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

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

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inverse oflaplace 1/((s^2+5))

inverse laplace

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

\frac{1}{5} sin (5 t)

Solution steps

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

= L^{- 1} {\frac{1}{5} \cdot \frac{5}{s ^{2} + ( 5 ) ^{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}{5} L^{- 1} {\frac{5}{s ^{2} + ( 5 ) ^{2}}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{5}{s ^{2} + ( 5 ) ^{2}}} = sin (5 t)

= \frac{1}{5} sin (5 t)

f (x) = \frac{1 - 4 x}{1 + 4 x}

\frac{\partial}{\partial y} (4 x^{3} y^{3} + 2)

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

\frac{\partial}{\partial x} (6 e^{x^{5} y})

\int_{0}^{1} \frac{x - 1}{x} d x

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