Mathematics

Mathematics

Problem

The solution to the differential equation yln⁡y+xy′=0y \ln y + xy' = 0 , where y(1)=ey(1) = e , is - (A) x(ln⁡y)=1x(\ln y) = 1 (B) xy(ln⁡y)=1xy(\ln y) = 1 (C) (ln⁡y)2=2(\ln y)^2 = 2 (D) ln⁡y+(x22)y=1\ln y + \left(\frac{x^2}{2}\right)y = 1

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