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

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

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

inverse laplace

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

e^{2 t} cos (t)

Solution steps

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

Apply inverse transform rule:

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

then

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

For

\frac{s - 2}{( s - 2 ) ^{2} + 1} : a = 2, F (s) = \frac{s}{s ^{2} + 1}

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

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

= e^{2 t} cos (t)

x^{9} d x + y^{9} d y = 0

\frac{1}{s ( s ^{2} + 2 s + 1 )}

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

y = f (x) = tan (x + 1)

f (x) = 7 x - 6 x, at x = 1

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