Given,
2yey2xdx+(y2−4xey2x)dy=0
2ey2x[ydx−2xdy]+y2dy=0
2ey2x[yy2dx−x⋅(2y)dy]+y2dy=0
Divide by y3
2ey2x[y4y2dx−x⋅(2y)dy]+y1dy=0
2ey2xd(y2x)+y1dy=0
Now integrating both side we get,
∫2ey2xd(y2x)+∫y1dy=0
2ey2x+lny+c=0
Given, (0,1) lies on it,
So, 2e0+ln1+c=0⇒c=−2
Hence required curve: 2ey2x+lny−2=0
For x(e)
2ee2x+lne−2=0⇒x=−e2loge2