What is inverse oflaplace (10)/(s(s+3)) ?

The answer to inverse oflaplace (10)/(s(s+3)) is 10/3 \H(t)-10/3 e^{-3t}

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inverse oflaplace (10)/(s(s+3))

inverse laplace

\frac{10}{s ( s + 3 )}

\frac{10}{3} H (t) - \frac{10}{3} e^{- 3 t}

Solution steps

L^{- 1} {\frac{10}{s ( s + 3 )}}

Take the partial fraction of

\frac{10}{s ( s + 3 )} : \frac{10}{3 s} - \frac{10}{3 ( s + 3 )}

= L^{- 1} {\frac{10}{3 s} - \frac{10}{3 ( s + 3 )}}

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}{3 s}} - L^{- 1} {\frac{10}{3 ( s + 3 )}}

L^{- 1} {\frac{10}{3 s}} : \frac{10}{3} H (t)

L^{- 1} {\frac{10}{3 ( s + 3 )}} : \frac{10}{3} e^{- 3 t}

= \frac{10}{3} H (t) - \frac{10}{3} e^{- 3 t}

2 x^{2} - 6 x

x \to 2 + lim (∣ x + 2 ∣ - 2)

\frac{s}{s - 3}

\frac{d}{d x} (4 x - 3 ln (x))

x \to - 2 lim (∣ ln ∣ x + 2 ∣ ∣)

What is inverse oflaplace (10)/(s(s+3)) ?
The answer to inverse oflaplace (10)/(s(s+3)) is 10/3 \H(t)-10/3 e^{-3t}