+2y‌tan‌x=sin‌x ‌ Here, ‌P=2‌tan‌x‌ and ‌Q=sin‌x ‌ So ‌ IF‌‌=e∫2‌tan‌x‌d‌x ‌‌=e2‌ln|sec‌x| ‌‌=eln‌sec2‌x ‌‌=sec2x Now the required solution is y×(IF)=∫Q(IF)dx+c Substitute values. y×sec2x=∫sin‌x‌sec2‌x‌d‌x+c ysec2x‌‌=∫sec‌x‌tan‌c‌d‌x+c ‌‌=sec‌x+c ‌ At ‌y=0,x=‌
Ï€
3
‌ so, ‌ 0=2+c c=−2 Thus, ysec2x‌‌=sec‌x−2 y‌‌=‌
sec‌x
sec2x
−‌
2
sec2x
‌‌=cos‌x−−2cos2x ‌‌=cos−2(1−sin2x) Further simplify the above, y=cos‌x−2+2sin2x y=2sin2x+cos‌x−2