Solving $-u''(x) = \delta(x)$ A question asks us to solve the differential equation
$-u''(x) = \delta(x)$
with boundary conditions 
$u(-2) = 0$ and $u(3) = 0$ where $\delta(x)$ is the Dirac delta function. But inside the same question, teacher gives the solution in two pieces as $u = A(x+2)$ for $x\le0$ and $u = B(x-3)$ for $x \ge 0$. I understand when we integrate the delta function twice the result is the ramp function $R(x)$. However elsewhere in his lecture the teacher had given the general solution of that DE as
$u(x) = -R(x) + C + Dx$
So I dont understand how he was able to jump from this solution to the two pieces. Are these the only two pieces possible, using the boundary conditions given, or can there be other solutions? 
Full solution is here (section 1.2 answer #2)
http://ocw.mit.edu/courses/mathematics/18-085-computational-science-and-engineering-i-fall-2008/assignments/pset1.pdf

This is a good example of a question to which one can answer at some very different levels of mathematical sophistication... Since you say nothing about this, let me try an elementary approach.
What you call the Dirac delta function (which is not a function, at least not in the sense of a function from $\mathbb R$ to $\mathbb R$) is a strange object but something about it is clear: 

One asks that $\displaystyle\int_y^z\delta(x)\mathrm dx=0$ if $y\leqslant z<0$ or if $0<y\leqslant z$ and that $\displaystyle\int_y^z\delta(x)\mathrm dx=1$ is $y<0<z$. 

We will not use anything else about the Dirac $\delta$.
If one also asks that $\displaystyle\int_y^zu''(x)\mathrm dx=u'(z)-u'(y)$ for every $y\leqslant z$, one can integrate once your equation $\color{red}{-u''=\delta}$, getting
that there exists $a$ such that 
$$
u'(x)=a-[x\geqslant0],
$$
where we used Iverson bracket notation. Now let us integrate this once again. 
Using the facts that $\displaystyle\int_y^zu'(x)\mathrm dx$ should be $u(z)-u(y)$ for every $y\leqslant z$, and the value of $\displaystyle\int_y^z[x\geqslant0]\mathrm dx$, one gets that for every fixed negative number $x_0$,
$$
u(x)=u(x_0)+a\cdot (x-x_0)-x\cdot[x\geqslant0].
$$
This means that $b=u(x_0)-a\cdot x_0$ does not depend on $x_0<0$, hence finally, for every $x$ in $\mathbb R$,
$$
\color{red}{u(x)=a\cdot x+b-x\cdot[x\geqslant0]}.
$$
(And, in the present case, the condition that $u(-2)=u(3)=0$ imposes that $a=3/5$ and $b=6/5$.)
This is the general solution of the equation $-u''=\delta$. Note that every solution $u$ is $C^\infty$ on $\mathbb R\setminus\{0\}$ but only $C^0$ at $0$ hence $u'$ and $u''$ do not exist in the rigorous sense usually meant in mathematics. Note finally that $u$ is also
$$
u(x)=a\cdot x+b-x\cdot[x\gt0].
$$
