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

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

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

inverse laplace

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

e^{3 t}

Solution steps

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

Expand

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

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

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

= - \frac{1}{2} e^{t} + \frac{3}{2} e^{3 t} - (- \frac{1}{2} e^{t} + \frac{1}{2} e^{3 t})

Simplify

- \frac{1}{2} e^{t} + \frac{3}{2} e^{3 t} - (- \frac{1}{2} e^{t} + \frac{1}{2} e^{3 t}) : e^{3 t}

= e^{3 t}

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

\int \frac{1 + cos ^{2} ( x )}{cos ^{2} ( x )} d x

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

\int 3 x^{2} + 2 x - 1 d x

\frac{d}{d x} (3 x^{2} - x + 5)

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