What is inverse oflaplace (e^{-2x})/(2x^2+1) ?

The answer to inverse oflaplace (e^{-2x})/(2x^2+1) is \H(t-2) 1/(sqrt(2))sin(sqrt(1/2)(t-2))

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inverse oflaplace (e^{-2x})/(2x^2+1)

inverse laplace

\frac{e ^{- 2 x}}{2 x ^{2} + 1}

H (t - 2) \frac{1}{2} sin (\frac{1}{2} (t - 2))

Solution steps

L^{- 1} {\frac{e ^{- 2 x}}{2 x ^{2} + 1}}

Apply inverse transform rule:

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

then

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

Where

H (t)

is Heaviside step function

= H (t - 2) L^{- 1} {\frac{1}{2 x ^{2} + 1}} (t - 2)

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

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

y^{^{''}} - 3 y^{^{'}} - 10 y = 4 e^{4 t}, y (0) = 0, y^{^{'}} (0) = 0

x \to - 3 lim (2 x^{2} + 4 x + 1)

n = 9 \sum \infty e^{5 - 6 n}

\int (\frac{1}{1 + 9 x ^{2}}) d x

\frac{d}{d x} (5 sin (- 4 x))

What is inverse oflaplace (e^{-2x})/(2x^2+1) ?
The answer to inverse oflaplace (e^{-2x})/(2x^2+1) is \H(t-2) 1/(sqrt(2))sin(sqrt(1/2)(t-2))