What is inverse oflaplace ((2s+3))/(s^2+6s+13) ?

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

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inverse oflaplace ((2s+3))/(s^2+6s+13)

inverse laplace

\frac{( 2 s + 3 )}{s ^{2} + 6 s + 13}

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

Solution steps

L^{- 1} {\frac{( 2 s + 3 )}{s ^{2} + 6 s + 13}}

Expand

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

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

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)}

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

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

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

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

Refine

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

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

\frac{1}{e ( s - 1 ) ^{3}}

sin (π x)

y = 3 + 5 x^{2} - 2 x^{3}

x \to 0 lim ((5 x + 2)^{3})

ln (25 sin^{2} (x))

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