# 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)= 4n2y2
[(y1/n + y-1/n)2- 4] (dy/dx)= 4n.n.y.y
From 1,
[(2x)- 4)] (dy/dx)= 4n.n.y.y
[(4x- 4)] (dy/dx)= 4n2y2
(x- 1) (dy/dx)= n2y2
again diff. w.r.t. x,
2.y1.y2 (1-x2) + 2xy1= 2y.y1n2
dividing both the sides by 2y1, we get
(1-x2)y2 + xy1=yn2
Thus option 2 is correct.

• 3
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)= 4n2y2
[(y1/n + y-1/n)2- 4] (dy/dx)= 4n.n.y.y

From 1,
[(2x)- 4)] (dy/dx)= 4n2y2
[(4x- 4)] (dy/dx)= 4n2y2

dividing both the sides by 4,
(x- 1) (dy/dx)= n2y2

again diff. w.r.t. x,
2.y1.y2 (1-x2) + 2xy1= 2y.y1n2

dividing both the sides by 2y1, we get
(1-x2)y2 + xy1=yn2
Thus option 2 is correct.
• 4
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