What is inverse oflaplace (4s+20)/(s^2+8s+16) ?

The answer to inverse oflaplace (4s+20)/(s^2+8s+16) is 4e^{-4t}+e^{-4t}*4t

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inverse oflaplace (4s+20)/(s^2+8s+16)

inverse laplace

\frac{4 s + 20}{s ^{2} + 8 s + 16}

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

Solution steps

L^{- 1} {\frac{4 s + 20}{s ^{2} + 8 s + 16}}

Take the partial fraction of

\frac{4 s + 20}{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

\frac{d ^{2}}{d x ^{2}} (\frac{8}{9 t ^{6}})

\frac{\partial}{\partial y} (x^{4})

\int 4 d x

f (x) = (5 x + 1)^{2}

32 - 8 x, [2, 4]

What is inverse oflaplace (4s+20)/(s^2+8s+16) ?
The answer to inverse oflaplace (4s+20)/(s^2+8s+16) is 4e^{-4t}+e^{-4t}*4t