Weird looking ODE solution: How to verify this is indeed a solution? I will try to express my question using an example.
Consider this homogeneous ODE: $y' = \frac{y-x}{x+y }$
Its solution is: $\boxed{\frac12 \log\left( \frac{y^2(x)}{x^2} +1\right) - \log(x) + \arctan\left( \frac{y(x)}{x}\right) = c} \quad  c\in \mathbb{R}$

As far as I am understanding solving a differential equation means finding a $y(x)$ that satisfies the differential equation.
But this particular solution, cannot be expressed in terms of $y(x)$ (at least I can't).

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*If the previous argument is true how could one verify the solution, given the fact that there is no actual $y$ to plug into the ODE?

*If the previous argument is false, how could the solution be expressed in terms of $y$?

P.S: A previous question I've asked is How do $\arctan$ and $\ln$ relate?. Given the fact that the solution only contains those two functions, I suspect that it may be helpful to note this down. Could there be some complex analysis involved?

Use the substitution $y = xv$
$$v+xv' = \frac{x(v-1)}{x(v+1)} \implies xv' = -\frac{v^2+1}{v+1}$$
which is separable
$$\int\frac{v}{v^2+1} + \frac{1}{v^2+1} \:dv = -\int\frac{dx}{x} \implies \frac{1}{2}\log\left(v^2+1\right)+ \tan^{-1}v = -\log|x| + C$$
which means the solution can be given as
$$\frac{1}{2}\log\left(x^2v^2+x^2\right) + \tan^{-1}v = C \implies \log(y^2+x^2) + 2 \tan^{-1}\left(\frac{y}{x}\right) = C$$
