What is inverse oflaplace 4/(s(s^2-4s+5)) ?

The answer to inverse oflaplace 4/(s(s^2-4s+5)) is 4/5 \H(t)-4/5 e^{2t}cos(t)+8/5 e^{2t}sin(t)

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inverse oflaplace 4/(s(s^2-4s+5))

inverse laplace

\frac{4}{s ( s ^{2} - 4 s + 5 )}

\frac{4}{5} H (t) - \frac{4}{5} e^{2 t} cos (t) + \frac{8}{5} e^{2 t} sin (t)

Solution steps

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

Take the partial fraction of

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

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

Expand

\frac{- 4 s + 16}{5 ( s ^{2} - 4 s + 5 )} : - \frac{4}{5} \cdot \frac{s - 2}{( s - 2 ) ^{2} + 1} + \frac{8}{5} \cdot \frac{1}{( s - 2 ) ^{2} + 1}

= L^{- 1} {\frac{4}{5 s} - \frac{4}{5} \cdot \frac{s - 2}{( s - 2 ) ^{2} + 1} + \frac{8}{5} \cdot \frac{1}{( s - 2 ) ^{2} + 1}}

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}{5 s}} - \frac{4}{5} L^{- 1} {\frac{s - 2}{( s - 2 ) ^{2} + 1}} + \frac{8}{5} L^{- 1} {\frac{1}{( s - 2 ) ^{2} + 1}}

L^{- 1} {\frac{4}{5 s}} : \frac{4}{5} H (t)

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

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

= \frac{4}{5} H (t) - \frac{4}{5} e^{2 t} cos (t) + \frac{8}{5} e^{2 t} sin (t)

\frac{- 5 x - 9}{2 x + 8}

sin (x), 1.5

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

\frac{d}{d x} (tan (e^{s i n (x)}))

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

What is inverse oflaplace 4/(s(s^2-4s+5)) ?
The answer to inverse oflaplace 4/(s(s^2-4s+5)) is 4/5 \H(t)-4/5 e^{2t}cos(t)+8/5 e^{2t}sin(t)