Let x=x(y) be the solution of the differential equation 2y2 dxdy-2xy+x2=0, y>1, x(e)=e. Then x (e2 ) is equal to:
(2) 23e2
dxdy = 2xy - x22y2 is homogeneous, so put x = vy, dxdy = v + y dvdy:
v + y dvdy = v - v22 y dvdy = - v22
2v2 dv = - dyy
- 2v = - e y - C 2yx = e y + C
Using x(e) = e, 2 = 1 + C, so C = 1 and
2yx = e y + 1
At y = e2: 2e2x = 3 x = 23e2
Answer: (2)
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