What is inverse oflaplace (30s+102)/(s^2+4s+40) ?

The answer to inverse oflaplace (30s+102)/(s^2+4s+40) is 30e^{-2t}cos(6t)+7e^{-2t}sin(6t)

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inverse oflaplace (30s+102)/(s^2+4s+40)

inverse laplace

\frac{30 s + 102}{s ^{2} + 4 s + 40}

30 e^{- 2 t} cos (6 t) + 7 e^{- 2 t} sin (6 t)

Solution steps

L^{- 1} {\frac{30 s + 102}{s ^{2} + 4 s + 40}}

Expand

\frac{30 s + 102}{s ^{2} + 4 s + 40} : 30 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 36} + 42 \cdot \frac{1}{( s + 2 ) ^{2} + 36}

= L^{- 1} {30 \cdot \frac{s + 2}{( s + 2 ) ^{2} + 36} + 42 \cdot \frac{1}{( s + 2 ) ^{2} + 36}}

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)}

= 30 L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 36}} + 42 L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 36}}

L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 36}} : e^{- 2 t} cos (6 t)

L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 36}} : e^{- 2 t} \frac{1}{6} sin (6 t)

= 30 e^{- 2 t} cos (6 t) + 42 e^{- 2 t} \frac{1}{6} sin (6 t)

Refine

30 e^{- 2 t} cos (6 t) + 42 e^{- 2 t} \frac{1}{6} sin (6 t) : 30 e^{- 2 t} cos (6 t) + 7 e^{- 2 t} sin (6 t)

= 30 e^{- 2 t} cos (6 t) + 7 e^{- 2 t} sin (6 t)

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What is inverse oflaplace (30s+102)/(s^2+4s+40) ?
The answer to inverse oflaplace (30s+102)/(s^2+4s+40) is 30e^{-2t}cos(6t)+7e^{-2t}sin(6t)