What is inverse oflaplace 1/(2s^2+4s+3) ?

The answer to inverse oflaplace 1/(2s^2+4s+3) is 1/(sqrt(2))e^{-t}sin(sqrt(1/2)t)

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inverse oflaplace 1/(2s^2+4s+3)

inverse laplace

\frac{1}{2 s ^{2} + 4 s + 3}

\frac{1}{2} e^{- t} sin (\frac{1}{2} t)

Solution steps

L^{- 1} {\frac{1}{2 s ^{2} + 4 s + 3}}

Expand

\frac{1}{2 s ^{2} + 4 s + 3} : \frac{1}{2} \cdot \frac{1}{( s + 1 ) ^{2} + \frac{1}{2}}

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

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

= \frac{1}{2} e^{- t} 2 sin (\frac{1}{2} t)

Refine

\frac{1}{2} e^{- t} 2 sin (\frac{1}{2} t) : \frac{1}{2} e^{- t} sin (\frac{1}{2} t)

= \frac{1}{2} e^{- t} sin (\frac{1}{2} t)

y^{^{'}} + 4 y = 3 sin (e^{4 x})

\int \frac{4 x ^{2} - 16}{x} d x

\frac{d}{d x} ((tan (x))^{- 1})

\int sec^{2} (x) tan^{4} (x) d x

x \to \infty lim (e^{\frac{- x}{100}})

What is inverse oflaplace 1/(2s^2+4s+3) ?
The answer to inverse oflaplace 1/(2s^2+4s+3) is 1/(sqrt(2))e^{-t}sin(sqrt(1/2)t)