Solving an electronic circuit I have come up with the following non-linear equation system, where to ease the reading, I will work with the coefficients $a_i$ and $b_i$ instead of circuit parameters, besides I do know that both variables $x$ and $y$ are independant from each other.
$$\tag{1} \left\{ \begin{array}{c} y^2*x^2*a_1 + y^2*x^3*a_2 - y*x*a_3+x^2*a_4+x^3a_5 = 0 \\ y^2*x^2*b_1 + y^2*x^3*b_2 + x^3*b_3 + x^2*b_4+x*b5 = 0 \end{array} \right. $$
I see that the trivial solution is $x=0$, so if I simplify both by dividing by $x$ and I re-arrange it a bit, I get $(2)$ system
$$\tag{2} \left\{ \begin{array}{c} y^2*x*a_1 + y^2*x^2*a_2 + x^2*a_5 + x*a_4 - y*a_3 = 0 \\ y^2*x*b_1 + y^2*x^2*b_2 + x^3*b_3 + x^2*b_4+b5 = 0 \end{array} \right. $$
So, I was wondering what would be the best approach to solve the equation system $(2)$? My first thought was an iterative process like trust region. But I was wondering what other approaches could be tackled. Any suggestions?
Thanks
Edit: I see that I made a mistake when simplifying in $x$ on the second equaion of $(1)$, the system should be $$\tag{2} \left\{ \begin{array}{c} y^2*x*a_1 + y^2*x^2*a_2 + x^2*a_5 + x*a_4 - y*a_3 = 0 \\ y^2*x*b_1 + y^2*x^2*b_2 + x^2*b_3 + x*b_4+b5 = 0 \end{array} \right. $$
If you discard the trivial $x=0$,from the equations you wrote, write them as $$y^2 x a_1 + y^2 x^2 a_2 + x^2 a_5 + x a_4 = y a_3 \tag 1$$ $$y^2 x b_1 + y^2 x^2 b_2 + x^3 b_3 + x^2 b_4+b_5 = 0\tag 2$$ Extract $y^2$ from $(2)$ to get $$y^2=-\frac{x^3 b_3 + x^2 b_4+b_5 } {x b_1 + x^2 b_2 }\tag 3$$ and rewrite $(1)$ as $$(y^2 x a_1 + y^2 x^2 a_2 + x^2 a_5 + x a_4)^2- y^2 a^2_3=0 \tag 4$$ Plug $(3)$ in $(4)$.
Assuming that $(b_1+x b_2)\neq 0$, you end with a polynomial of degree $4$ in $x$
$$a_3^2 b_5+x \left(a_3^2 b_4-a_1 b_5\right)+x^2 \left(a_3^2 b_3+a_4 b_1-a_1 b_4-a_2 b_5\right)+$$ $$x^3 (a_5 b_1+a_4 b_2-a_1 b_3-a_2 b_4)+x^4 (a_5 b_2-a_2 b_3)=0$$ Compute the roots $x_i$; but the squaring process introduced extraneous solutions. You need to discard all of those which would not make $y^2$ (as given by $(3)$) positive.
So, you have the appropriate $x$'s and the corresponding $y$'s.
Edit
After your mofifications, consider you last equations as $$y^2 x a_1 + y^2 x^2 a_2 + x^2 a_5 + x a_4 = y a_3 \tag 5 $$ $$y^2 x b_1 + y^2 x^2*b_2 + x^2 b_3 + x b_4+b_5 = 0 \tag 6$$ and, for time being, let $z=y^2$ to make $$z x a_1 + z x^2 a_2 + x^2 a_5 + x a_4 = y a_3 \tag 7 $$ $$z x b_1 + z x^2*b_2 + x^2 b_3 + x b_4+b_5 = 0 \tag 8$$ From $(8)$ $$z=-\frac{b_3 x^2+b_4 x+b_5}{x(b_1+b_2 x)}\tag 9$$ which was already $(3)$.
Now,we have (rewriting $(7)$) $$zx( a_1+ a_2 x)+x(a_4+a_5x)=a_3 y$$ and squaring $$\color{blue}{x^2\big(z( a_1+ a_2 x)+(a_4+a_5x)\big)^2=a_2^2 z} \tag {10}$$ where $z$ is given by $(9)$.
I strongly recommend to avoid any expansion of $(10)$; this would lead to an almost impraticable polynomial equation in $x$ of very high degree (probably $7$) with messy coefficients.
Keep the equation as it is and apply a solver