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

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

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inverse oflaplace 1/(s^2+16)

inverse laplace

\frac{1}{s ^{2} + 16}

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

Solution steps

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

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

Use Inverse Laplace Transform table:

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

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

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

\frac{1}{s ( s ^{2} + s + \frac{5}{4} )}

e^{x - 1}

\frac{\partial}{\partial x} (2 x y + 5 x^{2})

f (x) = \frac{x ^{2}}{x + 9}, at x = 1

x \to 0 lim (\frac{x}{x ^{2} + 5 x})

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