Question

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

Answer 1

Answer:

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

Answer 2

Answer:

see below

Step-by-step explanation:

y' = -2e⁻ˣcos(x) -sin(x)2e⁻ˣ +e⁻ˣsin(x)-cos(x)e⁻ˣ = e⁻ˣ(-2cos(x)-2sin(x)+sin(x) -cos(x))

=e⁻ˣ(-3cos(x)-sin(x))

y'' = -e⁻ˣ(-3cos(x)-sin(x)) + (3sin(x)-cos(x))e⁻ˣ

y'' = e⁻ˣ[3cos(x)+sin(x) + 3sin(x)-cos(x)] = e⁻ˣ[2cos(x)+4sin(x)]

y''+2y'+2y = 0

e⁻ˣ[2cos(x)+4sin(x)] + 2[e⁻ˣ(-3cosx-sinx)] + 2[2e⁻ˣ cos(x)−e⁻ˣ sin(x)]

e⁻ˣ[2cos(x)+4sin(x)-6cos(x)-2sin(x)+4cos(x)-2sin(x)]

e⁻ˣ[-4cos(x)+2sin(x)+4cos(x)-2sin(x)]

e⁻ˣ[0cos(x)+0sin(x)] = 0