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}

\frac{- x ^{2} - 4}{( x ^{2} - 4 ) ^{2}}

n = 0 \sum \infty \frac{1}{( 1 + n ^{2} )}

t^{- 2}

n = 0 \sum \infty \frac{1}{7 ^{n}}

\frac{d y}{d x} = \frac{( x + sec ^{2} ( x ) )}{y}

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