What is inverse oflaplace 5/((s-2)(s-3)(s-6)) ?

The answer to inverse oflaplace 5/((s-2)(s-3)(s-6)) is 5/4 e^{2t}-5/3 e^{3t}+5/12 e^{6t}

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inverse oflaplace 5/((s-2)(s-3)(s-6))

inverse laplace

\frac{5}{( s - 2 ) ( s - 3 ) ( s - 6 )}

\frac{5}{4} e^{2 t} - \frac{5}{3} e^{3 t} + \frac{5}{12} e^{6 t}

Solution steps

L^{- 1} {\frac{5}{( s - 2 ) ( s - 3 ) ( s - 6 )}}

Take the partial fraction of

\frac{5}{( s - 2 ) ( s - 3 ) ( s - 6 )} : \frac{5}{4 ( s - 2 )} - \frac{5}{3 ( s - 3 )} + \frac{5}{12 ( s - 6 )}

= L^{- 1} {\frac{5}{4 ( s - 2 )} - \frac{5}{3 ( s - 3 )} + \frac{5}{12 ( s - 6 )}}

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{5}{4 ( s - 2 )}} - L^{- 1} {\frac{5}{3 ( s - 3 )}} + L^{- 1} {\frac{5}{12 ( s - 6 )}}

L^{- 1} {\frac{5}{4 ( s - 2 )}} : \frac{5}{4} e^{2 t}

L^{- 1} {\frac{5}{3 ( s - 3 )}} : \frac{5}{3} e^{3 t}

L^{- 1} {\frac{5}{12 ( s - 6 )}} : \frac{5}{12} e^{6 t}

= \frac{5}{4} e^{2 t} - \frac{5}{3} e^{3 t} + \frac{5}{12} e^{6 t}

\int cot (\frac{θ}{5}) d θ

\int \frac{48 x ^{2}}{( x - 24 ) ( x + 8 ) ^{2}} d x

\int_{7}^{\infty} e^{- 2 x} + x^{- 3} d x

x^{3} + y^{3} - 9 x y = 0, (2, 4)

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

What is inverse oflaplace 5/((s-2)(s-3)(s-6)) ?
The answer to inverse oflaplace 5/((s-2)(s-3)(s-6)) is 5/4 e^{2t}-5/3 e^{3t}+5/12 e^{6t}