What is inverse oflaplace 5/(8s) ?

The answer to inverse oflaplace 5/(8s) is 5/8 \H(t)

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inverse oflaplace 5/(8s)

inverse laplace

\frac{5}{8 s}

\frac{5}{8} H (t)

Solution steps

L^{- 1} {\frac{5}{8 s}}

= L^{- 1} {\frac{1}{8} \cdot \frac{5}{s}}

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}{8} L^{- 1} {\frac{5}{s}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{a}{s}} = a H (t)

Where

H (t)

is Heaviside step function

= \frac{1}{8} \cdot 5 H (t)

= \frac{5}{8} H (t)

x \to 3 + lim ((x^{2} - 9)^{x - 3})

\int \frac{t ^{2} + 8}{t ^{2} - 5 t + 6} d t

3 x^{2} + 4 x

\int ∣ x - 5 ∣ d x

x \to 0 lim (\frac{1}{ln ∣ x ∣})

What is inverse oflaplace 5/(8s) ?
The answer to inverse oflaplace 5/(8s) is 5/8 \H(t)