If x3dy+xydx=x2dy+2ydx;y(2)=e and x>1, then y(4) is equal to

The given differential equation is x3dy+xydx=x2dy+2ydx⇒x3-x2dy=y(2-x)dx⇒dyy=2-xx2(x-1)dx=1x-1-1x-2x2dx∫dyy=∫  1x-1dx-∫ 1x-1dx-∫2x2dx⇒logy=log(x-1)-logx+2x+C(∵x>1) Given that y(2)=e ⇒1=0-log2+1+C⇒C=log2⇒logy=logx-1x+2x+log2 For x=4⇒log(y(4))=log34+12+log2log34+12log e+log 2=log3e2 Therefore, y(4)=3e2