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)

\frac{d}{d x} (- 5 x^{\frac{1}{3}})

\frac{d}{d t} (e^{t} (sin (t) + cos (t)))

y^{^{'''}} + 5 y^{^{''}} - 8 y^{^{'}} - 48 y = 0

\int e - x cos (2 x) d x

\int \frac{e ^{a x + b} - e ^{b x + a}}{e ^{a x}} 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)