What is inverse oflaplace 1/((s^2+3)) ?

The answer to inverse oflaplace 1/((s^2+3)) is 1/(sqrt(3))sin(sqrt(3)t)

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inverse oflaplace 1/((s^2+3))

inverse laplace

\frac{1}{( s ^{2} + 3 )}

\frac{1}{3} sin (3 t)

Solution steps

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

= L^{- 1} {\frac{1}{3} \cdot \frac{3}{s ^{2} + ( 3 ) ^{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}{3} L^{- 1} {\frac{3}{s ^{2} + ( 3 ) ^{2}}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{3}{s ^{2} + ( 3 ) ^{2}}} = sin (3 t)

= \frac{1}{3} sin (3 t)

4 - x^{2}, x

- 2 y^{^{''}} + 2 y^{^{'}} = 0, y (1) = - 2, y^{^{'}} (- 2) = 3

x \to 0 lim (\frac{x}{3 + x - 3})

\int \frac{( x - 4 )}{x ^{2} - 5 x + 6} d x

\int x^{4} e^{1 - x^{5}} d x

What is inverse oflaplace 1/((s^2+3)) ?
The answer to inverse oflaplace 1/((s^2+3)) is 1/(sqrt(3))sin(sqrt(3)t)