Trying to solve by substitution...getting stuck near the end

Hello, fractal5!

Quote:

Hi, I have a differential equation that I'm trying to solve my substitution.
I've got most of it, but I can't figure out how to end it properly.

The equation: . $x\frac{dy}{dx} - y \:=\: y \left[1 - \ln^2\!\left(\tfrac{y}{x}\right)\right]$

I divide by $x\!:\;\;\frac{dy}{dx} -\frac{y}{x} \:=\:\frac{y}{x}\left[1 - \ln^2\!\left(\tfrac{y}{x}\right)\right]$

Let $v \,=\, \frac{y}{x} \quad\Rightarrow\quad \frac{dy}{dx} \:=\:v + x\frac{dv}{dx}$

Substitute: . $\frac{dv}{1-\ln^2v} \:=\:\frac{dx}{x}$

Let $u \,=\,\ln v \quad\Rightarrow\quad du \,=\, \frac{dv}{v}$

Substitute: . $\frac{du}{1-u^2} \:=\:\frac{dx}{x}$

Integrating: . $\tfrac{1}{2}\ln\left|\frac{1-u}{1+u}\right| \:=\:\ln|x| + c_1$

Back-substitute: . $\tfrac{1}{2}\ln\left|\frac{1-\ln v}{1 + \ln v }\right| \;=\;\ln|x| + \ln c_2$

Back-substitute: . $\tfrac{1}{2}\ln\left|\frac{1-\ln\frac{y}{x}}{1+\ln\frac{y}{x}}\right| \;=\;\ln|c_2x|$

How do I proceed after this? . You could stop here.
Is there any way to get the solution in terms of $y$ ? . Yes, if you dare.

We have: . $\tfrac{1}{2}\ln\left|\frac{1-\ln\frac{y}{x}}{1+\ln\frac{y}{x}}\right| \;=\;\ln|c_2x|$

. . . . . . . . . $\ln\left|\frac{1-\ln\frac{y}{x}}{1 + \ln\frac{y}{x}}\right| \;=\;2\ln|c_2x| \;=\;\ln|c_2x|^2 \;=\;\ln|c_2^2x^2| \;=\;\ln|Cx^2|$

. . . . . . . . . . . . $\frac{1-\ln\frac{y}{x}}{1+\ln\frac{y}{x}} \;=\;Cx^2$

. . . . . . . . . . . . $1 - \ln\tfrac{y}{x} \;=\;Cx^2\left(1 + \ln\tfrac{y}{x}\right)$

. . . . . . . . . . . . $1 - \ln\tfrac{y}{x} \;=\;Cx^2 + Cx^2\ln\tfrac{y}{x}$

. . . . . . . $Cx^2\ln\tfrac{y}{x} + \ln\tfrac{y}{x} \;=\;1 - Cx^2$

.n. . . . . . $\left(Cx^2 + 1\right)\ln\tfrac{y}{x} \;=\;1 - Cx^2$

n . . . . . . . . . . . . . $\ln\tfrac{y}{x} \;=\;\frac{1-Cx^2}{1+Cx^2}$

n . . . . . . . . . . . . . . $\frac{y}{x} \;=\;e^{\frac{1-Cx^2}{1+Cx^2}}$

n . . . . . . . . . . . . . . $y \;=\;x\cdot e^{\frac{1-Cx^2}{1+Cx^2}}$

This time we were lucky (if you can call it that) . . . we could solve for y.

But 99% of the solutions cannot be solved for y.
. . And we are not expected to.