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

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

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

inverse laplace

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

8 e^{- 4 t} - e^{- 4 t} \cdot 16 t

Solution steps

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

Take the partial fraction of

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

= L^{- 1} {\frac{8}{s + 4} - \frac{16}{( 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{8}{s + 4}} - L^{- 1} {\frac{16}{( s + 4 ) ^{2}}}

L^{- 1} {\frac{8}{s + 4}} : 8 e^{- 4 t}

L^{- 1} {\frac{16}{( s + 4 ) ^{2}}} : e^{- 4 t} \cdot 16 t

= 8 e^{- 4 t} - e^{- 4 t} \cdot 16 t

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