What is inverse oflaplace 8/(s^2+1) ?

The answer to inverse oflaplace 8/(s^2+1) is 8sin(t)

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inverse oflaplace 8/(s^2+1)

inverse laplace

\frac{8}{s ^{2} + 1}

8 sin (t)

Solution steps

L^{- 1} {\frac{8}{s ^{2} + 1}}

= L^{- 1} {8 \cdot \frac{1}{s ^{2} + 1 ^{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)}

= 8 L^{- 1} {\frac{1}{s ^{2} + 1 ^{2}}}

Use Inverse Laplace Transform table:

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

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

= 8 sin (t)

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

y = 4 x^{4} + \frac{1}{3} x^{3} + x + 1

\int \frac{1}{x ^{2} - 3 x + 2} d x

\int \frac{4 x + 4}{2 x + 1} d x

\frac{d y}{d t} + 2 t (y - 5) = 0

What is inverse oflaplace 8/(s^2+1) ?
The answer to inverse oflaplace 8/(s^2+1) is 8sin(t)