What is inverse oflaplace 5/(s^2+4) ?

The answer to inverse oflaplace 5/(s^2+4) is 5/2 sin(2t)

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inverse oflaplace 5/(s^2+4)

inverse laplace

\frac{5}{s ^{2} + 4}

\frac{5}{2} sin (2 t)

Solution steps

L^{- 1} {\frac{5}{s ^{2} + 4}}

= L^{- 1} {\frac{5}{2} \cdot \frac{2}{s ^{2} + 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{5}{2} L^{- 1} {\frac{2}{s ^{2} + 2 ^{2}}}

Use Inverse Laplace Transform table:

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

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

= \frac{5}{2} sin (2 t)

\frac{\partial}{\partial x} (- 4 x^{5} y^{- 5})

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

\frac{d}{d x} (ln (x^{11}))

\frac{\partial}{\partial x} (x^{2} y^{3} (6 - x - y))

\frac{d}{d x} (x^{15})

What is inverse oflaplace 5/(s^2+4) ?
The answer to inverse oflaplace 5/(s^2+4) is 5/2 sin(2t)