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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

\int \frac{25 x ^{2} - 1}{x} d x

\int 2 (x + 20) e^{\frac{x}{20}} d x

\frac{d}{d x} (x cos (4 x))

\int \frac{t ^{4} + 1}{t ^{5} + 4 t ^{3}} d t

(x^{2} + 1) y^{^{'}} + 9 x (y - 1) = 0, y (0) = 4

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