What is inverse oflaplace s/(6s^2+4) ?

The answer to inverse oflaplace s/(6s^2+4) is 1/6 cos(sqrt(2/3)t)

Popular >

inverse oflaplace s/(6s^2+4)

inverse laplace

\frac{s}{6 s ^{2} + 4}

\frac{1}{6} cos (\frac{2}{3} t)

Solution steps

L^{- 1} {\frac{s}{6 s ^{2} + 4}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

\frac{s}{6 s ^{2} + 4} : F (s) = \frac{1}{6 s ^{2} + 4}

= \frac{d}{d t} (L^{- 1} {\frac{1}{6 s ^{2} + 4}}) + L^{- 1} {\frac{1}{6 s ^{2} + 4}} (0)

L^{- 1} {\frac{1}{6 s ^{2} + 4}} : \frac{1}{2 6} sin (\frac{2}{3} t)

\frac{d}{d t} (\frac{1}{2 6} sin (\frac{2}{3} t)) = \frac{1}{6} cos (\frac{2}{3} t)

Evaluate

\frac{1}{2 6} sin (\frac{2}{3} t)

at point

0 = 0

= \frac{1}{6} cos (\frac{2}{3} t)

\frac{d}{d x} (\frac{1}{sin ( x + 1 )})

\int_{0}^{\frac{π}{2}} 3 sin^{2} (3 x) d x

\int \frac{3 x ^{2} + 5}{( x ^{2} + 4 ) ^{2}} d x

\int_{0}^{3} \frac{x}{36 - x ^{2}} d x

\int \frac{1 + x}{2 x} d x

What is inverse oflaplace s/(6s^2+4) ?
The answer to inverse oflaplace s/(6s^2+4) is 1/6 cos(sqrt(2/3)t)