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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

\frac{1}{s ^{2} ( 1 + 7 s )} : - \frac{7}{s} + \frac{1}{s ^{2}} + \frac{49}{7 s + 1}

= L^{- 1} {- \frac{7}{s} + \frac{1}{s ^{2}} + \frac{49}{7 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{7}{s}} + L^{- 1} {\frac{1}{s ^{2}}} + L^{- 1} {\frac{49}{7 s + 1}}

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

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

L^{- 1} {\frac{49}{7 s + 1}} : 7 e^{- \frac{t}{7}}

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

\frac{d}{d x} (x (1 - \frac{x}{10}))

\frac{d}{d x} (5 x cos (x^{2}))

x \to + 0 lim (\frac{5}{x ^{2} + x} - \frac{5}{x})

ln (\frac{x - 1}{x + 1})

x \to 3 lim (\frac{sin ( x )}{x})

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