What is inverse oflaplace (2s-44)/(7s^2-14s+70) ?

The answer to inverse oflaplace (2s-44)/(7s^2-14s+70) is 2/7 e^tcos(3t)-2e^tsin(3t)

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inverse oflaplace (2s-44)/(7s^2-14s+70)

inverselaplace

\frac{2}{7} e^{t} cos (3 t) - 2 e^{t} sin (3 t)

Solution steps

L^{- 1} {\frac{2 s - 44}{7 s ^{2} - 14 s + 70}}

Expand

\frac{2 s - 44}{7 s ^{2} - 14 s + 70} : \frac{2}{7} \cdot \frac{s - 1}{( s - 1 ) ^{2} + 9} - 6 \cdot \frac{1}{( s - 1 ) ^{2} + 9}

= L^{- 1} {\frac{2}{7} \cdot \frac{s - 1}{( s - 1 ) ^{2} + 9} - 6 \cdot \frac{1}{( s - 1 ) ^{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)}

= \frac{2}{7} L^{- 1} {\frac{s - 1}{( s - 1 ) ^{2} + 9}} - 6 L^{- 1} {\frac{1}{( s - 1 ) ^{2} + 9}}

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

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

= \frac{2}{7} e^{t} cos (3 t) - 6 e^{t} \frac{1}{3} sin (3 t)

Refine

\frac{2}{7} e^{t} cos (3 t) - 6 e^{t} \frac{1}{3} sin (3 t) : \frac{2}{7} e^{t} cos (3 t) - 2 e^{t} sin (3 t)

= \frac{2}{7} e^{t} cos (3 t) - 2 e^{t} sin (3 t)

What is inverse oflaplace (2s-44)/(7s^2-14s+70) ?
The answer to inverse oflaplace (2s-44)/(7s^2-14s+70) is 2/7 e^tcos(3t)-2e^tsin(3t)