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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

Refine

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

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

\frac{1 - x}{1 + x}

\frac{d}{d x} ((cos^{2} (x)) x)

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

\frac{10}{( x ^{2} - 7 ) ^{4}}

y = sec (x), (\frac{π}{3}, 2)

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