What is inverse oflaplace 1/(2s-5) ?

The answer to inverse oflaplace 1/(2s-5) is 1/2 e^{(5t)/2}

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inverse oflaplace 1/(2s-5)

inverse laplace

\frac{1}{2 s - 5}

\frac{1}{2} e^{\frac{5 t}{2}}

Solution steps

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

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

Use Inverse Laplace Transform table:

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

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

= \frac{1}{2} e^{\frac{5 t}{2}}

\frac{d}{d x} (ln (sin (3 x)))

\frac{d}{d x} (x^{\frac{2}{3}} - x^{\frac{1}{3}} + 4)

\frac{d}{d y} (- y)

\int x e^{12 x} d x

\frac{\partial}{\partial x} (- 4 x)

What is inverse oflaplace 1/(2s-5) ?
The answer to inverse oflaplace 1/(2s-5) is 1/2 e^{(5t)/2}