What is inverse oflaplace 3/(s^2+6s+10) ?

The answer to inverse oflaplace 3/(s^2+6s+10) is 3e^{-3t}sin(t)

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inverse oflaplace 3/(s^2+6s+10)

inverse laplace

\frac{3}{s ^{2} + 6 s + 10}

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

Solution steps

L^{- 1} {\frac{3}{s ^{2} + 6 s + 10}}

Expand

\frac{3}{s ^{2} + 6 s + 10} : 3 \cdot \frac{1}{( s + 3 ) ^{2} + 1}

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

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)}

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

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

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

2 cos^{3} (t)

\int e^{\frac{1}{3} x} d x

x \to - 2 + lim (x + 2)

\frac{d}{d x} (\frac{x}{1 - ln ( x - 1 )})

f (x) = 2 + 4 x^{2}, (0, 2)

What is inverse oflaplace 3/(s^2+6s+10) ?
The answer to inverse oflaplace 3/(s^2+6s+10) is 3e^{-3t}sin(t)