I am trying to solve the following ODE with boundary conditions in the set $C^1[-2,1]$, with $u(-2)=-1/2$ and $u(1)=1$. that is I must tell if there exist a solution $u$ in $C^1[-2,1]$
$$0= \frac{d}{dx}(x^2u'(x))\\ x^2u'(x)=c$$
I think I can't just put $x^2 $to the right in the denominator and and integrate because $0$ is in the domain, how does one proceed in this cases? And when is that the solution is admissible?
Assume that such $u$ exists. As you wrote, $\frac d{dx}(x^2u'(x)) = 0$ implies $x^2u'(x) = c$, for all $x\in[-2,1]$. Plugging in $x=0$ gives us $c = 0$, so we have $x^2u'(x) = 0$ for all $x\in [-2,1]$, so $u'(x) = 0$ for all $x\neq 0$, but due to continuity, $u'(x) = 0$ for all $x\in [-2,1]$. Thus, $u(x)$ is constant function. However, $u(-2)\neq u(1)$. Contradiction.