What is inverse oflaplace 6/(s^2(7s+1)) ?

The answer to inverse oflaplace 6/(s^2(7s+1)) is -42\H(t)+6t+42e^{-t/7}

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inverse oflaplace 6/(s^2(7s+1))

inverse laplace

\frac{6}{s ^{2} ( 7 s + 1 )}

- 42 H (t) + 6 t + 42 e^{- \frac{t}{7}}

Solution steps

L^{- 1} {\frac{6}{s ^{2} ( 7 s + 1 )}}

Take the partial fraction of

\frac{6}{s ^{2} ( 7 s + 1 )} : - \frac{42}{s} + \frac{6}{s ^{2}} + \frac{294}{7 s + 1}

= L^{- 1} {- \frac{42}{s} + \frac{6}{s ^{2}} + \frac{294}{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{42}{s}} + L^{- 1} {\frac{6}{s ^{2}}} + L^{- 1} {\frac{294}{7 s + 1}}

L^{- 1} {\frac{42}{s}} : 42 H (t)

L^{- 1} {\frac{6}{s ^{2}}} : 6 t

L^{- 1} {\frac{294}{7 s + 1}} : 42 e^{- \frac{t}{7}}

= - 42 H (t) + 6 t + 42 e^{- \frac{t}{7}}

What is inverse oflaplace 6/(s^2(7s+1)) ?
The answer to inverse oflaplace 6/(s^2(7s+1)) is -42\H(t)+6t+42e^{-t/7}