What is inverse oflaplace (2s+9)/((s+4)(s+6)) ?

The answer to inverse oflaplace (2s+9)/((s+4)(s+6)) is 1/2 e^{-4t}+3/2 e^{-6t}

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inverse oflaplace (2s+9)/((s+4)(s+6))

inverse laplace

\frac{2 s + 9}{( s + 4 ) ( s + 6 )}

\frac{1}{2} e^{- 4 t} + \frac{3}{2} e^{- 6 t}

Solution steps

L^{- 1} {\frac{2 s + 9}{( s + 4 ) ( s + 6 )}}

Take the partial fraction of

\frac{2 s + 9}{( s + 4 ) ( s + 6 )} : \frac{1}{2 ( s + 4 )} + \frac{3}{2 ( s + 6 )}

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

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

L^{- 1} {\frac{3}{2 ( s + 6 )}} : \frac{3}{2} e^{- 6 t}

= \frac{1}{2} e^{- 4 t} + \frac{3}{2} e^{- 6 t}

\int_{0}^{2} x^{3} e^{x^{2}} d x

\int_{- 1}^{3} ∣ x - 2 ∣ d x

\frac{\partial}{\partial y} (2 x^{4} y z^{3})

cos (2 x)

\frac{d}{d x} (\frac{x y}{x ^{2} + y ^{2} + z ^{2}})

What is inverse oflaplace (2s+9)/((s+4)(s+6)) ?
The answer to inverse oflaplace (2s+9)/((s+4)(s+6)) is 1/2 e^{-4t}+3/2 e^{-6t}