The solution of  y2x2y+exdx−ex+y3dy=0, if y0=1 is

yexdx−exdy+2x2y2dx−y3dy=0⇒yexdx−exdyy2+2x2dx−ydy=0dexy+2x2dx−ydy=0 Integrating we get, exy+2x33−y22=Cy(0)=1⇒1+0−12=C given C=12∴exy+2x33−y22=12 Gives 6ex+4x3y−3y3−3y=0