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Popular >

inverse oflaplace 1/(2s+1)

Solution

inverse laplace

\frac{1}{2 s + 1}

Solution

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

Solution steps

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

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

Popular Examples

integral of (-x)/(sqrt(x+1))

\int \frac{- x}{x + 1} d x

integral of (sin(x)-2cos(x))

\int (sin (x) - 2 cos (x)) d x

inverse oflaplace (2s)/(s^2+4)inverse laplace

\frac{2 s}{s ^{2} + 4}

derivative of y=ln(8-3x)derivative of

y = ln (8 - 3 x)

integral of e^{-u}u

\int e^{- u} u d u

Frequently Asked Questions (FAQ)

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