What is inverse oflaplace-(20)/(s(s+10)) ?

The answer to inverse oflaplace-(20)/(s(s+10)) is -2\H(t)+2e^{-10t}

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inverse oflaplace-(20)/(s(s+10))

inverse laplace

- \frac{20}{s ( s + 10 )}

- 2 H (t) + 2 e^{- 10 t}

Solution steps

L^{- 1} {- \frac{20}{s ( s + 10 )}}

Take the partial fraction of

\frac{20}{s ( s + 10 )} : \frac{2}{s} - \frac{2}{s + 10}

= L^{- 1} {- \frac{2}{s} + \frac{2}{s + 10}}

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} {\frac{2}{s}} + L^{- 1} {\frac{2}{s + 10}}

L^{- 1} {\frac{2}{s}} : 2 H (t)

L^{- 1} {\frac{2}{s + 10}} : 2 e^{- 10 t}

= - 2 H (t) + 2 e^{- 10 t}

y = 2 x^{3} + 3 x^{2} - 12 x + 3

\frac{d}{d x} (5 x^{6} - 12)

\int e^{5 x} cos (8 x) d x

\int_{\frac{11 π}{2}}^{6 π} 9 sin (x) d x

y^{^{'}} = (1 - y) cos (x)

What is inverse oflaplace-(20)/(s(s+10)) ?
The answer to inverse oflaplace-(20)/(s(s+10)) is -2\H(t)+2e^{-10t}