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

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

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

inverse laplace

\frac{s}{s ^{2} + 6 s + 34}

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

Solution steps

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

Expand

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

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

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

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

Refine

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

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

\frac{\partial}{\partial y} (\frac{( x + y )}{x - y})

\int \frac{1}{sin ( x ) cos ( x ) + cos ^{2} ( x )} d x

\int \frac{cos ( x )}{1 - sin ( x )} d x

n = 0 \sum \infty \frac{1}{( 4 n ^{2} - 1 )}

\int_{8}^{\infty} x e^{- 2 x} d x

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