What is inverse oflaplace 5/(s^2+1) ?

The answer to inverse oflaplace 5/(s^2+1) is 5sin(t)

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inverse oflaplace 5/(s^2+1)

inverse laplace

\frac{5}{s ^{2} + 1}

5 sin (t)

Solution steps

L^{- 1} {\frac{5}{s ^{2} + 1}}

= L^{- 1} {5 \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)}

= 5 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)

= 5 sin (t)

\frac{d}{d x} ((x + \frac{1}{x ^{3}})^{2})

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

y = \frac{4 x ^{2} + 8 x + 4}{x}

x \to 5 lim (5 - x)

k = 0 \sum \infty \frac{1}{2} \cdot (- 1)^{k}

What is inverse oflaplace 5/(s^2+1) ?
The answer to inverse oflaplace 5/(s^2+1) is 5sin(t)