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

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

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

inverse laplace

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

H (t) - e^{- 2 t} cos (t) - 2 e^{- 2 t} sin (t)

Solution steps

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

Take the partial fraction of

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

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

Expand

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

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

L^{- 1} {\frac{1}{s}} : 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)

= H (t) - e^{- 2 t} cos (t) - 2 e^{- 2 t} sin (t)

\int \frac{6 x + 7}{( x + 2 ) ^{2}} d x

\int \frac{1}{x ( ln ( x ) ) ^{3}} d x

y^{^{'}} = y (y + 1) x

\int \frac{1 + ln ( x )}{x ln ( x )} d x

\frac{d}{d x} (3 sin^{2} (x) cos (2 x))

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