What is inverse oflaplace (2s-1)/(s^2-6s+18) ?

The answer to inverse oflaplace (2s-1)/(s^2-6s+18) is 2e^{3t}cos(3t)+5/3 e^{3t}sin(3t)

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inverse oflaplace (2s-1)/(s^2-6s+18)

inverse laplace

\frac{2 s - 1}{s ^{2} - 6 s + 18}

2 e^{3 t} cos (3 t) + \frac{5}{3} e^{3 t} sin (3 t)

Solution steps

L^{- 1} {\frac{2 s - 1}{s ^{2} - 6 s + 18}}

Expand

\frac{2 s - 1}{s ^{2} - 6 s + 18} : 2 \cdot \frac{s - 3}{( s - 3 ) ^{2} + 9} + 5 \cdot \frac{1}{( s - 3 ) ^{2} + 9}

= L^{- 1} {2 \cdot \frac{s - 3}{( s - 3 ) ^{2} + 9} + 5 \cdot \frac{1}{( s - 3 ) ^{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)}

= 2 L^{- 1} {\frac{s - 3}{( s - 3 ) ^{2} + 9}} + 5 L^{- 1} {\frac{1}{( s - 3 ) ^{2} + 9}}

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

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

= 2 e^{3 t} cos (3 t) + 5 e^{3 t} \frac{1}{3} sin (3 t)

Refine

2 e^{3 t} cos (3 t) + 5 e^{3 t} \frac{1}{3} sin (3 t) : 2 e^{3 t} cos (3 t) + \frac{5}{3} e^{3 t} sin (3 t)

= 2 e^{3 t} cos (3 t) + \frac{5}{3} e^{3 t} sin (3 t)

\frac{d}{d t} (12 t)

\int_{0}^{1} e^{- s t} d t

\int (x^{x}) d x

\frac{d y}{d x} = x^{5} (x - x y)

n = 0 \sum \infty \frac{1}{9}

What is inverse oflaplace (2s-1)/(s^2-6s+18) ?
The answer to inverse oflaplace (2s-1)/(s^2-6s+18) is 2e^{3t}cos(3t)+5/3 e^{3t}sin(3t)