What is laplacetransform 1+e^{4t} ?

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

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laplacetransform 1+e^{4t}

laplace transform

1 + e^{4 t}

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

Solution steps

L {1 + 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 {1} + L {e^{4 t}}

L {1} : \frac{1}{s}

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

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

x \to - 4 lim (8 x + 3)

y^{^{'}} + y = t e^{t}

x \frac{d y}{d x} + 4 y = 2 - \frac{7}{x}

\int (x^{3} + 4 x - 3) d x

5 x^{- 15}

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