If x=f(y) is the solution of the differential equation (1+y2 )+ (x-2e -1y ) dydx=0, y (- 2, 2 ) with f(0)=1, then f ( 1 3 ) is equal to :
(3) e /6
Treat x as the dependent variable: dxdy+ x1+y2= 2e -1y1+y2, I.F. =e -1y
Put t= -1y: x et= 2e2tdt=e2t+c
y=0, x=1 c=0 x=e -1y
y= 1 3 f=e /6
Answer: (3)
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