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

• June 30th 2012, 02:40 PM
fractal5
Trying to solve by substitution...getting stuck near the end
Hi, I have a differential equation that I'm trying to solve my substitution. I've got most of it, but I can figure out how to end it properly.

The equation: x y' - y = y [1 - ln2(y/x)]

I divide by x
let v = y/x
y' = v + xv'

sub v and y' in the equation to get:

dx/x = dv/[v(1-ln2v)]

let u = ln v
du = dv/v

sub these to get:
dx/x = du/(1-u2)

Integrating this I get:
ln |x| = 1/2 * ln | (1+u)/(1-u) |

reverse substituting v:
ln |x| = 1/2 * ln | (1+ ln v) / (1 - ln v ) |

reverse substituting y:
ln |x| = 1/2 * ln | [1+ ln (y/x)] / [1- ln (y/x)] |

How do I proceed after this. Is there any way to get the solution in terms of y?

Or should I use a different approach from the beginning itself?

Any help is appreciated, thanks :)
• June 30th 2012, 04:26 PM
Soroban
Re: 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.
• June 30th 2012, 04:44 PM
fractal5
Re: Trying to solve by substitution...getting stuck near the end
Thanks Soroban, seems easy once I've seen the solution. I'd just like to point out for anyone using this in the future, after integrating it comes to ln|(1+u)/(1-u)|. I believe you reciprocated it by mistake, but no worries I got my answer. Thanks again.