Solve the differential equation which is HDE(Homogeneous differential equation): (3xy + y2)dx + (x2 + xy)dy = 0 Plzzzz answer to this question as soon as possible!!!! Share with your friends Share 0 Vipin Verma answered this (3xy+y2)dx+(x2+xy)dy=0So dxdy=-(x2+xy)(3xy+y2)Let x=vy, then dxdy=v+ydvdySo v+ydvdy=-(v2y2+vy2)(3vy2+y2)Or v+ydvdy=-(v2+v)(3v+1) Or ydvdy=-(v2+v)(3v+1)-v Or ydvdy=-v2-v-3v2-v3v+1=-4v2-2v3v+1 Or 3v+1-4v2-2vdv=dyy Or -3v4v2+2vdv-14v2+2vdv=dyy Or -32(2v+1)dv-14v2+2vdv=dyy And 14v2+2v=12v(2v+1)=12v-12v+1, so Hence -32(2v+1)dv-12vdv+12v+1dv=dyy Or -12v+1dv-12vdv=dyy Lets 2v+1 =t, then 2dv=dt, hence -dt2t- 12vdv=dyy Or -12lnt-12lnv =lny+lnC Or -12ln(2v+1)-12lnv =lny+lnC Or -12(ln(2v+1)+lnv) =lny+lnC Or -ln(2v2+v)=2lny+2lnC Or ln12v2+v=2lny+2lnC And v =xy, So ln12(xy)2+xy=lny +lnC2 =lnyC2 Taking antilog on both side ,we get Or 12(xy)2+xy=yC2 (Ans) 0 View Full Answer
