What is inverse oflaplace 6/(2s+1) ?

The answer to inverse oflaplace 6/(2s+1) is 3e^{-t/2}

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inverse oflaplace 6/(2s+1)

inverse laplace

\frac{6}{2 s + 1}

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

Solution steps

L^{- 1} {\frac{6}{2 s + 1}}

= L^{- 1} {3 \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)}

= 3 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}}

= 3 e^{- \frac{t}{2}}

\int cot (x) ln (sin (x)) d x

\frac{d}{d x} (\frac{x - 2}{x ^{2} + 5 x + 4})

\frac{d y}{d x} = - y^{\frac{1}{3}}

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

y = 2 x^{3}, y = 18 x

What is inverse oflaplace 6/(2s+1) ?
The answer to inverse oflaplace 6/(2s+1) is 3e^{-t/2}