First-order non-linear differential equation and transcendental equation

Question

I'm trying to solve this differential equation : $$ \frac{dy}{dx}= \frac{-2 y^3}{(y+1)^2(y+2)^2} $$ with the boundary condition $y(x_0)=x_0$, $x>0$, and $y(x)$ being a positive function. The integration of the equation is straightforward, however after integration, one gets a transcendental equation of the form $$a y(x)+by^2(x)+c \log(y)+\frac{d}{y}+\frac{e}{y^2}+ g(x_0)= z (x-x_0)$$ where $a,b,c , d, e,z $ are constants, and $g(x_0) $ is a function of $x_0$. I tried to solve it with Lagrange inversion theorem, however due to the non triviality of the LHS, the computation of the $n$'th derivative is very complicated, is there any other way to solve it ?

Answer 1

This is a purely numerical problem which does not require much.

What you obtained is $$x=-\frac{y^2}{2}-6 y+\frac{12}{y}+\frac{2}{y^2}-13 \log (y)+ C \tag1$$ and the initial conditions make $$C=\frac{y_0^2}{2}+6 y_0-\frac{12}{y_0}-\frac{2}{y_0^2}+13 \log (y_0)+x_0\tag2$$ For a given value of $x$, you want to compute $y$ that is to say to find the zero of function $$f(y)=-\frac{y^2}{2}-6 y+\frac{12}{y}+\frac{2}{y^2}-13 \log (y)+(C-x)\tag3$$

$f(y)$ is quite nice since its derivative never cancels and the solution is unique which means that any equation solver will do the job, the only potential problem being to find a "reasonable" starting guess. To get it, I should use a marching process (or a golden search) until I find a value $y_*$ such that $f(y_*)\, f''(y_*) >0$ in order, by Darboux theorem, to avoid any overshoot of the solution if Newton method (the simplest one) has to be used.

Notice that, $y$ being supposed to be positive (as in must be because of the logarithmic term) $$f''(y)=\frac{12}{y^4}+\frac{24}{y^3}+\frac{13}{y^2}-1$$ is positive for $0< y <\frac{1}{2} \left(3+\sqrt{33}\right)$ and negative if $y >\frac{1}{2} \left(3+\sqrt{33}\right)$.