What is laplacetransform 5+4e-2t ?

The answer to laplacetransform 5+4e-2t is 5/s+(4e)/s-2/(s^2)

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laplacetransform 5+4e-2t

laplace transform

5 + 4 e - 2 t

\frac{5}{s} + \frac{4 e}{s} - \frac{2}{s ^{2}}

Solution steps

L {5 + 4 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 {5} + L {4 e} - 2 L {t}

L {5} : \frac{5}{s}

L {4 e} : \frac{4 e}{s}

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

= \frac{5}{s} + \frac{4 e}{s} - 2 \cdot \frac{1}{s ^{2}}

Refine

\frac{5}{s} + \frac{4 e}{s} - 2 \frac{1}{s ^{2}} : \frac{5}{s} + \frac{4 e}{s} - \frac{2}{s ^{2}}

= \frac{5}{s} + \frac{4 e}{s} - \frac{2}{s ^{2}}

\frac{\partial}{\partial x} (x^{3} y + x y^{2} + x y + x)

\frac{d}{d x} (\frac{- 200 ln ( \frac{x}{40} )}{x + 20})

\int_{0}^{15} 2 π x (15 x - x^{2}) d x

\frac{\partial}{\partial x} (ln (\frac{2 x y}{5 z}))

\int - \frac{9}{x ^{10}} d x

What is laplacetransform 5+4e-2t ?
The answer to laplacetransform 5+4e-2t is 5/s+(4e)/s-2/(s^2)