y''=sin^(2)3x, x_(0)=π/12, y(0)=-π^(2)/16, y'(0)=0.
Помогите решить пожалуйста
y`= ∫ (1-cos2x)dx/2=(1/2)x-(1/4)sin2x+C_(1)
y= ∫ ((1/2)x-(1/4)sin2x+C_(1))dx=
=(1/2)*(x^2/2) +(1/8)cos2x+C_(1)x+C_(2)
y(0)=–π^2/16, y'(0)=0 ⇒
{-π^2/16=(1/2)*0+(1/8)cos0+C_(1)*0+C_(2) ⇒ C_(2)=(-π^2/16)-(1/8)
{0=(1/2)*0-(1/4)sin0+C_(1) ⇒ C_(1)=0
[b]y=(1/2)*(x^2/2) +(1/8)cos2x- (π^2/16)-(1/8)[/b]
y(π/12)=(1/2)*(π^2/288) +(1/8)cos(π/6)- (π^2/16)-(1/8) - считайте