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

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

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laplacetransform 2-2e^{-4t}

laplace transform

2 - 2 e^{- 4 t}

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

Solution steps

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

L {2} : \frac{2}{s}

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

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

Simplify

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

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

sin (t) sin (t)

\frac{\partial}{\partial x} (x^{3} - 3 x y + y^{3})

x \to - 2 lim (\frac{1}{∣ x + 2 ∣} + x^{2})

y^{^{'}} = 2.5 - \frac{y}{2000}

\frac{d y}{d t} + \frac{1}{2} y = 17 sin (2 t), y (0) = - 1

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