What is inverse oflaplace 2/(s^2-2s+9) ?

The answer to inverse oflaplace 2/(s^2-2s+9) is 1/(sqrt(2))e^tsin(2sqrt(2)t)

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inverse oflaplace 2/(s^2-2s+9)

inverse laplace

\frac{2}{s ^{2} - 2 s + 9}

\frac{1}{2} e^{t} sin (22 t)

Solution steps

L^{- 1} {\frac{2}{s ^{2} - 2 s + 9}}

Expand

\frac{2}{s ^{2} - 2 s + 9} : 2 \cdot \frac{1}{( s - 1 ) ^{2} + 8}

= L^{- 1} {2 \cdot \frac{1}{( s - 1 ) ^{2} + 8}}

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)}

= 2 L^{- 1} {\frac{1}{( s - 1 ) ^{2} + 8}}

L^{- 1} {\frac{1}{( s - 1 ) ^{2} + 8}} : e^{t} \frac{1}{2 2} sin (22 t)

= 2 e^{t} \frac{1}{2 2} sin (22 t)

Refine

2 e^{t} \frac{1}{2 2} sin (22 t) : \frac{1}{2} e^{t} sin (22 t)

= \frac{1}{2} e^{t} sin (22 t)

y^{^{''}} + 2 y^{^{'}} + y = x^{- 4} e^{- x}

y = \frac{x}{x ^{2} + 9}

x \to \infty lim (1 + \frac{2}{x})

\int (5 x - 2) d x

\int 4 x - 7 d x

What is inverse oflaplace 2/(s^2-2s+9) ?
The answer to inverse oflaplace 2/(s^2-2s+9) is 1/(sqrt(2))e^tsin(2sqrt(2)t)