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

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

Popular >

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

inverse laplace

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

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

Solution steps

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

= L^{- 1} ⎩ ⎨ ⎧ \frac{1}{2} \cdot \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)}

= \frac{1}{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)

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

\int_{0}^{2} \frac{1}{x ^{2} - 7 x + 6} d x

\int \frac{36 x ^{2} - 49}{x ^{3}} d x

x \to \frac{π}{2} - lim (sec (x))

\frac{d y}{d x} = e^{8 x - 3 y}, y (0) = 4

f (x) = \frac{x ^{2}}{( 4 + x )}

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