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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

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)}

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

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}} \cdot 2 sin (\frac{t}{2})

Refine

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

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

arctan (x^{\frac{1}{2}})

\int (x + 3) (x^{2} + 6 x - 7) d x

\frac{\partial}{\partial x} (x^{2} + z^{2})

\frac{1}{x ^{2} ( x ^{2} + 4 x + 13 )}

\int_{0}^{2} 3 x d x

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