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)

x \to 9 + lim (\frac{7}{x - 9})

\int e^{y} cos (x) d x

\frac{sin ( 2 x )}{2}

\int_{0}^{3} x^{2} - 2 x d x

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

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)