What is inverse oflaplace 2/(s^2+6s+13) ?

The answer to inverse oflaplace 2/(s^2+6s+13) is e^{-3t}sin(2t)

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inverse oflaplace 2/(s^2+6s+13)

inverse laplace

\frac{2}{s ^{2} + 6 s + 13}

e^{- 3 t} sin (2 t)

Solution steps

L^{- 1} {\frac{2}{s ^{2} + 6 s + 13}}

Expand

\frac{2}{s ^{2} + 6 s + 13} : 2 \cdot \frac{1}{( s + 3 ) ^{2} + 4}

= L^{- 1} {2 \cdot \frac{1}{( s + 3 ) ^{2} + 4}}

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)}

= 2 L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 4}}

L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 4}} : e^{- 3 t} \frac{1}{2} sin (2 t)

= 2 e^{- 3 t} \frac{1}{2} sin (2 t)

Refine

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

= e^{- 3 t} sin (2 t)

What is inverse oflaplace 2/(s^2+6s+13) ?
The answer to inverse oflaplace 2/(s^2+6s+13) is e^{-3t}sin(2t)