What is inverse oflaplace (25)/(s(s^2+4s+5)) ?

The answer to inverse oflaplace (25)/(s(s^2+4s+5)) is 5\H(t)-5e^{-2t}cos(t)-10e^{-2t}sin(t)

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inverse oflaplace (25)/(s(s^2+4s+5))

inverse laplace

\frac{25}{s ( s ^{2} + 4 s + 5 )}

5 H (t) - 5 e^{- 2 t} cos (t) - 10 e^{- 2 t} sin (t)

Solution steps

L^{- 1} {\frac{25}{s ( s ^{2} + 4 s + 5 )}}

Take the partial fraction of

\frac{25}{s ( s ^{2} + 4 s + 5 )} : \frac{5}{s} + \frac{- 5 s - 20}{s ^{2} + 4 s + 5}

= L^{- 1} {\frac{5}{s} + \frac{- 5 s - 20}{s ^{2} + 4 s + 5}}

Expand

\frac{- 5 s - 20}{s ^{2} + 4 s + 5} : - 5 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 1} - 10 \cdot \frac{1}{( s + 2 ) ^{2} + 1}

= L^{- 1} {\frac{5}{s} - 5 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 1} - 10 \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{5}{s}} - 5 L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 1}} - 10 L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 1}}

L^{- 1} {\frac{5}{s}} : 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)

= 5 H (t) - 5 e^{- 2 t} cos (t) - 10 e^{- 2 t} sin (t)

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What is inverse oflaplace (25)/(s(s^2+4s+5)) ?
The answer to inverse oflaplace (25)/(s(s^2+4s+5)) is 5\H(t)-5e^{-2t}cos(t)-10e^{-2t}sin(t)