What is inverse oflaplace (2s-4)/(s^2-4s+13) ?

The answer to inverse oflaplace (2s-4)/(s^2-4s+13) is e^{2t}*2cos(3t)

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inverse oflaplace (2s-4)/(s^2-4s+13)

inverse laplace

\frac{2 s - 4}{s ^{2} - 4 s + 13}

e^{2 t} \cdot 2 cos (3 t)

Solution steps

L^{- 1} {\frac{2 s - 4}{s ^{2} - 4 s + 13}}

Expand

\frac{2 s - 4}{s ^{2} - 4 s + 13} : \frac{2 ( s - 2 )}{( s - 2 ) ^{2} + 9}

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

Apply inverse transform rule:

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

then

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

For

\frac{2 ( s - 2 )}{( s - 2 ) ^{2} + 9} : a = 2, F (s) = \frac{2 s}{s ^{2} + 9}

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

L^{- 1} {\frac{2 s}{s ^{2} + 9}} : 2 cos (3 t)

= e^{2 t} \cdot 2 cos (3 t)

\frac{d}{d x} (- 0.5 x^{2} + 64 x - 25)

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

y^{^{''}} + y^{^{'}} = 6 sin (2 x)

x \to 8 lim (\frac{5 x - 2}{x + 8})

\int_{- 2}^{2} x^{2} 4 - x^{2} d x

What is inverse oflaplace (2s-4)/(s^2-4s+13) ?
The answer to inverse oflaplace (2s-4)/(s^2-4s+13) is e^{2t}*2cos(3t)