What is inverse oflaplace (10)/((s*(s+1))) ?

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

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inverse oflaplace (10)/((s*(s+1)))

inverse laplace

\frac{10}{( s \cdot ( s + 1 ) )}

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

Solution steps

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

Take the partial fraction of

\frac{10}{s ( s + 1 )} : \frac{10}{s} - \frac{10}{s + 1}

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

L^{- 1} {\frac{10}{s}} : 10 H (t)

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

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

x^{2} - 2 x, 4 x + 7

\frac{( x ^{2} - 1 ) d y}{d x} + 3 x (y - 1) = 0, y (0) = 7

(\frac{2 x - 1}{x + 1}), 2

\int 8 arctan (x) d x

z \to 2 lim ((- \frac{2}{z + 2})^{3})

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