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Question

Let y=y(x) be the solution of the differential equation, (x2+1)2dydx+2x(x2+1)y=1 such that y(0)=0. If a y(1)=π32, then the value of a is:
  1. 116
  2. 1
  3. 12
  4. 14


Solution

The correct option is A 116
Given,
(x2+1)2dydz+2x(x2+1)y=1
dydx+(2xx2+1)y=1(x2+1)2

Integrating Factor(I.F.)=e2x1+x2dx=(1+x2)
Solution of differential equation:-
y(1+x2)=1(1+x2)2×(1+x2)dx
y(1+x2)=tan1(x)+c

Given that
y(0)=0
c=0
y(1+x2)=tan1x
Put x=1
y×2=π4
y=π8
14 y=π32
a=14
a=116

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