= Solution
{c}
Let $V_h$ be the continuous piecewise linear space including both endpoint <hat functions>. Robin conditions are natural conditions and do not restrict the endpoint values of this space. The <finite element> problem is $u_h\in V_h$, $a(u_h,v_h)=\ell(v_h)$ for all $v_h\in V_h$. <coercivity> restricted to $V_h$ gives <existence> and <uniqueness>.
Expanding $u_h=\sum_{j=0}^NU_j\phi_j$ gives $\mathsf A U=F$, where
$$
\boxed{\mathsf A_{ij}=\int_a^b(\phi_j'\phi_i'+x^2\phi_j\phi_i)\,dx+\delta_{i0}\delta_{j0}+\delta_{iN}\delta_{jN},\quad
F_i=\int_a^bf\phi_i\,dx+A\delta_{i0}+B\delta_{iN}.}
$$
The stiffness contribution has diagonal $2/h$ at interior nodes, $1/h$ at endpoints, and adjacent entries $-1/h$. For completeness, the weighted mass contribution has interior diagonal $2hx_i^2/3+h^3/15$, left endpoint diagonal $ha^2/3+ah^2/6+h^3/30$, right endpoint diagonal $hb^2/3-bh^2/6+h^3/30$, and adjacent entry $hx_i^2/6+h^2x_i/6+h^3/20$. All more distant entries vanish. The Robin contribution adds one to each endpoint diagonal.
Symmetry follows from symmetry of the <bilinear form>. If $V\neq0$, nodal independence makes $v_h=\sum_iV_i\phi_i$ nonzero, and $V^T\mathsf AV=a(v_h,v_h)\geq\gamma\|v_h\|_{H^1}^2>0$. Thus the <matrix> is \b[symmetric positive definite].
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