What is laplacetransform f(t)=e^{2t+1} ?

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

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laplacetransform f(t)=e^{2t+1}

laplace transform

f (t) = e^{2 t + 1}

\frac{e}{s - 2}

Solution steps

L {e^{2 t + 1}}

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= e^{1} L {e^{2 t}}

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

= e^{1} \frac{1}{s - 2}

Refine

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

= \frac{e}{s - 2}

\frac{d}{d x} (\frac{- 3 x + 8}{2 4 - x})

\frac{d}{d x} (\frac{x - 5}{4 x ^{2} e ^{x}})

x \to 0 lim (x^{4} cos (\frac{1}{x}))

(e^{- 2 t})^{^{'}}

\int \frac{- 6 x ^{4} + 6}{x ^{2} + 1} d x

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