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

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

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

inverse laplace

\frac{s + 2}{s ^{2} + s}

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

Solution steps

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

Take the partial fraction of

\frac{s + 2}{s ^{2} + s} : \frac{2}{s} - \frac{1}{s + 1}

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

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

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

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

\int sin^{7} (2 x) cos (2 x) d x

\int \frac{1}{x ^{2} - 2 x + 3} d x

\frac{d y}{d x} = 3 y + a

f (x) = 13440 (2 x + 5)^{3}

\int \frac{3 x - 1}{x x ^{2} - 25} d x

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