What is inverse oflaplace (2s)/(s^2+6s+9) ?

The answer to inverse oflaplace (2s)/(s^2+6s+9) is 2e^{-3t}-e^{-3t}*6t

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inverse oflaplace (2s)/(s^2+6s+9)

inverse laplace

\frac{2 s}{s ^{2} + 6 s + 9}

2 e^{- 3 t} - e^{- 3 t} \cdot 6 t

Solution steps

L^{- 1} {\frac{2 s}{s ^{2} + 6 s + 9}}

Take the partial fraction of

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

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

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

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

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

= 2 e^{- 3 t} - e^{- 3 t} \cdot 6 t

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\int (x^{2})^{2} d (x^{2})^{2}

What is inverse oflaplace (2s)/(s^2+6s+9) ?
The answer to inverse oflaplace (2s)/(s^2+6s+9) is 2e^{-3t}-e^{-3t}*6t