Solution of 2ysinxdydx=2sinx⋅cosx−y2cosx,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 Ay2=sinx 2ysinxdydx=2sinx⋅cosx−y2cosx ⇒2ysinxdydx+y2cosx=2sinx⋅cosx ⇒sinxddx(y2)+y2ddxsinx=2sinx⋅cosx ⇒ddx(y2sinx)=2sinx⋅cosx ⇒∫d(y2sinx)=∫sin2x⋅dx ⇒y2sinx=−cos2x2+c At x=π2,y=1 12⋅sinπ2=−cosπ2+c⇒c=12 So, y2sinx=−cos2x2+12 ⇒y2=1−cos2x2sinx ⇒y2=sinx