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Question

If cosxdydx+ysinx=1 and y(π4)=2, then y(π3) is

A
132
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B
1+32
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C
13
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D
1+3
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Solution

The correct option is A 132
cosxdydx+ysinx=1dydx+ytanx=secx
which is a linear differential equation.
I.F.=etanx dx=secx

The general solution is
y×secx=sec2x dxysecx=tanx+c
Given y(π4)=2
2=1+cc=1
Hence, the solution of the given equation is ysecx=tanx+1

Putting x=π3, we get
2y=3+1y=132

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