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

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

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inverse oflaplace 2/(2s+1)

inverse laplace

\frac{2}{2 s + 1}

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

Solution steps

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

= L^{- 1} {1 \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)}

= 1 \cdot 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}}

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

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

x \to 0 lim ((1 + 4 x) \frac{1}{x})

f (x) = x^{4}, at x = 4

x \to \infty lim (\frac{ln ( x ) + x}{x n x})

f (x) = \frac{3 x ^{2} + 1}{x ^{\frac{3}{2}}}

\int e^{i 8 x} cos (4 x) d x

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