We have, y=(ax+b)‌cos‌x ‌ysecx=ax+b ‌ysecx‌tan‌x+secxy1=a Again, differentiating w.r.t. x, we get ‌⇒ysecxsec2x+secx‌tan‌x⋅y1+y‌tan‌x‌‌s‌e‌c‌x‌tan‌x+secxy2+y1secx‌tan‌x=0 ‌⇒ysec3x+y1secx‌tan‌x+ysecxtan2x+y1secx‌tan‌x+secx⋅y2=0 ‌⇒y2secx+2y1secx‌tan‌x+y(sec2x+tan2x)secx=0
On dividing both sides by sec3x, we get y2cos2x+2y1sin‌x⋅cos‌x+y(1+sin‌2x)=0 ‌y2−y1sin‌2x+y1sin‌2x+y(1+sin‌2x)=0 ‌∴y2+y1sin‌2x+y(1+sin‌2x)=y2sin‌2x