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

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

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inverse oflaplace s/(2s^2+2s+1)

inverse laplace

\frac{s}{2 s ^{2} + 2 s + 1}

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

Solution steps

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

Expand

\frac{s}{2 s ^{2} + 2 s + 1} : \frac{1}{2} \cdot \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}} - \frac{1}{4} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}}

= L^{- 1} {\frac{1}{2} \cdot \frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}} - \frac{1}{4} \cdot \frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}}}

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{1}{2} L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}}} - \frac{1}{4} L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}}}

L^{- 1} {\frac{s + \frac{1}{2}}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}}} : e^{- \frac{t}{2}} cos (\frac{t}{2})

L^{- 1} {\frac{1}{( s + \frac{1}{2} ) ^{2} + \frac{1}{4}}} : e^{- \frac{t}{2}} \cdot 2 sin (\frac{t}{2})

= \frac{1}{2} e^{- \frac{t}{2}} cos (\frac{t}{2}) - \frac{1}{4} e^{- \frac{t}{2}} \cdot 2 sin (\frac{t}{2})

Refine

\frac{1}{2} e^{- \frac{t}{2}} cos (\frac{t}{2}) - \frac{1}{4} e^{- \frac{t}{2}} 2 sin (\frac{t}{2}) : \frac{1}{2} e^{- \frac{t}{2}} cos (\frac{t}{2}) - \frac{1}{2} e^{- \frac{t}{2}} sin (\frac{t}{2})

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

\int_{1}^{2} 4^{- x} d x

y^{^{'}} + 2 y = 2 t

\int 2 x + 7 e^{x} d x

x \to 2 lim (x)

\frac{d}{d x} ((10 x - 2 x^{2}) (16 - 2 x))

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