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}}

x \to + 0 lim (cos^{2} (x))

x \to - \infty lim (- 3)

\frac{\partial}{\partial x} (\frac{4}{3} π (y^{3} - x^{3}))

x^{6 c o s (x)}

e^{- 11 x}

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