What is inverse oflaplace 1/(2s+3) ?

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

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inverse oflaplace 1/(2s+3)

inverse laplace

\frac{1}{2 s + 3}

\frac{1}{2} e^{- \frac{3 t}{2}}

Solution steps

L^{- 1} {\frac{1}{2 s + 3}}

= L^{- 1} {\frac{1}{2} \cdot \frac{1}{s + \frac{3}{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{3}{2}}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{1}{s + \frac{3}{2}}} = e^{- \frac{3 t}{2}}

= \frac{1}{2} e^{- \frac{3 t}{2}}

\frac{\partial}{\partial y} (x y e^{2 y})

x \to 1 lim (arcsec (x))

\int cos (3 t) + sin (4 t) d t

\frac{d}{d x} (x^{3} + \frac{1}{x})

\int_{0}^{\infty} 70 \frac{e ^{x}}{e ^{2 x} + 3} d x

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