Given 2x2dxdy−2xy+3y2=0
Dividing by 2x2y2, we get,
−y21dxdy+xy1=2x23
Let y1=p,y2−1dxdy=dxdp
⇒dxdp+x1(p)=2x23
General solution will be p⋅e∫x1dx=∫2x23⋅e∫x1dxdx+c
⇒yx=∫2x3dx+c⇒yx=23lnx+c
∵y(e)=3e⇒c=23⇒yx=23lnx+23 is the particular solution.
when x=1;y(1)=32