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

1 Answer
Sep 18, 2017

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 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