What is inverse oflaplace 3/((s^2+4)) ?

The answer to inverse oflaplace 3/((s^2+4)) is 3/2 sin(2t)

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inverse oflaplace 3/((s^2+4))

inverse laplace

\frac{3}{( s ^{2} + 4 )}

\frac{3}{2} sin (2 t)

Solution steps

L^{- 1} {\frac{3}{( s ^{2} + 4 )}}

= L^{- 1} {\frac{3}{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{3}{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{3}{2} sin (2 t)

27 n + 3 q^{2} n

y = \frac{4 x - 1}{2 x + 5}

\frac{d}{d x} (sin (x^{3}) cos (x^{2}))

\frac{d y}{d x} = x^{2} - y - 2, y (0) = - 1

\int_{0}^{5} 9800 π (5 - y) d y

What is inverse oflaplace 3/((s^2+4)) ?
The answer to inverse oflaplace 3/((s^2+4)) is 3/2 sin(2t)