What is inverse oflaplace 4(((1))/(s^2+16)) ?

The answer to inverse oflaplace 4(((1))/(s^2+16)) is sin(4t)

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inverse oflaplace 4(((1))/(s^2+16))

inverse laplace

4 (\frac{( 1 )}{s ^{2} + 16})

sin (4 t)

Solution steps

L^{- 1} {4 (\frac{( 1 )}{s ^{2} + 16})}

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)}

= 4 L^{- 1} {\frac{1}{s ^{2} + 16}}

L^{- 1} {\frac{1}{s ^{2} + 16}} : \frac{1}{4} sin (4 t)

= 4 \cdot \frac{1}{4} sin (4 t)

Refine

4 \frac{1}{4} sin (4 t) : sin (4 t)

= sin (4 t)

y = \frac{2}{x}, y = 7 - x

\int x 4 - x^{2} - x^{2} \frac{3}{3} d x

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

x 9 - x

\int tan (- 6 x + 3) d x

What is inverse oflaplace 4(((1))/(s^2+16)) ?
The answer to inverse oflaplace 4(((1))/(s^2+16)) is sin(4t)