What is laplacetransform e^{-4t}+5 ?

The answer to laplacetransform e^{-4t}+5 is 1/(s+4)+5/s

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laplacetransform e^{-4t}+5

laplace transform

e^{- 4 t} + 5

\frac{1}{s + 4} + \frac{5}{s}

Solution steps

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

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

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

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

= \frac{1}{s + 4} + \frac{5}{s}

x^{\frac{1}{2}}, 1

\frac{d}{d x} (sin (x^{2}) + \frac{1 - x}{1 + x})

\frac{d}{d x} (2 (x - sin (x)))

\frac{d}{d x} (\frac{4 x}{9 - tan ( x )})

\frac{\partial}{\partial z} (x e^{w (x, z)})

What is laplacetransform e^{-4t}+5 ?
The answer to laplacetransform e^{-4t}+5 is 1/(s+4)+5/s