What is inverse oflaplace 5/(s^2+12) ?

The answer to inverse oflaplace 5/(s^2+12) is 5/(2sqrt(3))sin(2sqrt(3)t)

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inverse oflaplace 5/(s^2+12)

inverse laplace

\frac{5}{s ^{2} + 12}

\frac{5}{2 3} sin (23 t)

Solution steps

L^{- 1} {\frac{5}{s ^{2} + 12}}

= L^{- 1} {\frac{5}{2 3} \cdot \frac{2 3}{s ^{2} + ( 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{5}{2 3} L^{- 1} {\frac{2 3}{s ^{2} + ( 2 3 ) ^{2}}}

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{2 3}{s ^{2} + ( 2 3 ) ^{2}}} = sin (23 t)

= \frac{5}{2 3} sin (23 t)

\int sin^{3} (2 a x) cos (2 a x) d x

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

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

\frac{d y}{d t} = y^{3} - y

\int \frac{1}{x 7 x - 1} d x

What is inverse oflaplace 5/(s^2+12) ?
The answer to inverse oflaplace 5/(s^2+12) is 5/(2sqrt(3))sin(2sqrt(3)t)