ( s - 3 )}} : \frac{3}{4} e^{3 t}

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

= \frac{3}{4} e^{3 t} - \frac{3}{4} e^{- t}

\frac{\partial}{\partial x} (r^{2} s t + r s^{2} t + rs t^{2})

\frac{1}{3} \frac{d x}{d t} - 2 e^{- t} = - x

\int_{l n (8)}^{l n (64)} e^{\frac{x}{3}} d x

x \to 0 lim (\frac{tan ( π ) x}{x})

x \to 0 + lim (x - x)

What is inverse oflaplace 3/((s-3)(s+1)) ?
The answer to inverse oflaplace 3/((s-3)(s+1)) is 3/4 e^{3t}-3/4 e^{-t}