What is inverse oflaplace (6(s+34))/(s(s^2+10s+34)) ?

The answer to inverse oflaplace (6(s+34))/(s(s^2+10s+34)) is 6\H(t)-6e^{-5t}cos(3t)-8e^{-5t}sin(3t)

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inverse oflaplace (6(s+34))/(s(s^2+10s+34))

inverse laplace

\frac{6 ( s + 34 )}{s ( s ^{2} + 10 s + 34 )}

6 H (t) - 6 e^{- 5 t} cos (3 t) - 8 e^{- 5 t} sin (3 t)

Solution steps

L^{- 1} {\frac{6 ( s + 34 )}{s ( s ^{2} + 10 s + 34 )}}

Take the partial fraction of

\frac{6 ( s + 34 )}{s ( s ^{2} + 10 s + 34 )} : \frac{6}{s} + \frac{- 6 s - 54}{s ^{2} + 10 s + 34}

= L^{- 1} {\frac{6}{s} + \frac{- 6 s - 54}{s ^{2} + 10 s + 34}}

Expand

\frac{- 6 s - 54}{s ^{2} + 10 s + 34} : - 6 \cdot \frac{s + 5}{( s + 5 ) ^{2} + 9} - 24 \cdot \frac{1}{( s + 5 ) ^{2} + 9}

= L^{- 1} {\frac{6}{s} - 6 \cdot \frac{s + 5}{( s + 5 ) ^{2} + 9} - 24 \cdot \frac{1}{( s + 5 ) ^{2} + 9}}

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{6}{s}} - 6 L^{- 1} {\frac{s + 5}{( s + 5 ) ^{2} + 9}} - 24 L^{- 1} {\frac{1}{( s + 5 ) ^{2} + 9}}

L^{- 1} {\frac{6}{s}} : 6 H (t)

L^{- 1} {\frac{s + 5}{( s + 5 ) ^{2} + 9}} : e^{- 5 t} cos (3 t)

L^{- 1} {\frac{1}{( s + 5 ) ^{2} + 9}} : e^{- 5 t} \frac{1}{3} sin (3 t)

= 6 H (t) - 6 e^{- 5 t} cos (3 t) - 24 e^{- 5 t} \frac{1}{3} sin (3 t)

Refine

6 H (t) - 6 e^{- 5 t} cos (3 t) - 24 e^{- 5 t} \frac{1}{3} sin (3 t) : 6 H (t) - 6 e^{- 5 t} cos (3 t) - 8 e^{- 5 t} sin (3 t)

= 6 H (t) - 6 e^{- 5 t} cos (3 t) - 8 e^{- 5 t} sin (3 t)

f (x) = x^{4} + 5 e^{x}, (0, 5)

f (x) = a x

\int_{- \infty}^{0} x \cdot e^{x} d x

\frac{d x}{d t} = x^{2} e^{- 2 t}

\int (\frac{3 - x}{x ^{2} - 6 x + 2}) d x

What is inverse oflaplace (6(s+34))/(s(s^2+10s+34)) ?
The answer to inverse oflaplace (6(s+34))/(s(s^2+10s+34)) is 6\H(t)-6e^{-5t}cos(3t)-8e^{-5t}sin(3t)