What is laplacetransform 1-2e^{-2t}+e^{-4t} ?

The answer to laplacetransform 1-2e^{-2t}+e^{-4t} is 8/(s(s+2)(s+4))

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laplacetransform 1-2e^{-2t}+e^{-4t}

laplace transform

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

\frac{8}{s ( s + 2 ) ( s + 4 )}

Solution steps

L {1 - 2 e^{- 2 t} + e^{- 4 t}}

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= L {1} - 2 L {e^{- 2 t}} + L {e^{- 4 t}}

L {1} : \frac{1}{s}

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

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

= \frac{1}{s} - 2 \cdot \frac{1}{s + 2} + \frac{1}{s + 4}

Simplify

\frac{1}{s} - 2 \frac{1}{s + 2} + \frac{1}{s + 4} : \frac{8}{s ( s + 2 ) ( s + 4 )}

= \frac{8}{s ( s + 2 ) ( s + 4 )}

\int \frac{1}{16 - 81 x ^{2}} d x

\int_{\frac{1}{2}}^{1} (x^{- 3} - 17) d x

\frac{d}{d x} (\frac{x ^{2} + a}{x ^{2} - a})

f (x) = - 3 sin (3 x)

\int ln (x) x^{2} d x

What is laplacetransform 1-2e^{-2t}+e^{-4t} ?
The answer to laplacetransform 1-2e^{-2t}+e^{-4t} is 8/(s(s+2)(s+4))