What is inverse oflaplace (s+1)/(0.1s+1) ?

The answer to inverse oflaplace (s+1)/(0.1s+1) is 10\delta(t)-90e^{-10t}

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inverse oflaplace (s+1)/(0.1s+1)

inverse laplace

\frac{s + 1}{0.1 s + 1}

10 δ (t) - 90 e^{- 10 t}

Solution steps

L^{- 1} {\frac{s + 1}{0.1 s + 1}}

Take the partial fraction of

\frac{s + 1}{0.1 s + 1} : 10 - \frac{9}{0.1 s + 1}

= L^{- 1} {10 - \frac{9}{0.1 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} {10} - L^{- 1} {\frac{9}{0.1 s + 1}}

L^{- 1} {10} : 10 δ (t)

L^{- 1} {\frac{9}{0.1 s + 1}} : 90 e^{- 10 t}

= 10 δ (t) - 90 e^{- 10 t}

\frac{d}{d x} (\frac{sin ( 4 x )}{x})

\int_{1}^{2} 12 x^{2} 2^{x^{3}} d x

y^{^{''}} + 26 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 0

f (x) = (x + 5) cos (x)

\frac{2 s + 9}{( s + 4 ) ( s + 6 )}

What is inverse oflaplace (s+1)/(0.1s+1) ?
The answer to inverse oflaplace (s+1)/(0.1s+1) is 10\delta(t)-90e^{-10t}