I am trying to solve the following problem with two point load:
$$ \frac{d^2u}{dx^2} = \delta(x-1/4) - \delta(x-3/4) $$ With boundary conditions $u'(0) = 0$ and where $u'(1) = 0$
From the definition, we can say that:
$$ u(x) = \begin{cases} \hfill ax+b \hfill & 0 \leq x < 1/4 \\ \hfill c(x- \frac{1}{4}) +d \hfill & 1/4 < x < 3/4 \\ \hfill e(x- \frac{3}{4}) +g \hfill & 3/4 < x \leq 1 \end{cases} $$
How do I go from here to find all solutions to the problem?
On
Assuming your beam has not broken, $u(x)$ must be continuous. The only threats to this are at $x=\frac 14, \frac34$. You can write two equations joining the solutions across those points. If you integrate $u''(x)$ across one of the delta functions, you should get a step in $u'(x)$ at those points. You can use that to get two more equations. Now you have four equations in six unknowns, so have a two parameter family of solutions before you consider the boundary conditions. The boundary conditions look like they give you two more equations, so there will be a unique solution, but they turn out to be redundant, so you have a one parameter family
You can really work out the solution to the problem by just thinking of moving left to right. You start out at $u_0$, and are constant until the first jump at $1/4$. Then the slope is $1$, and stays that way until the second jump at $3/4$. Then the slope is $0$ again, and stays that way. These slopes are consistent with the boundary conditions. Putting what I've said into symbols and taking into account the continuity requirement, you get
$$u(x) = \begin{cases} u_0 & x \in [0,1/4) \\ (x-1/4) + u_0 & x \in [1/4,3/4) \\ 1/2 + u_0 & x \in [3/4,1]\end{cases}.$$
Note that $u_0$ is a free parameter, as should be expected with a pure Neumann boundary condition.