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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

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

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

Refine

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

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

3 (3 x^{2} + y^{2}) d x - 2 x y d y = 0

t + y e^{2 t y} + t e^{2 t y} \frac{d y}{d t} = 0

\frac{\partial}{\partial x} (2 e^{- 2 x y} + x y^{3})

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

x + \frac{3}{x}, at x = 2

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