dy/dx+xsin2y=x3cos2y Share with your friends Share 8 Vipin Verma answered this dydx+xsin2y =x3cos2ydividing both sides by cos2y, we get1cos2ydydx+xsin2ycos2y =x3sec2ydydx+2xtanx =x3Let tanx = t, sec2ydydx=dtdxHence dtdx +2xt = x3this is linear form, so integrating factor will be e∫2xdx=ex2So t×ex2 =∫x3×ex2dxLet ex2=u, then 2xex2dx =du, and logu = x2Hence t×ex2 = 12∫loguduOr t×ex2 =12(ulogu -u) =12(x2 ex2 - ex2)Or tanx × ex2=12(x2 ex2 - ex2) +C 31 View Full Answer Aswin answered this DIVIDE THROUGHOUT BY COS2Y AND THEN PUT TANY=T,and differentiate w.r.t xTHEN YOU WILL GET SEC2Ydy/dx=dt/dx and then proceed as usual 18