What is inverse oflaplace 1/((s-2)(s-4)) ?

The answer to inverse oflaplace 1/((s-2)(s-4)) is -1/2 e^{2t}+1/2 e^{4t}

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inverse oflaplace 1/((s-2)(s-4))

inverse laplace

\frac{1}{( s - 2 ) ( s - 4 )}

- \frac{1}{2} e^{2 t} + \frac{1}{2} e^{4 t}

Solution steps

L^{- 1} {\frac{1}{( s - 2 ) ( s - 4 )}}

Take the partial fraction of

\frac{1}{( s - 2 ) ( s - 4 )} : - \frac{1}{2 ( s - 2 )} + \frac{1}{2 ( s - 4 )}

= L^{- 1} {- \frac{1}{2 ( s - 2 )} + \frac{1}{2 ( s - 4 )}}

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}{2 ( s - 2 )}} + L^{- 1} {\frac{1}{2 ( s - 4 )}}

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

L^{- 1} {\frac{1}{2 ( s - 4 )}} : \frac{1}{2} e^{4 t}

= - \frac{1}{2} e^{2 t} + \frac{1}{2} e^{4 t}

y^{^{'''}} - 12 y^{^{''}} + 52 y^{^{'}} - 96 y = 0

y = e^{x}, y = 7 e^{- x} + 6, x = 0

\int tan^{9} (x) sec^{4} (x) d x

\frac{d}{d x} (1 0^{x^{2}} + lo g_{10} (∣ 2 + 3 x ∣))

What is inverse oflaplace 1/((s-2)(s-4)) ?
The answer to inverse oflaplace 1/((s-2)(s-4)) is -1/2 e^{2t}+1/2 e^{4t}