What is inverse oflaplace ((6+6s))/(2s^2+1) ?

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

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inverse oflaplace ((6+6s))/(2s^2+1)

inverse laplace

\frac{( 6 + 6 s )}{2 s ^{2} + 1}

32 sin (\frac{1}{2} t) + 3 cos (\frac{1}{2} t)

Solution steps

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

Expand

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

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

L^{- 1} {\frac{s}{2 s ^{2} + 1}} : \frac{1}{2} cos (\frac{1}{2} t)

= 6 \cdot \frac{1}{2} sin (\frac{1}{2} t) + 6 \cdot \frac{1}{2} cos (\frac{1}{2} t)

Refine

6 \frac{1}{2} sin (\frac{1}{2} t) + 6 \frac{1}{2} cos (\frac{1}{2} t) : 32 sin (\frac{1}{2} t) + 3 cos (\frac{1}{2} t)

= 32 sin (\frac{1}{2} t) + 3 cos (\frac{1}{2} t)

x \to 1 + lim (\frac{x - 3}{x - 1})

9 x^{\frac{1}{2}}

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

\frac{d}{d x} (7 x^{6})

x \to 0 lim (\frac{( sin ( 3 x ) )}{x})

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