1. ## First-order homogeneous DE

Hi,
I got the next DE on a lesson 2 weeks ago. After I wasn't able to find any solution my teacher assigned it to be my homework. Thrice. Last time on a lecture he said it will be my prerequisite for exam (not quite sure if it's in conflict with forum rules...). But because of it being my homework three times I can't hand over any other homework and I lose opportunities for activity points... That 'simple' DE is:

$y'=e^{\frac{x}{y}}+e^{\frac{y}{x}}$

The facts are:

(1)Simple techniques we've learnt up to now doesn't seem to work. I am unable to efficiently separate variables, the substitution $y=xu$ leads to probably unsolvable(?) integral. Also I should mention that we started this DE topic on lectures only recently.

(2)I groped a bit in an accessible literature recommended by teacher and luckily I figured out where does the DE comes from (Elias J., Horvath J., Kajan J.,- Zbierka uloh z vyssej matematiky). It's a problem no.1194, but result listed at the end of the book isn't correct ( $ln|Cx|=-e^{-\frac{y}{x}}$ ). Neither problem no.1193 nor no.1195 have correctly listed solutions, but the former one matches correct result for problem no.1193. Given solutions at the end slightly resemble those correct ones, like if they were incorrectly rewritten from somewhere else...

(3)Wolfram can't solve neither that DE nor the derived integral.

(4)On the last lesson I told my teacher the book has incorrect results and that my DE possibly can't be solved but he insisted on it can be done. So now it's my homework for a third time

Because it's my homework I just ask for hints what technique of DE-solving should I use, not the entire flat solution... Also it would be of great appeasment to me if anyone could confirm it can be really solved.

2. ## Re: First-order homogeneous DE

Originally Posted by Iustus
Hi,
I got the next DE on a lesson 2 weeks ago. After I wasn't able to find any solution my teacher assigned it to be my homework. Thrice. Last time on a lecture he said it will be my prerequisite for exam (not quite sure if it's in conflict with forum rules...). But because of it being my homework three times I can't hand over any other homework and I lose opportunities for activity points... That 'simple' DE is:

$y'=e^{\frac{x}{y}}+e^{\frac{y}{x}}$

The facts are:

(1)Simple techniques we've learnt up to now doesn't seem to work. I am unable to efficiently separate variables, the substitution $y=xu$ leads to probably unsolvable(?) integral. Also I should mention that we started this DE topic on lectures only recently.

(2)I groped a bit in an accessible literature recommended by teacher and luckily I figured out where does the DE comes from (Elias J., Horvath J., Kajan J.,- Zbierka uloh z vyssej matematiky). It's a problem no.1194, but result listed at the end of the book isn't correct ( $ln|Cx|=-e^{-\frac{y}{x}}$ ). Neither problem no.1193 nor no.1195 have correctly listed solutions, but the former one matches correct result for problem no.1193. Given solutions at the end slightly resemble those correct ones, like if they were incorrectly rewritten from somewhere else...

(3)Wolfram can't solve neither that DE nor the derived integral.

(4)On the last lesson I told my teacher the book has incorrect results and that my DE possibly can't be solved but he insisted on it can be done. So now it's my homework for a third time

Because it's my homework I just ask for hints what technique of DE-solving should I use, not the entire flat solution... Also it would be of great appeasment to me if anyone could confirm it can be really solved.
\displaystyle \begin{align*} \frac{dy}{dx} &= e^{\frac{x}{y}} + e^{\frac{y}{x}} \\ \frac{dy}{dx} &= e^{\left(\frac{y}{x}\right)^{-1}} + e^{\frac{y}{x}} \end{align*}

Now make the substitution

\displaystyle \begin{align*} u &= \frac{y}{x} \\ y &= u\,x \\ \frac{dy}{dx} &= u + x\,\frac{du}{dx} \end{align*}

and the integral becomes

\displaystyle \begin{align*} \frac{dy}{dx} &= e^{\left(\frac{y}{x}\right)^{-1}} + e^{\frac{y}{x}} \\ u + x\,\frac{du}{dx} &= e^{u^{-1}} + e^u \\ x\,\frac{du}{dx} &= e^{u^{-1}} + e^u - u \\ \frac{1}{e^{u^{-1}} + e^u - u}\,\frac{du}{dx} &= \frac{1}{x} \end{align*}

I'm afraid I don't know how to continue...

3. ## Re: First-order homogeneous DE

Originally Posted by Prove It
<etc>
$\displaystyle \frac{1}{e^{u^{-1}} + e^u - u}\,\frac{du}{dx}= \frac{1}{x}$
I'm afraid I don't know how to continue...
Indeed that's the integral I mentioned before. Could anyone else confirm, that this task is inappropriately hard? I also tried to get rid of those exponentials by searching for such $f$, $g$ that $\frac{df}{dx}=e^{\frac{x}{f+g}}$ and $\frac{dg}{dx}=e^{\frac{f+g}{x}}$. But it's lead me to this not very appealing system of equations:
\begin{align*}\\ (f+g)^2\frac{d^2f}{dx^2}&=(f+g)\frac{df}{dx}-(\frac{df}{dx}+\frac{dg}{dx})\frac{df}{dx}x\\ \frac{d^2g}{dx^2}x^2&=\frac{dg}{dx}(\frac{df}{dx}+ \frac{dg}{dx})x-(f+g)\frac{dg}{dx}\\ \end{align*}

Positive fact is there are no exponentials anymore. But I don't know how to solve this non-linear system of equations of higher degree. Does anyone have a knowledge of some useful techniques?
It's slightly surprising to me that so simply looking DE wouldn't have a generally known approach.

4. ## Re: First-order homogeneous DE

OK, I asked one friend, who's occupation relates to solving DE, to look at it and his opinion is that it could be solved only with numeric methods. But my teacher likes to tease people and he demands solution (but I suspect him he just wants to keep me busy so he wouldn't have to check my homeworks- he is unbelievably lazy...). So now my new question is if anyone knows how to prove that a solution of this DE cannot be expressed in a form of finite term. If I remeber right one should somehow convert it to be able to use Galois groups or something like that. But I never studied this stuff too much Any hints? Or should this question be put in a different forum?