1) With $\beta' = 0$ this reduces to $L''= 0$. Integrating this twice and applying the other two conditions I get $L = c_1 \cdot u + c_2 - 1$. Fairly easy so far.
2) Doesn't $\beta = 0$ for all $u$ imply that $\beta' = 0$? Wouldn't this also reduce the differential equation to $L'' = 0$? Integrating this twice and applying the other conditions gives me. $L = (c_1 + 1) \cdot u + c_2 - 1$.