Let y = y(x) be the solution of the differential equation, (x^{2} + 1)^{2} + 2x (x^{2}+ 1) y = 1 such that y(0) = 0. If y(1) = , then the value of ‘a’ is 

(a) 

(b) 

(c) 
1 
(d) 

Let y = y(x) be the solution of the differential equation, (x^{2} + 1)^{2} + 2x (x^{2}+ 1) y = 1 such that y(0) = 0. If y(1) = , then the value of ‘a’ is 

(a) 

(b) 

(c) 
1 
(d) 

(1 + x^{2}) + 2x(1 + x^{2}) y = 1
⇒
It is a linear differential equation
I.F. = = 1 + x^{2}
⇒ y’ (1 + x^{2}) = + c
⇒ y (= tan^{ − 1} x + c
If x = 0 then y = 0
So, 0 = 0 + c
⇒ c = 0
⇒ y (1 + x^{2}) = tan^{−1} x
put x = 1
2y =
⇒ 2 =
⇒ =
⇒ a =