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

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

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inverse oflaplace 4/(s^2+1)

inverse laplace

\frac{4}{s ^{2} + 1}

4 sin (t)

Solution steps

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

= L^{- 1} {4 \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)}

= 4 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)

= 4 sin (t)

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