$\int_a^b M(x)\eta''(x)dx=0$ for all $\eta$ implies $M(x) = c_0 + c_1x$ proof

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Here's a lemma that my book proves:

If $M(x)\in C[a,b]$ and $$\int_a^b M(x)\eta'(x)dx = 0$$ for all $\eta(x)\in C^1[a,b]$ such that $\eta(a)=\eta(b)=0$, then $$M(x)=c,$$ a constant, for all $x\in [a,b]$.

It then asks the reader to prove a related lemma:

Let $M(x)\in C[a,b]$ be a continuous function on the closed interval $a\le x\le b$ that satisfies $$ \int_a^b M(x)\eta''(x)dx = 0$$ for all $\eta(x)\in C^2[a,b]$ satisfying $$\eta(a)=\eta(b)=\eta'(a)=\eta'(b)=0.$$ Then $M(x)=c_0+c_1x$ for suitable constants $c_0$ and $c_1$.

My thoughts: If $\eta(x) \in C^2[a,b]$, then $\eta'(x)\in C^1[a,b]$ and thus the exact conditions of the former lemma applies (with the role of $\eta$ in the first lemma being played by $\eta'$ in the second) meaning that $M(x)$ should just be constant, not $c_0+c_1x$. Why is this reasoning false?

Also hints on how that second lemma would be proven appreciated. ;)