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 cos (\frac{1}{2}) x d x

t y^{^{'}} + 2 y = sin (t), y (\frac{π}{2}) = 5

\frac{\partial}{\partial x} (6 x - 4 y)

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

\int \frac{1}{8 - 2 x ^{2}} d x

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