In my FuncAna exam I had the following problem, but I was not able to do anything. Right now I am still overstrained in fining a proof... It cannot be that hard ;)
Let $I=(0,1)$, $b>0$ and $f\in L^2(I)$ be given. We had to show that there exists one and only one $u\in W^{2,2}(I)$ s.t. \begin{align} \int_0^1 u''\phi'' + bu'\phi'+u\phi \mathrm{d}x = \int_0^1 f\phi\mathrm{d}x \end{align} $\forall \phi \in W^{2,2}(I)$.
For any $b>0$ $$ \langle u,v\rangle=\int_0^1 (u''v'' + b\,u'v'+u\,v)\,dx $$ is an inner product on $W^{2,2}$ equivalent to the usual one. On the other hand $$ \phi\mapsto\int_0^1 f\,\phi\,dx $$ is a bounded linear functional on $W^{2,2}$. The result now follows from the Riesz representation theorem.
Another possibility would be to use the Lax-Milgram theorem on the bilinear form $\langle u,v\rangle$.