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

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

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laplacetransform 1-2e^{-2t}

laplace transform

1 - 2 e^{- 2 t}

\frac{1}{s} - \frac{2}{s + 2}

Solution steps

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

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

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

Refine

\frac{1}{s} - 2 \frac{1}{s + 2} : \frac{1}{s} - \frac{2}{s + 2}

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

\int e^{4 x} 1 + e^{2 x} d x

x \to 3 lim (2 x^{2} + 4 x + 1)

y^{^{''}} - 2 y^{^{'}} + 5 y = 25 x^{2} + 12

\int_{0}^{2} \frac{3 t}{( 3 - t ) ^{2}} d t

y = 3 x^{2} + 5 x, (- 3, - 13)

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