Show that y=2e^(−x) cos(x)−e^(−x) sin(x) is a solution to y''+2y'+2y=0.

y=2e^(−x)cosx−e^(−x)sinx

y = e^(-x)[2cosx - sinx]

y': product law

y' = -e^(-x)[2cosx - sinx] + e^(-x)[-2sinx - cosx]

y' = -e^(-x)[2cosx - sinx + 2sinx + cosx]

y' = -e^(-x)[3cosx + sinx]

y" = e^(-x)[3cosx + sinx] - e^(-x)[-3sinx + cosx]

y" = e^(-x)[3cosx - cosx + sinx + 3sinx]

y" = e^(-x)[2cosx + 4sinx]

y" + 2y' + 2y

e^(-x)[2cosx + 4sinx] - 2e^(-x)[3cosx + sinx] +2e^(-x)[2cosx - sinx]

e^(-x)[4sinx - 2sinx - 2sinx + 2cosx - 6 cosx + 4cosx]

= e^(-x) × 0

= 0

Hence a solution

lol

step-by-step explanation: