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Question

An equation of the curve satisfying xdyydx=x2y2dx and y(1)=0 is

A
y=xsin(logx)+C
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B
y=xsin(log|sinx|)
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C
y=y2=x(x1)2
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D
y=2x2(x1)
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Solution

The correct option is A y=xsin(logx)+C
xdyydx=x2y2dx

xdy=ydx+x2y2dx

xdy=(y+x2y2)dx

dydx=y+x2y2x

put y=vxdydx=v+xdvdx

v+xdvdx=vx+x2(vx)2x

v+xdvdx=1v2+v

xdvdx=1v2

dv1v2=dxx

sin1v=logx+C

sin1(yx)=logx+C

yx=sin(logx)+C

y=xsin(logx)+C

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