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

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

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inverse oflaplace s/(2s+1)

inverse laplace

\frac{s}{2 s + 1}

\frac{1}{2} δ (t) - \frac{1}{4} e^{- \frac{t}{2}}

Solution steps

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

Take the partial fraction of

\frac{s}{2 s + 1} : \frac{1}{2} - \frac{1}{2 ( 2 s + 1 )}

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

L^{- 1} {\frac{1}{2}} : \frac{1}{2} δ (t)

L^{- 1} {\frac{1}{2 ( 2 s + 1 )}} : \frac{1}{4} e^{- \frac{t}{2}}

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

\frac{d y}{d x} + \frac{3}{x} y = 16 y^{3}, y (1) = 1

\int_{- 2}^{2} 2 - 4 - x^{2} d x

\frac{1}{s + 5}

f (x) = - 5 x^{2}, at x = 8

x (x + 1) (\frac{d y}{d x}) + x y = 1, y (e) = 1

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