What is inverse oflaplace ((2s+5))/(s^2+6s+34) ?

The answer to inverse oflaplace ((2s+5))/(s^2+6s+34) is 2e^{-3t}cos(5t)-1/5 e^{-3t}sin(5t)

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inverse oflaplace ((2s+5))/(s^2+6s+34)

inverse laplace

\frac{( 2 s + 5 )}{s ^{2} + 6 s + 34}

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

Solution steps

L^{- 1} {\frac{( 2 s + 5 )}{s ^{2} + 6 s + 34}}

Expand

\frac{2 s + 5}{s ^{2} + 6 s + 34} : 2 \cdot \frac{s + 3}{( s + 3 ) ^{2} + 25} - 1 \cdot \frac{1}{( s + 3 ) ^{2} + 25}

= L^{- 1} {2 \cdot \frac{s + 3}{( s + 3 ) ^{2} + 25} - 1 \cdot \frac{1}{( s + 3 ) ^{2} + 25}}

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} + 25}} - 1 \cdot L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 25}}

L^{- 1} {\frac{s + 3}{( s + 3 ) ^{2} + 25}} : e^{- 3 t} cos (5 t)

L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 25}} : e^{- 3 t} \frac{1}{5} sin (5 t)

= 2 e^{- 3 t} cos (5 t) - 1 \cdot e^{- 3 t} \frac{1}{5} sin (5 t)

Refine

2 e^{- 3 t} cos (5 t) - 1 e^{- 3 t} \frac{1}{5} sin (5 t) : 2 e^{- 3 t} cos (5 t) - \frac{1}{5} e^{- 3 t} sin (5 t)

= 2 e^{- 3 t} cos (5 t) - \frac{1}{5} e^{- 3 t} sin (5 t)

\int x e^{x + 2} d x

n = 1 \sum \infty \frac{( x - 1 ) ^{n}}{n}

\int cos^{2} (x) sin^{3} (x) d x

\int sin^{4} (y) cos^{7} (y) d y

What is inverse oflaplace ((2s+5))/(s^2+6s+34) ?
The answer to inverse oflaplace ((2s+5))/(s^2+6s+34) is 2e^{-3t}cos(5t)-1/5 e^{-3t}sin(5t)