First-order nonlinear ordinary differential equation
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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$\begingroup$$x\dfrac{dy}{dx}=y+x\dfrac{e^x}{e^y}$
$x\dfrac{dy}{dx}=y+xe^{x-y}$
Let $u=y-x$ ,
Then $y=u+x$
$\dfrac{dy}{dx}=\dfrac{du}{dx}+1$
$\therefore x\left(\dfrac{du}{dx}+1\right)=u+x+xe^{-u}$
$x\dfrac{du}{dx}+x=u+x+xe^{-u}$
$x\dfrac{du}{dx}=xe^{-u}+u$
$(xe^{-u}+u)\dfrac{dx}{du}=x$
Let $v=x+ue^u$ ,
Then $x=v-ue^u$
$\dfrac{dx}{du}=\dfrac{dv}{du}-(u+1)e^u$
$\therefore e^{-u}v\left(\dfrac{dv}{du}-(u+1)e^u\right)=v-ue^u$
$e^{-u}v\dfrac{dv}{du}-(u+1)v=v-ue^u$
$e^{-u}v\dfrac{dv}{du}=(u+2)v-ue^u$
$v\dfrac{dv}{du}=(u+2)e^uv-ue^{2u}$
This belongs to an Abel equation of the second kind.
Let $t=(u+1)e^u$ ,
Then $u=W(et)-1$
$\dfrac{dv}{du}=\dfrac{dv}{dt}\dfrac{dt}{du}=(u+2)e^u\dfrac{dv}{dt}$
$\therefore(u+2)e^uv\dfrac{dv}{dt}=(u+2)e^uv-ue^{2u}$
$v\dfrac{dv}{dt}=v-\dfrac{ue^u}{u+2}$
$v\dfrac{dv}{dt}-v=-\dfrac{t(W(et)-1)}{W(et)(W(et)+1)}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in
$\endgroup$ $\begingroup$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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