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

The answer to inverse oflaplace-5/(s^2+2s+4) is -5/(sqrt(3))e^{-t}sin(sqrt(3)t)

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

inverselaplace

- \frac{5}{3} e^{- t} sin (3 t)

Solution steps

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

Expand

\frac{5}{s ^{2} + 2 s + 4} : 5 \cdot \frac{1}{( s + 1 ) ^{2} + 3}

= L^{- 1} {- 5 \cdot \frac{1}{( s + 1 ) ^{2} + 3}}

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

= - 5 L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 3}}

L^{- 1} {\frac{1}{( s + 1 ) ^{2} + 3}} : e^{- t} \frac{1}{3} sin (3 t)

= - 5 e^{- t} \frac{1}{3} sin (3 t)

Refine

- 5 e^{- t} \frac{1}{3} sin (3 t) : - \frac{5}{3} e^{- t} sin (3 t)

= - \frac{5}{3} e^{- t} sin (3 t)

What is inverse oflaplace-5/(s^2+2s+4) ?
The answer to inverse oflaplace-5/(s^2+2s+4) is -5/(sqrt(3))e^{-t}sin(sqrt(3)t)