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

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

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

inverse laplace

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

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

Solution steps

L^{- 1} {\frac{1}{4 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)

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

\int \frac{1}{4 + 5 x ^{2}} d x

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

y^{^{''}} - 4 y^{^{'}} + 4 y = 6 e^{2 t} - 6 t e^{2 t} - 12 t + 12

\int tan^{2} (2 x) sec^{4} (2 x) d x

\frac{d y}{d x} + y = \frac{1}{1 + e ^{x}}

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