What is inverse oflaplace ((s+9))/((s^2+6s+34)) ?

The answer to inverse oflaplace ((s+9))/((s^2+6s+34)) is e^{-3t}cos(5t)+6/5 e^{-3t}sin(5t)

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inverse oflaplace ((s+9))/((s^2+6s+34))

inverse laplace

\frac{( s + 9 )}{( s ^{2} + 6 s + 34 )}

e^{- 3 t} cos (5 t) + \frac{6}{5} e^{- 3 t} sin (5 t)

Solution steps

L^{- 1} {\frac{( s + 9 )}{( s ^{2} + 6 s + 34 )}}

Expand

\frac{s + 9}{s ^{2} + 6 s + 34} : \frac{s + 3}{( s + 3 ) ^{2} + 25} + 6 \cdot \frac{1}{( s + 3 ) ^{2} + 25}

= L^{- 1} {\frac{s + 3}{( s + 3 ) ^{2} + 25} + 6 \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}} + 6 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) + 6 e^{- 3 t} \frac{1}{5} sin (5 t)

Refine

e^{- 3 t} cos (5 t) + 6 e^{- 3 t} \frac{1}{5} sin (5 t) : e^{- 3 t} cos (5 t) + \frac{6}{5} e^{- 3 t} sin (5 t)

= e^{- 3 t} cos (5 t) + \frac{6}{5} e^{- 3 t} sin (5 t)

\int x^{2} x^{3} + 19 d x

\frac{d y}{d x} - \frac{2}{x} y = \frac{3}{x ^{2}} y^{4}

\int_{1}^{2} 13 4 x^{2} - 3 d x

\frac{d}{d x} (- 5 x + 2)

\frac{\partial}{\partial y} (2 x - y^{2})

What is inverse oflaplace ((s+9))/((s^2+6s+34)) ?
The answer to inverse oflaplace ((s+9))/((s^2+6s+34)) is e^{-3t}cos(5t)+6/5 e^{-3t}sin(5t)