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How to solve this differential equation:

$$x\frac{dy}{dx} = y + x\frac{e^x}{e^y}?$$

I tried to rearrange the equation to the form $f\left(\frac{y}{x}\right)$ but I couldn't thus I couldn't use $v = \frac{y}{x}$ to solve it.

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This differential equation is not homogeneous (and so you can't rearrange it in the form $f(y/x)$.) –  Lukas Geyer Oct 6 '12 at 16:15
    
what is it then? how can you solve it? –  David Hoffman Oct 6 '12 at 19:21
    
If you assume $y>0$, something like $u=\log y$ may help. I haven't checked, so it may also be a waste of time. –  Daryl Oct 6 '12 at 20:18
    
The clue is that $\dfrac{e^x}{e^y}$ should rewrite to $e^{x-y}$ and thus we get the idea that let $u=x-y$ or let $u=y-x$ will convert the ODE whose the terms are not contain any composite functions. –  doraemonpaul Oct 7 '12 at 1:31
    
The substitution $u = e^y$ leads to $$\frac{du}{dx} = \frac{dy}{dx}e^y = \frac yx e^y + e^x = \frac ux \log u + e^x.$$ May be this could lead somewhere. –  Sam Oct 7 '12 at 2:06
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2 Answers

$x\dfrac{dy}{dx}=y+x\dfrac{e^x}{e^y}$

$x\dfrac{dy}{dx}=y+xe^{x-y}$

Let $u=y-x$ ,

Then $y=u+x$

$\dfrac{dy}{dx}=\dfrac{du}{dx}+1$

$\therefore x\left(\dfrac{du}{dx}+1\right)=u+x+xe^{-u}$

$x\dfrac{du}{dx}+x=u+x+xe^{-u}$

$x\dfrac{du}{dx}=xe^{-u}+u$

$(xe^{-u}+u)\dfrac{dx}{du}=x$

This belongs to an Abel equation of the second kind

Let $v=x+ue^u$ ,

Then $x=v-ue^u$

$\dfrac{dx}{du}=\dfrac{dv}{du}-(u+1)e^u$

$\therefore e^{-u}v\left(\dfrac{dv}{du}-(u+1)e^u\right)=v-ue^u$

$e^{-u}v\dfrac{dv}{du}-(u+1)v=v-ue^u$

$e^{-u}v\dfrac{dv}{du}=(u+2)v-ue^u$

$v\dfrac{dv}{du}=(u+2)e^uv-ue^{2u}$

In fact, all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $v=\dfrac{1}{w}$ ,

Then $\dfrac{dv}{du}=-\dfrac{1}{w^2}\dfrac{dw}{du}$

$\therefore-\dfrac{1}{w^3}\dfrac{dw}{du}=\dfrac{(u+2)e^u}{w}-ue^{2u}$

$\dfrac{dw}{du}=ue^{2u}w^3-(u+2)e^uw^2$

Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2

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Rewrite this equation in the form:

\begin{equation} M(x,y)dx + N(x,y)dy = (xe^x+ye^y)dx-xe^ydy = 0 \end{equation}

Both $\frac{\partial M}{\partial y} = e^y(1+y)$ and $\frac{\partial N}{\partial x}=-e^y$ are depend on $y$ only. In this case some multiplier $\mu(x,y)$ can be simply found so that

\begin{equation} \dfrac{\partial (\mu M)}{\partial y} = 0,\quad \dfrac{\partial (\mu N)}{\partial x}=0 \end{equation}

and you get exact differential equation in form $du(x,y)=0$.

For $\mu$ depending only on $y$ we have $d\ln\mu=\dfrac{dy}{M}\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)$. In our case:

\begin{equation} \mu(y) = \exp\left(-\int\dfrac{e^y(y+2)}{xe^x+ye^y}dy\right) \end{equation} Solution of our DE is: \begin{equation} \int \mu M dx + \int \mu N dy = C \end{equation}

I do not substitute $\mu$ in the last equation because $\mu$ as I see cannot be expressed in elementary functions and complete solution will be cumbersome.

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