The fundamental lemma of the calculus of variations is often presented as: If $M(x) \in C[a,b]$ such that $\int_{a}^{b}{M(x)\eta(x)} = 0 ~~\forall\eta\in C^1[a,b],\eta(a)=\eta(b)=0$, then $M(x)=0$ for all $x\in[a,b]$.
I am trying to prove the specific higher order case when $M$ is multiplied not by $\eta$ but by $\eta''$. The case for $\eta'$ is easy enough to understand (Lemma of du Bois-Reymond). So, the problem is this: If $M(x) \in C[a,b]$ such that $\int_{a}^{b}{M(x)\eta''(x)} = 0 ~~\forall\eta\in C^2[a,b],\eta(a)=\eta(b)=\eta'(a)=\eta'(b)=0$, then $M(x)=px+q$ for all $x\in[a,b]$.
I have tried to work backward from what I think will be the last statement in the proof : $\int_{a}^{b}{[M(x) - (px+q)]^2}=0$. But I can't quite get it to work.