Let y=y(x) be the solution of the differential equation x 1-x^2 ,dy + (y 1-x^2 - x ⁻¹x )dx = 0 , x (0, 1) , _ x 1^- y(x) = 1 . Then y ( 1 2 ) equals:

Step-by-step solution

The given differential equation is: x 1-x^2 ,dy + (y 1-x^2 - x ⁻¹x )dx = 0 Rearranging the terms, we get: x 1-x^2 ,dy + y 1-x^2 ,dx = x ⁻¹x ,dx Dividing the entire equation by 1-x^2 : x ,dy + y ,dx = x ⁻¹x 1-x^2 ,dx The left side is the exact differential of xy : d(xy) = x ⁻¹x 1-x^2 ,dx Integrating both sides: xy = x ⁻¹x 1-x^2 ,dx + C To evaluate the integral, use integration by parts. Let u = ⁻¹x and dv = x 1-x^2 ,dx . du = - 1 1-x^2 ,dx and v = - 1-x^2 u ,dv = uv - v ,du x ⁻¹x 1-x^2 ,dx = - 1-x^2 ⁻¹x - (- 1-x^2 )