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

The answer to inverse oflaplace (s^2)/(s^2+6s+34) is \delta(t)-6e^{-3t}cos(5t)-16/5 e^{-3t}sin(5t)

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

inverse laplace

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

δ (t) - 6 e^{- 3 t} cos (5 t) - \frac{16}{5} e^{- 3 t} sin (5 t)

Solution steps

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

Take the partial fraction of

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

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

Expand

\frac{- 6 s - 34}{s ^{2} + 6 s + 34} : - 6 \cdot \frac{s + 3}{( s + 3 ) ^{2} + 25} - 16 \cdot \frac{1}{( s + 3 ) ^{2} + 25}

= L^{- 1} {1 - 6 \cdot \frac{s + 3}{( s + 3 ) ^{2} + 25} - 16 \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} {1} - 6 L^{- 1} {\frac{s + 3}{( s + 3 ) ^{2} + 25}} - 16 L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 25}}

L^{- 1} {1} : δ (t)

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)

= δ (t) - 6 e^{- 3 t} cos (5 t) - 16 e^{- 3 t} \frac{1}{5} sin (5 t)

Refine

δ (t) - 6 e^{- 3 t} cos (5 t) - 16 e^{- 3 t} \frac{1}{5} sin (5 t) : δ (t) - 6 e^{- 3 t} cos (5 t) - \frac{16}{5} e^{- 3 t} sin (5 t)

= δ (t) - 6 e^{- 3 t} cos (5 t) - \frac{16}{5} e^{- 3 t} sin (5 t)

\frac{d}{d x} (\frac{x - 3}{x + 1})

f (x) = \frac{1}{x ^{3}}, (1, 1)

\int sin^{2} (10 x) d x

\int (x + 3) (2 x - 1) d x

\frac{d}{d x} (3 x^{2} - 7)

What is inverse oflaplace (s^2)/(s^2+6s+34) ?
The answer to inverse oflaplace (s^2)/(s^2+6s+34) is \delta(t)-6e^{-3t}cos(5t)-16/5 e^{-3t}sin(5t)