What is inverse oflaplace (s+3.3)/(s+1) ?

The answer to inverse oflaplace (s+3.3)/(s+1) is \delta(t)+2.3e^{-t}

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inverse oflaplace (s+3.3)/(s+1)

inverse laplace

\frac{s + 3.3}{s + 1}

δ (t) + 2.3 e^{- t}

Solution steps

L^{- 1} {\frac{s + 3.3}{s + 1}}

Take the partial fraction of

\frac{s + 3.3}{s + 1} : 1 + \frac{2.3}{s + 1}

= L^{- 1} {1 + \frac{2.3}{s + 1}}

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{2.3}{s + 1}}

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{2.3}{s + 1}} : 2.3 e^{- t}

= δ (t) + 2.3 e^{- t}

\frac{d}{d x} (lo g_{10} (x^{2} - 5 x))

y = x (x - 6)^{2}

e^{- 5 x} + e^{3 x}

\frac{d}{d x} (e^{x} + 19)

x \to 2 + lim (2 - x)

What is inverse oflaplace (s+3.3)/(s+1) ?
The answer to inverse oflaplace (s+3.3)/(s+1) is \delta(t)+2.3e^{-t}