# Second order non-linear differential equation involving log

1. Dec 13, 2013

### 5hassay

EDIT: my problem is solved, thank you to those who helped

1. The problem statement, all variables and given/known data

Solve:

$x y^{\prime \prime} = y^{\prime} \log (\frac{y^{\prime}}{x})$

Note: This is the first part of an undergraduate applications course in differential equations. We were taught to solve second order non-linear equations by doing tricks such as substituting $u = y^{\prime}$, factoring, and then dealing with the two cases that result.

2. Relevant equations

None.

3. The attempt at a solution

I tried $u := y^{\prime}$, so $u^{\prime} = y^{\prime \prime}$, and so substitution gives

$x u^{\prime} = u \log(\frac{u}{x})$

but I can't seem to be able to do anything with that.

I also I noticed that

$y^{\prime \prime} = y^{\prime} x^{-1} \log(y^{\prime} x^{-1})$

which I thought was interesting. But I couldn't really do anything with it.

Thank you.

Last edited: Dec 13, 2013
2. Dec 13, 2013

### SteamKing

Staff Emeritus
How about log (y'/x) = log (y') - log (x)?

3. Dec 13, 2013

### Dick

You want to try and find a variable change that will make it separable. Since you have u/x in the log, substituting xv=u might be a good thing to try.

4. Dec 13, 2013

### Staff: Mentor

You have $x u^{\prime} = u \log(\frac{u}{x})$. This can be rewritten as
$$\frac{d \ln u}{d\ln x}=\ln(u)-\ln(x)$$
Let ln(u) = w and ln(x) = z

5. Dec 13, 2013

### 5hassay

thanks for the help.

OK, that was productive. Here's what I did:

$v := \frac{y^{\prime}}{x} \Longrightarrow v^{\prime} = \frac{x y^{\prime \prime} - y^{\prime}}{x^2} \Longrightarrow x^2 v^{\prime} + y^{\prime} = x y^{\prime \prime}$

Substituting,

$x^2 v^{\prime} + y^{\prime} = y^{\prime} \log(v) \Longrightarrow x v^{\prime} + v = v \log(v)$
$\Longrightarrow \frac{dv}{dx} = \frac{v \log(v) - v}{x} \Longrightarrow (v - v \log(v)) dx + x dv = 0$

$M := v - v \log(v)$
$N := x$

$M_v = 1 - \log(v) - 1 = -\log(v) \neq 1 = N_x$

so the equation is not exact. But

$\displaystyle{\frac{N_x - M_v}{M} = \frac{1 + log(v)}{v(1 - \log(v))} \neq f(v)}$
$\displaystyle{\frac{M_v - N_x}{N} = -\frac{\log(v) + 1}{x} \neq f(x)}$
$\displaystyle{\frac{N_x - M_v}{x M - v N} = -\frac{1 + \log(v)}{x v \log(v)} \neq f(xv)}$

so I am stuck (one of those works when we are given a separable equation in my course).

6. Dec 13, 2013

### Dick

xv'+v=vlog(v) is already separable. Just separate it and go from there.

7. Dec 13, 2013

### 5hassay

thanks for the help.

I don't know how to evaluate

$\frac{d \log(u)}{d \log(x)}$

I understand that I would be differentiating $\log(u)$ as a function of $\log(x)$, but I don't see how that is a function of $\log(x)$.

8. Dec 13, 2013

### 5hassay

true. OK, then

$\displaystyle{\frac{dv}{dx} = \frac{v \log(v) - v}{x} \Longrightarrow \int{\frac{1}{v} \cdot \frac{1}{\log(v) - 1}} dv = \int{\frac{1}{x}}} dx$
$\displaystyle{\Longrightarrow \log(\log(v) - 1) = \log(x) + C \Longrightarrow \log(v) - 1 = C x \Longrightarrow v = \frac{y^{\prime}}{x} = e^{C x + 1}}$
$\displaystyle{\Longrightarrow y = \int{x e^{C x + 1}} dx = x \frac{1}{C} e^{C x + 1} - \frac{1}{C} \int{e^{C x + 1}} dx = \frac{1}{C^2} e^{C x + 1} (C x - 1) + D}$

where I just re-used $C$ for $e^C$.

I haven't verified it yet by substituting

Last edited: Dec 13, 2013
9. Dec 13, 2013

### Dick

Looks fine for v, but I think you dropped a C when you were integrating by parts.

10. Dec 13, 2013

### Staff: Mentor

If u is a function of x, then it is a function of ln(x).

11. Dec 13, 2013

### 5hassay

awesome. Also, I also think you are correct about the missing $C$, thank you--I shall update that post. Thank you for your help!

12. Dec 13, 2013

### Dick

Very welcome! If you want some extra practice you should try doing it Chestermiller's way as well, it's very nice if you don't mind keeping track of a few extra substitution variables.