What is the general solution of the differential equation? : #(ylny)dx-xdy=0 #
1 Answer
# y = e^(Ax) #
Explanation:
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
# |lny| = Ax #
And after again taking exponents, and noting that the exponential function is positive over its entire domain (
# y = e^(Ax) #
Which is the Explicit General Solution
