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Question

A curve passes thorugh the point (x=1,y=0) and satisfies the differential equation dydx=x2+y22y+yx
The equation that decribes the curve is

A
ln(1+y2x2)=x1
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B
12ln(1+y2x2)=x1
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C
ln(1+yx)=x1
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D
12ln(1+yx)=x1
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Solution

The correct option is A ln(1+y2x2)=x1
dydx=x2+y22y+yx ..... (1)
Put y=vx
dydx=v+xdvdx
v+xdvdx=x2+v2x22vx+v
xdvdx=x2v+vx2
dvdx=12[1v+v]
(2v1+v2)dv=dx
log(1+v2)=v+c
log[1+y2x2]=x+c
Now, y(1)=0
C=1
so log(1+y2x2)=x1

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