What is inverse oflaplace 8/((9s^2+12s+4)(s)) ?

The answer to inverse oflaplace 8/((9s^2+12s+4)(s)) is -2e^{-(2t)/3}-(4e^{-(2t)/3}t)/(3)+2\H(t)

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inverse oflaplace 8/((9s^2+12s+4)(s))

inverse laplace

\frac{8}{( 9 s ^{2} + 12 s + 4 ) ( s )}

- 2 e^{- \frac{2 t}{3}} - \frac{4 e ^{- \frac{2 t}{3}} t}{3} + 2 H (t)

Solution steps

L^{- 1} {\frac{8}{( s ) ( 9 s ^{2} + 12 s + 4 )}}

Take the partial fraction of

\frac{8}{( 9 s ^{2} + 12 s + 4 ) s} : - \frac{6}{3 s + 2} - \frac{12}{( 3 s + 2 ) ^{2}} + \frac{2}{s}

= L^{- 1} {- \frac{6}{3 s + 2} - \frac{12}{( 3 s + 2 ) ^{2}} + \frac{2}{s}}

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{6}{3 s + 2}} - L^{- 1} {\frac{12}{( 3 s + 2 ) ^{2}}} + L^{- 1} {\frac{2}{s}}

L^{- 1} {\frac{6}{3 s + 2}} : 2 e^{- \frac{2 t}{3}}

L^{- 1} {\frac{12}{( 3 s + 2 ) ^{2}}} : \frac{4 e ^{- \frac{2 t}{3}} t}{3}

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

= - 2 e^{- \frac{2 t}{3}} - \frac{4 e ^{- \frac{2 t}{3}} t}{3} + 2 H (t)

\int x \cdot e^{x^{2}} d x

\frac{d ^{2} x}{d t ^{2}} + 9 x = - sin (3 t)

\frac{\partial}{\partial x} (ln (\frac{2 x y}{z}))

\int \frac{5 x}{3 x ^{\frac{1}{3}}} d x

What is inverse oflaplace 8/((9s^2+12s+4)(s)) ?
The answer to inverse oflaplace 8/((9s^2+12s+4)(s)) is -2e^{-(2t)/3}-(4e^{-(2t)/3}t)/(3)+2\H(t)