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Question

Solution of 2ysinxdydx=2sinxcosxy2cosx,x=π2,y=1 is given by

A
y2=sinx
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B
y=sin2x
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C
y2=cosx+1
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D
None of these
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Solution

The correct option is A y2=sinx
2ysinxdydx=2sinxcosxy2cosx
2ysinxdydx+y2cosx=2sinxcosx
sinxddx(y2)+y2ddxsinx=2sinxcosx
ddx(y2sinx)=2sinxcosx
d(y2sinx)=sin2xdx
y2sinx=cos2x2+c
At x=π2,y=1
12sinπ2=cosπ2+cc=12
So, y2sinx=cos2x2+12
y2=1cos2x2sinx
y2=sinx

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