If y^1/n + y^-1/n = 2x, the (1-x^2)y2 + xy1 =
1) -n^2y 2) n^2y 3) 0 4) none of these
2x = y1/n + y-1/n ----1
diff. both the sides w.r.t. x
2 = [1/n .y1/n-1 + (-1/n).y-1/n] dy/dx
[y1/n - y-1/n] dy/dx = 2ny
sq. both the sides,
[(y1/n - y-1/n-1)2] (dy/dx)2 = 4n2y2
[(y1/n + y-1/n)2- 4] (dy/dx)2 = 4n.n.y.y
From 1,
[(2x)2 - 4)] (dy/dx)2 = 4n.n.y.y
[(4x2 - 4)] (dy/dx)2 = 4n2y2
(x2 - 1) (dy/dx)2 = n2y2
again diff. w.r.t. x,
2.y1.y2 (1-x2) + 2xy12 = 2y.y1n2
dividing both the sides by 2y1, we get
(1-x2)y2 + xy1=yn2
Thus option 2 is correct.
diff. both the sides w.r.t. x
2 = [1/n .y1/n-1 + (-1/n).y-1/n] dy/dx
[y1/n - y-1/n] dy/dx = 2ny
sq. both the sides,
[(y1/n - y-1/n-1)2] (dy/dx)2 = 4n2y2
[(y1/n + y-1/n)2- 4] (dy/dx)2 = 4n.n.y.y
From 1,
[(2x)2 - 4)] (dy/dx)2 = 4n.n.y.y
[(4x2 - 4)] (dy/dx)2 = 4n2y2
(x2 - 1) (dy/dx)2 = n2y2
again diff. w.r.t. x,
2.y1.y2 (1-x2) + 2xy12 = 2y.y1n2
dividing both the sides by 2y1, we get
(1-x2)y2 + xy1=yn2
Thus option 2 is correct.