What is inverse oflaplace (20)/(-5s+3) ?

The answer to inverse oflaplace (20)/(-5s+3) is -4e^{(3t)/5}

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inverse oflaplace (20)/(-5s+3)

inverse laplace

\frac{20}{- 5 s + 3}

- 4 e^{\frac{3 t}{5}}

Solution steps

L^{- 1} {\frac{20}{- 5 s + 3}}

= L^{- 1} {- 4 \cdot \frac{1}{s - \frac{3}{5}}}

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)}

= - 4 L^{- 1} {\frac{1}{s - \frac{3}{5}}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{1}{s - \frac{3}{5}}} = e^{\frac{3 t}{5}}

= - 4 e^{\frac{3 t}{5}}

\frac{d}{d x} (\frac{1}{x} + 2 x)

sin (t - \frac{π}{2})

f (x) = 3 - 4 x

n = 0 \sum \infty \frac{1}{3} \cdot (\frac{1}{3})^{n}

\int x^{2} sin (z x) d x

What is inverse oflaplace (20)/(-5s+3) ?
The answer to inverse oflaplace (20)/(-5s+3) is -4e^{(3t)/5}