What is inverse oflaplace 3/(s^2+16) ?

The answer to inverse oflaplace 3/(s^2+16) is 3/4 sin(4t)

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inverse oflaplace 3/(s^2+16)

inverse laplace

\frac{3}{s ^{2} + 16}

\frac{3}{4} sin (4 t)

Solution steps

L^{- 1} {\frac{3}{s ^{2} + 16}}

= L^{- 1} {\frac{3}{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{3}{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{3}{4} sin (4 t)

n = 2 \sum \infty (\frac{1}{2})^{n}

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

\frac{d}{d x} (2 x ln (x) + x)

1500 x + 3000 (3 x + 3)^{- 1} - 1000

x^{\frac{2}{7}}, 1, x = 0

What is inverse oflaplace 3/(s^2+16) ?
The answer to inverse oflaplace 3/(s^2+16) is 3/4 sin(4t)