What is laplacetransform f(t)=1+e^{4t} ?

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

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laplacetransform f(t)=1+e^{4t}

laplace transform

f (t) = 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}

y = 2 x, (18, 6)

\frac{\partial}{\partial x} (\frac{y e ^{- \frac{y}{x}}}{x ^{2}})

\int - 6 y d y

\int \frac{x}{1 + cos ( x )} d x

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

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