What is a particular solution to the differential equation #dy/dx=Lnx/(xy)# and y(1)=2?

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Answer 1 · 2016-06-30T12:03:15

#y^2 = (ln x)^2 + 4#

first separate it

so #y dy/dx = (ln x)/x#

#int dy \ y = int dx \ (ln x)/x #

#y^2 /2 = (ln x)^2/2 + C#

put in the IV

#2^2 /2 = (ln 1)^2/2 + C \ implies C = 2#

#y^2 /2 = (ln x)^2/2 + 2#

#y^2 = (ln x)^2 + 4#