Laplacian operator I need to find A,$\alpha$,C $\in \mathbb{R}$ so that $u:\mathbb{R}^n \to \mathbb{R}$ of the form $u(x) = A||x||^\alpha + C $
satisfies $\Delta u(x) = ||x||^2$ and $u(0) = -1$.
This question gets very messy. I end up with terms that can potentially cancel but they both depend on A and $\alpha$ and i cannot satisfy what is required. 
 A: Let $v \colon \def\R{\mathbb R}\R^n \to \R$ be defined by $\def\norm#1{\left\lVert#1\right\rVert}v(x) = \norm x^\alpha = (\sum_i x_i^2)^{\alpha/2}$ we have for $1 \le j \le n$
\begin{align*}
  \partial_j v(x) &= \frac{\alpha}2\cdot\left(\sum_i x_i^2\right)^{\alpha/2-1}\cdot 2x_j\\
    &= \alpha x_j \cdot \left(\sum_i x_i^2\right)^{\alpha/2-1}\\
  \partial_j^2 v(x) &= \alpha \left(\sum_i x_i^2\right)^{\alpha/2 - 1}
         + \alpha x_j^2 (\alpha - 2)\left(\sum_i x_i^2\right)^{\alpha/2 - 2} 
\end{align*}
Hence
$$ \Delta v(x) = \sum_j \partial_j^2 v(x) = \alpha\norm{x}^{\alpha - 2} + \alpha(\alpha - 2)\norm x^2 \norm x^{\alpha - 4} = \alpha(\alpha - 1)\norm x^{\alpha - 2}$$
Hence, as $u = Av + C$, we have
$$ \Delta u(x) = A\alpha(\alpha - 1) \norm x^{\alpha - 2}$$
So $\alpha = 2$, that is $\alpha(\alpha - 1) = 12$, which gives $A = \frac 1{12}$, and as $\norm 0 = 0$ we must have $C = -1$.
A: This is a straightforward exercise in differentiation:
$$\Delta u = \nabla \cdot \nabla u = \nabla \cdot A\alpha\|x\|^{\alpha-1}\frac{x}{\|x\|}=\nabla \cdot A\alpha\|x\|^{\alpha-2}x$$
Now by the chain rule
\begin{align*}
\Delta u &= A\alpha (\alpha-2)\|x\|^{\alpha-3}\frac{x}{\|x\|}\cdot x + A\alpha \|x\|^{\alpha-2}n\\
&= A\alpha(\alpha-2)\|x\|^{\alpha-2} + A\alpha n\|x\|^{\alpha-2}\\
&= A\alpha(\alpha+n-2)\|x\|^{\alpha-2}.
\end{align*}
To get $\|x\|^2$ we meed $\alpha=4$, and $4A(2+n)=1$. Therefore
$$A = \frac{1}{8+4n}.$$
Lastly we easily get $u(0)=-1 \Rightarrow C=-1.$
