Find initial condition so that ODE has multiple solutions

Question

Given Cauchy problem $$y'=\dfrac{x^2}{y(1+x^3)}, \;y(x_0)=y_0.$$ Is it possible to find $(x_0,y_0)$ such that given Cauhcy problem has multiple solutions? It is obvious that equation is with separable variables and I got solution $y^2=\dfrac{2}{3}\ln |c(1+x^3)|$. If we use theorem of existence and uniqueness I conclude equation have unique solution for $(x_0,y_0)\in \big[(\infty,-1)\cup(-1,\infty)\big]\times\big[(\infty,0)\cup(0,\infty)\big]$. For any other $(x_0,y_0)$ ODE is not defined. Is my conclusion wrong?

Answer 1

Multiply by $y$: $$yy'=\frac{x^2}{1+x^3}$$Integrate both sides by $x$: $$\frac{y^2}{2}=\frac{\ln(|x^3+1|)}{3}+ C$$The integral of $yy'$ is obtained by making a $u$-substitution. The other integral is obtained by making a $u$-substitution of $x^3$. Now solve for $y$: $$y=\pm\sqrt{\frac{2}{3}\ln(|x^3+1|)+C}$$So there are at least two solutions.