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Question

Let y=y(x) be a solution of the differential equation, 1x2dydx+1y2=0,|x|<1.
If y(12)=32, then y(12) is equal to:

A
12
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B
32
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C
12
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D
32
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Solution

The correct option is C 12
1x2dydx+1y2=0
dy1y2+dx1x2=0
sin1y+sin1x=c

If x=12,y=32
then sin132+sin112=c
π3+π6=cc=π2

Therefore, the solution is
sin1y=π2sin1x=cos1x
sin1y=cos1x

Now, put x=12
sin1y=cos112
sin1y=3π4
y(12)=sin(3π4)=12

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