What is inverse oflaplace (60)/(2-5s) ?

The answer to inverse oflaplace (60)/(2-5s) is -12e^{(2t)/5}

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inverse oflaplace (60)/(2-5s)

inverse laplace

\frac{60}{2 - 5 s}

- 12 e^{\frac{2 t}{5}}

Solution steps

L^{- 1} {\frac{60}{2 - 5 s}}

= L^{- 1} {- 12 \cdot \frac{1}{s - \frac{2}{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)}

= - 12 L^{- 1} {\frac{1}{s - \frac{2}{5}}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{1}{s - \frac{2}{5}}} = e^{\frac{2 t}{5}}

= - 12 e^{\frac{2 t}{5}}

k = 1 \sum \infty \frac{1}{k ^{1 + \frac{1}{k}}}

f (x) = x^{\frac{5}{2}}

x \to - 1 lim (x (x + 3))

n = 0 \sum \infty (\frac{2}{7})^{n}

y = x + 2, y = 2, y = \frac{1}{x + 1}

What is inverse oflaplace (60)/(2-5s) ?
The answer to inverse oflaplace (60)/(2-5s) is -12e^{(2t)/5}