What is inverse oflaplace (s+1)/((s+2)(s+6)^2) ?

The answer to inverse oflaplace (s+1)/((s+2)(s+6)^2) is -1/16 e^{-2t}+1/16 e^{-6t}+(5e^{-6t}t)/4

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inverse oflaplace (s+1)/((s+2)(s+6)^2)

inverse laplace

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

- \frac{1}{16} e^{- 2 t} + \frac{1}{16} e^{- 6 t} + \frac{5 e ^{- 6 t} t}{4}

Solution steps

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

Take the partial fraction of

\frac{s + 1}{( s + 2 ) ( s + 6 ) ^{2}} : - \frac{1}{16 ( s + 2 )} + \frac{1}{16 ( s + 6 )} + \frac{5}{4 ( s + 6 ) ^{2}}

= L^{- 1} {- \frac{1}{16 ( s + 2 )} + \frac{1}{16 ( s + 6 )} + \frac{5}{4 ( s + 6 ) ^{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{1}{16 ( s + 2 )}} + L^{- 1} {\frac{1}{16 ( s + 6 )}} + L^{- 1} {\frac{5}{4 ( s + 6 ) ^{2}}}

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

L^{- 1} {\frac{1}{16 ( s + 6 )}} : \frac{1}{16} e^{- 6 t}

L^{- 1} {\frac{5}{4 ( s + 6 ) ^{2}}} : \frac{5 e ^{- 6 t} t}{4}

= - \frac{1}{16} e^{- 2 t} + \frac{1}{16} e^{- 6 t} + \frac{5 e ^{- 6 t} t}{4}

f (x) = x^{3} - 2 x + 7, at x = - 2

x^{4} - 2 x^{2} + 5

n = 1 \sum \infty 8 e^{- 6 n}

\frac{y ^{^{'}} - e ^{- t} + 4}{y} = - 4

\int sec (x) tan (x) - 5 e^{x} d x

What is inverse oflaplace (s+1)/((s+2)(s+6)^2) ?
The answer to inverse oflaplace (s+1)/((s+2)(s+6)^2) is -1/16 e^{-2t}+1/16 e^{-6t}+(5e^{-6t}t)/4