Let y = y(x) be the solution of the differential equation, (x^{2}+1 )^{2} \frac{d y}{d x}+2 x (x^{2}+1 ) y=1 such that y(0)=0 . If \sqrt{a} y(1)= \frac{\\pi}{32}, then the value of ‘a’ is: [8 Apr 2019 Shift 1]

JEE Main 8 Apr 2019 Paper 1 Solved Paper

Section: Mathematics

Question : 68 of 90

Marks: +1, -0

Let y = y(x) be the solution of the differential equation, (x2+1)2

+2x(x2+1)y=1 such that y(0)=0. If √ay(1)=

, then the value of ‘a’ is:

[8 Apr 2019 Shift 1]

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