y=∫(-1/2)*ctg2x+C_(1))dx=-(1/4)ln|sin2x|+C_(1)x+C_(2)
y=-(1/4)ln|sin2x|+C_(1)x+C_(2)
y(x_(o))=y(5π/4)=(-1/4)ln|sin(5π/2)+C_(1)*(5π/4)+C_(2)
y(5π/4)=C_(1)*(5π/4)+C_(2)
y(π/4)=1 ⇒ 1=-(1/4)ln|sin2*(π/4)|+C_(1)*(π/4)+C_(2)
y`(π/4)=1 ⇒ 1=(1/2)*(-ctg(2*(π/4))+C_(1)
Из системы уравнений находим C_(1) и С_(2):
{1=-(1/4)ln|sin2*(π/4)|+C_(1)*(π/4)+C_(2)
{1=(1/2)*(-ctg(2*(π/4))+C_(1) ⇒ [b] C_(1)=1[/b] ( ctg(π/2)=0)
1=-(1/4)ln1+(π/4)+C_(2) ⇒ C_(2)=1-(π/4)
Частное решение :
y=-(1/4)ln|sin2x|+x+1-(π/4)