What is inverse oflaplace (2s+5)/(s^2+s) ?

The answer to inverse oflaplace (2s+5)/(s^2+s) is 5\H(t)-3e^{-t}

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inverse oflaplace (2s+5)/(s^2+s)

inverse laplace

\frac{2 s + 5}{s ^{2} + s}

5 H (t) - 3 e^{- t}

Solution steps

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

Take the partial fraction of

\frac{2 s + 5}{s ^{2} + s} : \frac{5}{s} - \frac{3}{s + 1}

= L^{- 1} {\frac{5}{s} - \frac{3}{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{5}{s}} - L^{- 1} {\frac{3}{s + 1}}

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

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

= 5 H (t) - 3 e^{- t}

\int 34 x + 20 d x

\frac{d}{d y} (ln (\frac{( 3 y + 1 ) ^{4}}{y ^{2} + 1}))

x \to 0 - lim (x^{2} - \frac{1}{x})

f (x) = \frac{x ^{4}}{4}

\int e^{5 x + 1} d x

What is inverse oflaplace (2s+5)/(s^2+s) ?
The answer to inverse oflaplace (2s+5)/(s^2+s) is 5\H(t)-3e^{-t}