What is inverse oflaplace 1/((s-4)(s-5)) ?

The answer to inverse oflaplace 1/((s-4)(s-5)) is -e^{4t}+e^{5t}

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inverse oflaplace 1/((s-4)(s-5))

inverse laplace

\frac{1}{( s - 4 ) ( s - 5 )}

- e^{4 t} + e^{5 t}

Solution steps

L^{- 1} {\frac{1}{( s - 4 ) ( s - 5 )}}

Take the partial fraction of

\frac{1}{( s - 4 ) ( s - 5 )} : - \frac{1}{s - 4} + \frac{1}{s - 5}

= L^{- 1} {- \frac{1}{s - 4} + \frac{1}{s - 5}}

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 - 4}} + L^{- 1} {\frac{1}{s - 5}}

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

L^{- 1} {\frac{1}{s - 5}} : e^{5 t}

= - e^{4 t} + e^{5 t}

\frac{\partial}{\partial x} (9 y)

\frac{d}{d x} (b x^{2})

\frac{d}{d x} ((2 x + 6)^{2})

y = (2 x - 2)^{4} (x^{2} + x + 1)^{5}

\int e^{x^{2} + l n (x)} d x

What is inverse oflaplace 1/((s-4)(s-5)) ?
The answer to inverse oflaplace 1/((s-4)(s-5)) is -e^{4t}+e^{5t}