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

Sep 18, 2017

$y = {e}^{A x}$

Explanation:

We have:

$\left(y \ln y\right) \mathrm{dx} - x \mathrm{dy} = 0$ ..... [A]

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

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus x \setminus \mathrm{dy} = y \ln y \setminus \mathrm{dx}$

$\therefore \int \setminus \frac{1}{y \ln y} \setminus \mathrm{dy} = \int \setminus \frac{1}{x} \setminus \mathrm{dx}$ ..... [B]

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

$u = \ln y \implies \frac{\mathrm{du}}{\mathrm{dy}} = \frac{1}{y}$

Substituting into [B], we find

$\int \setminus \frac{1}{y \ln y} \setminus \mathrm{dy} = \int \setminus \frac{1}{y} \frac{1}{\ln} y \setminus \mathrm{dy}$
$\text{ } = \int \setminus \frac{1}{u} \setminus \mathrm{du}$
$\text{ } = \ln | u |$
$\text{ } = \ln | \ln y |$, after restoring the substitution.

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

$\therefore \ln | \ln y | = \ln x + C$
$\therefore \ln | \ln y | = \ln x + \ln A$, say
$\therefore \ln | \ln y | = \ln A x$

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

$| \ln y | = A x$

And after again taking exponents, and noting that the exponential function is positive over its entire domain (${e}^{x} > 0 \forall x \in \mathbb{R}$) we can write:

$y = {e}^{A x}$

Which is the Explicit General Solution