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

The answer to inverse oflaplace-(s-2)/(2(s^2-2s+2)) is -e^t(1/2 cos(t)-1/2 sin(t))

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inverse oflaplace-(s-2)/(2(s^2-2s+2))

inverse laplace

- \frac{s - 2}{2 ( s ^{2} - 2 s + 2 )}

- e^{t} (\frac{1}{2} cos (t) - \frac{1}{2} sin (t))

Solution steps

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

Expand

\frac{s - 2}{2 ( s ^{2} - 2 s + 2 )} : \frac{1}{2} \cdot \frac{s - 1}{( s - 1 ) ^{2} + 1} - \frac{1}{2} \cdot \frac{1}{( s - 1 ) ^{2} + 1}

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

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

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

\int \frac{2 x}{x ^{2} + 1} d x

(12 - x) y^{2} = x^{3}, (6, 6)

\int (4 x^{3} + 2 x y^{2}) d x

x^{2} y^{^{'}} + 2 x y = ln (x)

\int \frac{x}{2 x + 3} d x

What is inverse oflaplace-(s-2)/(2(s^2-2s+2)) ?
The answer to inverse oflaplace-(s-2)/(2(s^2-2s+2)) is -e^t(1/2 cos(t)-1/2 sin(t))