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