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

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

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

inverse laplace

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

2 sin (\frac{1}{2} t)

Solution steps

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

= L^{- 1} ⎩ ⎨ ⎧ 2 \frac{\frac{1}{2}}{s ^{2} + ( \frac{1}{2} ) ^{2}} ⎭ ⎬ ⎫

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{\frac{1}{2}}{s ^{2} + ( \frac{1}{2} ) ^{2}} ⎭ ⎬ ⎫

Use Inverse Laplace Transform table:

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

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

= 2 sin (\frac{1}{2} t)

\frac{\partial}{\partial y} (y - 2 x - 2)

y = \frac{1}{x - 3}

\int \frac{( x ^{2} + 1 )}{( x - 9 ) ( x - 8 ) ^{2}} d x

\int x cos (8 x^{2}) d x

f (x) = x^{2} - 2 x + 3, at x = 2

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