If x=f(y) is the solution of the differential equation (1+y^2 )+ (x-2 e ^ ⁻¹ y ) d y ~d x =0, y (- 2 , 2 ) with f(0)=1 , then f ( 1 3 ) is equal to :

Step-by-step solution

aligned & dx dy + x 1+ y ^2 = 2 e ^ ⁻¹ y 1+ y ^2 & I.F. = e ^ ⁻¹ y & xe ^ ⁻¹ y = 2 ( e ^ ⁻¹ y )^2 dy 1+ y ^2 & Put ⁻¹ y = t , dy 1+ y ^2 = dt & xe ^ ⁻¹ y = 2 e ^ 2 t dt & xe ^ ⁻¹ y = e ^ 2 ⁻¹ y + c & x = e ^ ⁻¹ y + ce ^ - ⁻¹ y & y =0, x =1 & 1=1+ c c =0 & y = 1 3 , x = e ^ / 6 aligned