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:
- 32e2
- 23e2
- e2
- 2e2
Answer
(2) 23e2
Solution
dxdy= 2xy-x22y2 (homogeneous). 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
x(e)=e: 2=1+C C=1
2yx= e y+1
At y=e2: 2e2x=3 x= 23e2
Answer: (2)