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

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

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

inverse laplace

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

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

Solution steps

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

Take the partial fraction of

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

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

Expand

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

= L^{- 1} {\frac{1}{2} - \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)}

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

L^{- 1} {\frac{1}{2}} : \frac{1}{2} δ (t)

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

Refine

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

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

f (x) = \frac{1}{( 3 x + 8 ) ^{2}}

\frac{d}{d x} (\frac{- x ^{2} + 2 x + 1}{( x ^{2} + 1 ) ^{2}})

\int cos (2 πn x) d x

n = 0 \sum \infty \frac{( x - 3 ) ^{n}}{n}

\frac{d}{d x} (\frac{1}{e ^{\frac{1}{x}}})

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