What is inverse oflaplace s/(s+1.3) ?

The answer to inverse oflaplace s/(s+1.3) is \delta(t)-1.3e^{-1.3t}

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inverse oflaplace s/(s+1.3)

inverse laplace

\frac{s}{s + 1.3}

δ (t) - 1.3 e^{- 1.3 t}

Solution steps

L^{- 1} {\frac{s}{s + 1.3}}

Take the partial fraction of

\frac{s}{s + 1.3} : 1 - \frac{1.3}{s + 1.3}

= L^{- 1} {1 - \frac{1.3}{s + 1.3}}

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} {1} - L^{- 1} {\frac{1.3}{s + 1.3}}

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{1.3}{s + 1.3}} : 1.3 e^{- 1.3 t}

= δ (t) - 1.3 e^{- 1.3 t}

\frac{\partial}{\partial x} (x y + \frac{x}{y})

f (z) = e^{\frac{z}{z - 1}}

\frac{d}{d x} (5 (x^{3} - 4 x)^{- 3})

\int_{0}^{3} \frac{8 t}{( t - 4 ) ^{2}} d t

\int_{0}^{\frac{π}{2}} sin^{3} (2 x) d x

What is inverse oflaplace s/(s+1.3) ?
The answer to inverse oflaplace s/(s+1.3) is \delta(t)-1.3e^{-1.3t}