What is the general solution of the differential equation? : #(ylny)dx-xdy=0 #

We have:

# (ylny)dx - xdy=0 # ..... [A]

This is a separable Differential Equation so we can collect terms and write it in separated differential form and integrate:

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ dy = ylny \ dx #

# :. int \ 1/(ylny) \ dy = int \ 1/x \ dx # ..... [B]

The RHS integral is trivial, and the LHS initially seems quite daunting, but if we perform, the substitution:

# u = ln y => (du)/dy = 1/y #

Substituting into [B], we find

# int \ 1/(ylny) \ dy = int \ 1/y 1/lny \ dy #
# " " = int \ 1/u \ du #
# " " = ln|u| #
# " " = ln|lny| #, after restoring the substitution.

Using this result, we can now integrate [B] to get an Implicit General Solution:

# :. ln|lny| = lnx + C #
# :. ln|lny| = lnx + lnA #, say
# :. ln|lny| = lnAx #

We would typically want an explicit solution (where possible), so we can take exponential to base #e# (or anti-logarithms), giving:

# |lny| = Ax #

And after again taking exponents, and noting that the exponential function is positive over its entire domain (#e^x gt 0 AA x in RR#) we can write:

# y = e^(Ax) #

Which is the Explicit General Solution