What is inverse oflaplace 4/(2s^2+2s+1) ?

The answer to inverse oflaplace 4/(2s^2+2s+1) is 4e^{-t/2}sin(t/2)

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inverse oflaplace 4/(2s^2+2s+1)

inverse laplace

\frac{4}{2 s ^{2} + 2 s + 1}

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

Solution steps

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

Expand

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

= L^{- 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)}

= 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})

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

Refine

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

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

d er i v a t i v e y = e^{4} x^{2} + 1

\frac{\partial}{\partial y} (3 x^{2} y^{2} + 1)

\frac{x}{( 1 + x ^{2} )} d x + \frac{y}{( 1 + y ^{2} )} d y = 0

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

63 x^{2} y^{^{''}} + 14 x (x - 9) y^{^{'}} - 14 (x - 9) y = - 2 x^{3}

What is inverse oflaplace 4/(2s^2+2s+1) ?
The answer to inverse oflaplace 4/(2s^2+2s+1) is 4e^{-t/2}sin(t/2)