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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

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

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

Refine

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

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

y^{^{'}} e^{9 x} = 0

y^{^{'}} + 2 y - y^{2} = 0

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

\int \frac{x ^{2}}{2 + x ^{3}} d x

x \to 3 lim (2 x^{2} + 3 x - 1)

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