Solve a system of non-linear eqations: $x\ln(x^2+y)+\frac{1}{x^2+y}=0$ and $y\ln(x^2+y)+\frac{2x}{x^2+y}=0$

Question

Solve a system of non-linear eqations: $$x\ln(x^2+y)+\frac{1}{x^2+y}=0$$ $$y\ln(x^2+y)+\frac{2x}{x^2+y}=0$$

My idea was to write the first equation like so: $$-x\ln(x^2+y)=\frac{1}{x^2+y}$$ and so from the second equation we would get: $$y\ln(x^2+y)-2x^2\ln(x^2+y)=0$$ $$\ln(x^2+y)(y-2x^2)=0$$ which means that $x^2+y = 1$ or $2x^2=y$

but that would mean I can choose any 2 values for $x$ and $y$ as long as one of the two last equation is satisfied and that would be the solution, which is not true. Could anybody please explain me where my mistakes are or whether I've done something I shouldn't have done?

And also, how should I approach this problem?

Thanks for help.

Answer 1

If equations 1 and 2 are true, then equation 3 is true, but equation 3 might also be true when neither 1 nor 2 is. So you can't just find all solutions to equation 3 and stop.

If $x^2+y=1$, you can use that to simplify equations 1 and 2. That gives you one set of solutions. If $y=2x^2$, equations 1 and 2 simplify in a different way, and that gives the other set of solutions.

Answer 2

$$\ln(x^2+y)=-\frac{1}{x(x^2+y)}$$ $$\ln(x^2+y)=-\frac{2x}{y(x^2+y)}$$ so $$y=2x^2$$ substituting this we get $$\ln(3x^2)+\frac{1}{3x^3}=0$$ $$\ln(3x^2)+\frac{1}{3x^3}=0$$ By a numerical method we get $$x\approx -0.79974035684145550579$$