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Questions  

A solution of the differential equation yxy1dx+xylnxdy=0, when y2=1  is

a
yx=1
b
xy+1=0
c
1y=log2x
d
x=log2y

detailed solution

Correct option is C

yxy−1+xylog⁡xdy=0⇒dxy=0⇒∫dxy=∫0dx⇒xy=cy(2)=1⇒c=2xy=2⇒yln⁡x=ln⁡2 ⇒1y=log2⁡x

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