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

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

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

inverse laplace

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

H (t) - e^{3 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= H (t) - e^{3 t}

\int (\frac{2 x}{1 + x ^{4}}) d x

\int cot^{2} (4 x) d x

f (x) = e^{x} + x^{e}

\frac{d}{d x} (sec (8 x))

y^{^{''}} - y = 2 e^{x}

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