What is inverse oflaplace 9/((s-2)(s-3)) ?

The answer to inverse oflaplace 9/((s-2)(s-3)) is -9e^{2t}+9e^{3t}

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inverse oflaplace 9/((s-2)(s-3))

inverse laplace

\frac{9}{( s - 2 ) ( s - 3 )}

- 9 e^{2 t} + 9 e^{3 t}

Solution steps

L^{- 1} {\frac{9}{( s - 2 ) ( s - 3 )}}

Take the partial fraction of

\frac{9}{( s - 2 ) ( s - 3 )} : - \frac{9}{s - 2} + \frac{9}{s - 3}

= L^{- 1} {- \frac{9}{s - 2} + \frac{9}{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{9}{s - 2}} + L^{- 1} {\frac{9}{s - 3}}

L^{- 1} {\frac{9}{s - 2}} : 9 e^{2 t}

L^{- 1} {\frac{9}{s - 3}} : 9 e^{3 t}

= - 9 e^{2 t} + 9 e^{3 t}

f (x) = (2 x) (4 x^{2} + 6 x^{7} + 8)

\int \frac{1}{b ^{2} + a ^{2} x ^{2}} d x

\int x^{2} ln (x) + e^{2 x} + 1 d x

\frac{d}{d x} (x^{2} - \frac{1}{x})

e^{- 5 x}

What is inverse oflaplace 9/((s-2)(s-3)) ?
The answer to inverse oflaplace 9/((s-2)(s-3)) is -9e^{2t}+9e^{3t}