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Question

The solution of the differential equation (1+x2y2)ydx+(x2y2−1)xdy=0 is

A
xy=logxy+C
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B
xy=2logyx+C
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C
x2y2=logyx+C
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D
x2y2=2logyx+C
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Solution

The correct option is D x2y2=2logyx+C
Given differential equation is (1+x2y2)ydx+(x2y21)xdy=0 .... (i)
(1+x2y2)ydx=(x2y21)xdy

(1+x2y2)xdx=(1x2y2)ydy
(1x+xy2) dx=(1yx2y) dy
Integrating both sides
(1x+xy2) dx=(1yx2y) dy
(dxxdyy)+(xy2dx+x2ydy)=0
d(logxy)+12d(x2y2)+C=0
logxy+12x2y2+C=0


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