What is inverse oflaplace ((2s+2.5))/((s+0.5)) ?

The answer to inverse oflaplace ((2s+2.5))/((s+0.5)) is 2\delta(t)+1.5e^{-0.5t}

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inverse oflaplace ((2s+2.5))/((s+0.5))

inverse laplace

\frac{( 2 s + 2.5 )}{( s + 0.5 )}

2 δ (t) + 1.5 e^{- 0.5 t}

Solution steps

L^{- 1} {\frac{( 2 s + 2.5 )}{( s + 0.5 )}}

Take the partial fraction of

\frac{2 s + 2.5}{s + 0.5} : 2 + \frac{1.5}{s + 0.5}

= L^{- 1} {2 + \frac{1.5}{s + 0.5}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {2} + L^{- 1} {\frac{1.5}{s + 0.5}}

L^{- 1} {2} : 2 δ (t)

L^{- 1} {\frac{1.5}{s + 0.5}} : 1.5 e^{- 0.5 t}

= 2 δ (t) + 1.5 e^{- 0.5 t}

ln (csc (x) - cot (x))

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

y = e^{2 x}, at x = ln (2)

\frac{\partial}{\partial y} (y cos (x y) + e^{x})

What is inverse oflaplace ((2s+2.5))/((s+0.5)) ?
The answer to inverse oflaplace ((2s+2.5))/((s+0.5)) is 2\delta(t)+1.5e^{-0.5t}