What is inverse oflaplace 1/(s^2+7s) ?

The answer to inverse oflaplace 1/(s^2+7s) is 1/7 \H(t)-1/7 e^{-7t}

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inverse oflaplace 1/(s^2+7s)

inverse laplace

\frac{1}{s ^{2} + 7 s}

\frac{1}{7} H (t) - \frac{1}{7} e^{- 7 t}

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 7 s}}

Take the partial fraction of

\frac{1}{s ^{2} + 7 s} : \frac{1}{7 s} - \frac{1}{7 ( s + 7 )}

= L^{- 1} {\frac{1}{7 s} - \frac{1}{7 ( s + 7 )}}

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{1}{7 s}} - L^{- 1} {\frac{1}{7 ( s + 7 )}}

L^{- 1} {\frac{1}{7 s}} : \frac{1}{7} H (t)

L^{- 1} {\frac{1}{7 ( s + 7 )}} : \frac{1}{7} e^{- 7 t}

= \frac{1}{7} H (t) - \frac{1}{7} e^{- 7 t}

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

x \to 0 lim (\frac{a}{x})

n = 0 \sum \infty (2 n)! (\frac{x}{7})^{n}

\frac{d}{d t} (5 ln (t))

n = 0 \sum \infty (\frac{5}{16})^{n}

What is inverse oflaplace 1/(s^2+7s) ?
The answer to inverse oflaplace 1/(s^2+7s) is 1/7 \H(t)-1/7 e^{-7t}