What is inverse oflaplace 1/(4s+2) ?

The answer to inverse oflaplace 1/(4s+2) is 1/4 e^{-t/2}

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inverse oflaplace 1/(4s+2)

inverse laplace

\frac{1}{4 s + 2}

\frac{1}{4} e^{- \frac{t}{2}}

Solution steps

L^{- 1} {\frac{1}{4 s + 2}}

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= \frac{1}{4} L^{- 1} {\frac{1}{s + \frac{1}{2}}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{1}{s - a}} = e^{a t}

L^{- 1} {\frac{1}{s + \frac{1}{2}}} = e^{- \frac{t}{2}}

= \frac{1}{4} e^{- \frac{t}{2}}

x \to - 3 lim (- x^{2} + 9 x - 1)

\int \frac{7}{x ^{2} + 16} d x

\frac{d}{d θ} (2 θ)

\frac{d}{d x} (x^{2} tan (x y))

\frac{d}{d x} (\frac{3 x + 1}{7 x + 5})

What is inverse oflaplace 1/(4s+2) ?
The answer to inverse oflaplace 1/(4s+2) is 1/4 e^{-t/2}