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

The answer to inverse oflaplace (s-2)/((s+1)(s-3)) is 3/4 e^{-t}+1/4 e^{3t}

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

inverse laplace

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

\frac{3}{4} e^{- t} + \frac{1}{4} e^{3 t}

Solution steps

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

Take the partial fraction of

\frac{s - 2}{( s + 1 ) ( s - 3 )} : \frac{3}{4 ( s + 1 )} + \frac{1}{4 ( s - 3 )}

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

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

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

= \frac{3}{4} e^{- t} + \frac{1}{4} e^{3 t}

(\frac{1}{1 - \frac{x}{2}})^{3}, 0

n = 2 \sum \infty \frac{7}{4 n + 2}

y = x^{3} - 4 x + 4

\int \frac{y}{y ^{2}} d y

\frac{d}{d x} (x + \frac{400}{x})

What is inverse oflaplace (s-2)/((s+1)(s-3)) ?
The answer to inverse oflaplace (s-2)/((s+1)(s-3)) is 3/4 e^{-t}+1/4 e^{3t}