What is laplacetransform 1-e^{-t/4} ?

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

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laplacetransform 1-e^{-t/4}

laplace transform

1 - e^{- \frac{t}{4}}

\frac{1}{s ( 4 s + 1 )}

Solution steps

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

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^{- \frac{1}{4} t}}

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

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

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

Simplify

\frac{1}{s} - \frac{4}{4 s + 1} : \frac{1}{s ( 4 s + 1 )}

= \frac{1}{s ( 4 s + 1 )}

\int x^{2} (9 - x^{3})^{\frac{2}{3}} d x

\int \frac{6 x - 12}{x ^{2} - 4 x - 5} d x

\int \frac{1}{u ^{\frac{1}{2}}} d u

n = 0 \sum \infty \frac{n}{3 ^{n + 1}}

\frac{\partial}{\partial x} (\frac{- y - x y}{( 1 + x ) ^{3}})

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