What is inverse oflaplace (4s+12)/(s^2+8s+16) ?

The answer to inverse oflaplace (4s+12)/(s^2+8s+16) is 4e^{-4t}-e^{-4t}*4t

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inverse oflaplace (4s+12)/(s^2+8s+16)

inverse laplace

\frac{4 s + 12}{s ^{2} + 8 s + 16}

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

Solution steps

L^{- 1} {\frac{4 s + 12}{s ^{2} + 8 s + 16}}

Take the partial fraction of

\frac{4 s + 12}{s ^{2} + 8 s + 16} : \frac{4}{s + 4} - \frac{4}{( s + 4 ) ^{2}}

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

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

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

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

x \to 0 lim (\frac{4}{x ^{2} + x} - \frac{4}{x})

\int (5 - x^{4} - e^{2 x}) d x

\frac{d}{d x} (2 cos (x) sin (x))

\frac{d}{d x} (g (x) (x)^{2} (x))

2 x e^{x} + x^{2} e^{x}

What is inverse oflaplace (4s+12)/(s^2+8s+16) ?
The answer to inverse oflaplace (4s+12)/(s^2+8s+16) is 4e^{-4t}-e^{-4t}*4t