I considered a generalization of the ODE from this question. The generalization depends on $4$ parameters and has the form:
$$y=axp+bx^rp^s \tag{1} $$ $$ \frac{dy}{dx}=p \tag{2}$$
Where $a,b,r,s \in \mathbb{R}$ with some appropriate conditions if needed.
The equation is not separable in the general case.
The most convenient way would be to reduce it to Clairaut's equation (as was done for the particular case in this answer), but I don't see how it could be done for general $r,s$, at least not with the same kind of substitution.
It also doesn't fit the description of Lagrange differential equation, because of the $x^r$ term.
The only method of solution I figured out so far is the one I attempted in the linked question, based on reducing the problem to exact differential equation.
Differentiate (1) w.r.t. $p$:
$$\frac{dy}{dp}=(ap+brx^{r-1}p^s) \frac{dx}{dp}+(ax+bsx^rp^{s-1})$$
Multiply by $\frac{dp}{dx}$ and substitute in (2):
$$p=(ap+brx^{r-1}p^s)+(ax+bsx^rp^{s-1})\frac{dp}{dx}$$
Now the equation has the form:
$$((a-1)p+brx^{r-1}p^s)dx+(ax+bsx^rp^{s-1})dp=0$$
Introducing the integrating factor in the form $\mu=x^\alpha p^\beta$ we can find:
$$\alpha=-\frac{r}{a(r-s)+s} \qquad \beta=-\frac{s}{a(r-s)+s}$$
And after integrating the implicit solution looks like this:
$$\frac{a}{\beta+1} x^{\alpha+1} p^{\beta+1}+\frac{bs}{\beta+s} x^{\alpha+r} p^{\beta+s}=C_1 \tag{3}$$
Together (1) and (3) give the solution in parametric form.
The equation (3) can be solved by Newton-Raphson for example.
However, there's likely a better method of solving this equation, which is why I am asking:
(a) Does (1), (2) reduce to some well known equation for general values of parameters?
(b) Is there a better way of solving it, again, for the general case? Preferably in explicit form, unlike the way I did it.