What is inverse oflaplace (1.6)/(3.2s+1) ?

The answer to inverse oflaplace (1.6)/(3.2s+1) is 0.5e^{-0.3125t}

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inverse oflaplace (1.6)/(3.2s+1)

inverse laplace

\frac{1.6}{3.2 s + 1}

0.5 e^{- 0.3125 t}

Solution steps

L^{- 1} {\frac{1.6}{3.2 s + 1}}

= L^{- 1} {0.5 \cdot \frac{1}{s + 0.3125}}

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)}

= 0.5 L^{- 1} {\frac{1}{s + 0.3125}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{1}{s + 0.3125}} = e^{- 0.3125 t}

= 0.5 e^{- 0.3125 t}

\int \frac{4 x}{1 + 2 x} d x

x = y^{2}, x = y + 6

\frac{\partial}{\partial s} (s cos (t - \frac{π}{2}))

\frac{d}{d x} (\frac{- 3}{x ^{2} + 9})

\int \frac{x ^{2}}{a - x ^{2}} d x

What is inverse oflaplace (1.6)/(3.2s+1) ?
The answer to inverse oflaplace (1.6)/(3.2s+1) is 0.5e^{-0.3125t}