I am working on a differential equation $\frac{dy}{dx} = e^{\frac{x}{y}} + \frac{y}{x}$. My intuition is to perform a change of variable by letting $u(x) = \frac{y}{x}$
However, I am having trouble figuring out what $\frac{du}{dx}$ would be. How do I evaluate that?
Also, out of curiosity, what would $\frac{du}{dx}$ be if instead I substituted $u(x) = \frac{x}{y}$
Based on the proposed substitution, the proposed ODE becomes: \begin{align*} y' = \exp\left(\frac{x}{y}\right) + \frac{y}{x} & \Longleftrightarrow \left(\frac{x}{u}\right)' = \exp(u) + \frac{1}{u}\\\\ & \Longleftrightarrow \frac{u - xu'}{u^{2}} = \exp(u) + \frac{1}{u}\\\\ & \Longleftrightarrow u - xu' = u^{2}\exp(u) + u\\\\ & \Longleftrightarrow \frac{u'}{u^{2}\exp(u)} = -\frac{1}{x} \end{align*}
I am afraid there is no solution in terms of elementary functions.
However, I am having trouble figuring out what $\tfrac{\mathrm du}{\mathrm dx}$ would be. How do I evaluate that?
Just apply the product rule (or quotient rule).
$\begin{align}u(x) &= \dfrac{y(x)}{x}\\u'(x) &= \dfrac{xy'(x)-y(x)}{x^2}\\&=\dfrac{x(\mathrm e^{x/y(x)}+y(x)/x)-y(x)}{x^2}\\\dfrac{\mathrm d u(x)}{\mathrm d x}&=\dfrac{\mathrm e^{1/u(x)}}{x}\end{align}$
Also, out of curiosity, what would $\tfrac{\mathrm du}{\mathrm dx}$ be if instead I substituted $u(x)=x/y$
Now that you know what to do, do that.
If you let first $$e^{\frac x y}=z \implies y=\frac{x}{\log (z)}$$ the equation becomes $$x\,z'+z^2\log(z)^2=0$$ Switch variables $$\frac x {x'}+z^2\log(z)^2=0\qquad \implies \qquad \frac {x'}{x}=-\frac 1 {z^2\log(z)^2}$$
$$\log(x)+C=-\int \frac {dz} {z^2\log(z)^2}$$ One integration by parts to get $$\log(x)+C=\text{Ei}(-\log (z))+\frac{1}{z \log(z)}$$ which is impossible to inverse.
You don't need $\frac{du}{dx}$ but $\frac{dy}{dx}$ to be substituted in the DE.
$$u:=\frac yx\to y=xu\to\frac{dy}{dx}=u+x\frac{du}{dx}.$$
The equation turns to
$$u+x\frac{du}{dx}=e^{1/u}+u$$
which is separable.
$$e^{-1/u}du=\frac{dx}{x}$$
and
$$\int e^{-1/u}du=\log(Cx).$$
