What is inverse oflaplace (10)/(s(s+2)) ?

The answer to inverse oflaplace (10)/(s(s+2)) is 5\H(t)-5e^{-2t}

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inverse oflaplace (10)/(s(s+2))

inverse laplace

\frac{10}{s ( s + 2 )}

5 H (t) - 5 e^{- 2 t}

Solution steps

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

Take the partial fraction of

\frac{10}{s ( s + 2 )} : \frac{5}{s} - \frac{5}{s + 2}

= L^{- 1} {\frac{5}{s} - \frac{5}{s + 2}}

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{5}{s}} - L^{- 1} {\frac{5}{s + 2}}

L^{- 1} {\frac{5}{s}} : 5 H (t)

L^{- 1} {\frac{5}{s + 2}} : 5 e^{- 2 t}

= 5 H (t) - 5 e^{- 2 t}

f (x) = 4 x^{2} + 5 x + 1, (0, 1)

y = \frac{4 - 3 x}{3 x ^{2} + x}

630 - 4.9 t^{2}

x \to \frac{π}{2} lim (8 sec (x))

y^{^{'}} = y (x y^{4} + 3)

What is inverse oflaplace (10)/(s(s+2)) ?
The answer to inverse oflaplace (10)/(s(s+2)) is 5\H(t)-5e^{-2t}