What is inverse oflaplace (7s+2)/(s^2+6s+34) ?

The answer to inverse oflaplace (7s+2)/(s^2+6s+34) is 7e^{-3t}cos(5t)-19/5 e^{-3t}sin(5t)

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inverse oflaplace (7s+2)/(s^2+6s+34)

inverse laplace

\frac{7 s + 2}{s ^{2} + 6 s + 34}

7 e^{- 3 t} cos (5 t) - \frac{19}{5} e^{- 3 t} sin (5 t)

Solution steps

L^{- 1} {\frac{7 s + 2}{s ^{2} + 6 s + 34}}

Expand

\frac{7 s + 2}{s ^{2} + 6 s + 34} : 7 \cdot \frac{s + 3}{( s + 3 ) ^{2} + 25} - 19 \cdot \frac{1}{( s + 3 ) ^{2} + 25}

= L^{- 1} {7 \cdot \frac{s + 3}{( s + 3 ) ^{2} + 25} - 19 \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)}

= 7 L^{- 1} {\frac{s + 3}{( s + 3 ) ^{2} + 25}} - 19 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)

= 7 e^{- 3 t} cos (5 t) - 19 e^{- 3 t} \frac{1}{5} sin (5 t)

Refine

7 e^{- 3 t} cos (5 t) - 19 e^{- 3 t} \frac{1}{5} sin (5 t) : 7 e^{- 3 t} cos (5 t) - \frac{19}{5} e^{- 3 t} sin (5 t)

= 7 e^{- 3 t} cos (5 t) - \frac{19}{5} e^{- 3 t} sin (5 t)

n \to \infty lim (e^{\frac{1}{n} l n (n)})

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

y^{^{'}} = \frac{y}{x}, y (1) = 2

\frac{d}{d x} (\frac{2 x ^{2} - 2 x + 2}{( 2 x - 1 ) ^{2}})

(4.5) (9.25)

What is inverse oflaplace (7s+2)/(s^2+6s+34) ?
The answer to inverse oflaplace (7s+2)/(s^2+6s+34) is 7e^{-3t}cos(5t)-19/5 e^{-3t}sin(5t)