What is inverse oflaplace (3s)/((s+2)(s-4)) ?

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

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inverse oflaplace (3s)/((s+2)(s-4))

inverse laplace

\frac{3 s}{( s + 2 ) ( s - 4 )}

e^{- 2 t} + 2 e^{4 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

= e^{- 2 t} + 2 e^{4 t}

x = 5 y^{2}, x = 2 + 3 y^{2}

\frac{d ^{2}}{d x ^{2}} (\frac{2 t - 1}{t ^{4}})

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

\frac{\partial}{\partial x} (\frac{5}{x})

e^{2 x} \cdot 2

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