What is inverse oflaplace (4s)/(s^2+6s+9) ?

The answer to inverse oflaplace (4s)/(s^2+6s+9) is 4e^{-3t}-e^{-3t}*12t

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inverse oflaplace (4s)/(s^2+6s+9)

inverse laplace

\frac{4 s}{s ^{2} + 6 s + 9}

4 e^{- 3 t} - e^{- 3 t} \cdot 12 t

Solution steps

L^{- 1} {\frac{4 s}{s ^{2} + 6 s + 9}}

Take the partial fraction of

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

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

L^{- 1} {\frac{4}{s + 3}} : 4 e^{- 3 t}

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

= 4 e^{- 3 t} - e^{- 3 t} \cdot 12 t

x \to 0 lim ((1 + 3 x)^{\frac{7}{x}})

5 sin (6 t)

\int t e^{- 4 t} d t

\frac{d}{d x} (arctan (- x))

8 e^{x} cos (x)

What is inverse oflaplace (4s)/(s^2+6s+9) ?
The answer to inverse oflaplace (4s)/(s^2+6s+9) is 4e^{-3t}-e^{-3t}*12t