# Approximation of a linear function with polynomials of degree 1

If I have the following problem $$-\mu u'' + u' = 1$$

with boundary conditions $u(0) = u'(1) = 1$ in the interval $\Omega = (0,1)$. The exact solution is $$u(x) = x + 1$$

Will the FEM approximation with piecewise linear functions ($\mathbb{P}^1$) have a very small (or no) error? If so, how to think about it to have an image in my mind?

By Céa's lemma, the finite element discretization error is bounded by the best approximation error: $$\|u-u_h\|_{H^1(0,1)} \leq C \min_{v_h\in \mathbb{P}^1} \|u-v_h\|_{H^1(0,1)},$$ where $u\in H^1(0,1)$ is the exact solution, $u_h\in\mathbb{P}^1$ is the finite element solution. Since in your case the exact solution $u$ is actually in $\mathbb{P}^1$, the minimum is attained for $v_h = u$, and so the error is zero.
EDIT: The key idea is the following. You are looking for a weak solution, i.e., $u\in V$ such that $$a(u,v) = f(v) \quad\text{for all} \quad v\in V.\tag{1}$$ (In your case, $a(u,v) = \int_0^1 \mu u'v' + \int_0^1 u' v$ and $f(v) = \int_0^1 v + v(1)$.) The finite element approximation consists in choosing $V_h\subset V$ (in your case, $V_h = \mathbb{P}^1$) and finding $u_h\in V_h$ such that $$a(u_h,v_h) = f(v_h) \quad\text{for all} \quad v_h\in V_h.\tag{2}$$ Setting $v=v_h\in V$ in $(1)$ and subtracting $(2)$ gives $$a(u-u_h,v_h) = 0 \quad \text{for all}\quad v_h\in V_h.$$ This tells you that your error $u-u_h$ is orthogonal to $V_h$ with respect to the "inner product" defined by $a$, or that $u_h$ is the orthogonal projection of $u$ on $V_h$ (again, with respect to $a$). And orthogonal projections on a subspace are characterized by best approximation properties with respect to appropriate norms (in this case, the $H^1$ norm since $a$ happens to be coercive in this norm).