General solution of differential equation of order 3 Please ,how to find that the general solution of  $u'''(t)=e(t) , t\in [0,1]$ is given by 
$u(t)=c_0+c_1t+c_2 t^2 +\frac12 \int_0^t (t-s)^2 e(s) ds$ 
$e:(0,1)\rightarrow \mathbb{R}$, and $e\in L(0,1)$.
i think that it is general homogeneous solution + particular solution ,
the general homogeneous is $c_0+c_1 t+ c_2 t^2$ , but i dont know how to finde that the paricular solution is $\frac12 \int_0^t (t-s)^2 e(s)ds $
Please;
Thank you 
 A: You can find the particular solution by doing an integration by parts. 
$\frac12 \int_0^t (t-s)^2 e(s)ds = [0-0] -2* \frac12 \int_0^t (t-s) E(s)ds $ where E(s) is a primitive of $e(s)$ such that $E(0)=0$
Similarly we have : $-\int_0^t (t-s) E(s)ds = - ( [0-0] - \int_0^t F(s)ds)$ where $F(s)$...
And u(t)=$\int_0^t F(s)ds$ is a solution of $u'''(t)=e(t)$ 
EDIT, to be clearer : 
u(t)=$\int_0^t u'(s)ds + u(0)$
u(t)=$-\int_0^t (t-s) u''(s)ds + u'(0)t + u(0)$
$u(t)=\frac12 \int_0^t (t-s)^2 u'''(s) ds + u''(0)/2*t^2 + u'(0)t + u(0) = \frac12 \int_0^t (t-s)^2 e(s) ds + u''(0)/2*t^2 + u'(0)t + u(0) $
A: It suffices to see that the strange-looking integral is a particular solution. You probably came up with the following type of expression, or at least you should agree it is an obvious way to go:
$$u_p(t)=\int_0^t\int_0^v\int_0^u e(s)dsdudv.\tag{1}$$
Now this is equal to
$$\int_0^te(s)\cdot{\rm Area}(\{(u,v):s\le u\le v\le t\})ds=\int_0^t\frac{(t-s)^2}{2}e(s)ds. \tag{2}$$
How did we get this? First off, the region of points $(u,v)$ such that $s\le u\le v\le t$ (where $s$ and $t$ are fixed) is a right triangle (try plotting some examples to see this) with sides each of length $t-s$, so its area is $\frac{1}{2}(t-s)^2$. But the more substantial formula is the following:
$$\iint\cdots\int_D f(x_0)dx_0dx_1\cdots dx_n=\int f(x_0)\cdot{\rm Vol}(\{(x_0,x_1,\cdots,x_n)\in D\})dx_0 \tag{3}$$
(under suitable hypotheses most likely). This is a "continuous" generalization of a discrete version
$$\sum_{(x,y)\in A}f(x)=\sum_{x\in X}f(x)\,\#\{(x,y)\in A\} \tag{4}$$
(Where $A\subseteq X\times Y$.)
Anyway I presume most of the above is not relevant to you. If you want a way to derive $(2)$ from scratch without knowing ahead of time what to look for, you will need to familiarize yourself with a technique to change double integrals like $\int_0^v\int_0^u e(s)dsdu$ into single integrals (using by-parts integration), and apply it twice to go from $(1)$ to $(2)$.
If you just want to check that $(2)$ is a particular solution, then you can straight-up differentiate the given function three times and check that the result is $e(t)$, using the general formula 
$$\frac{d}{dt}\int_0^t f(t,s)ds=f(t,t)+\int_0^tf_t(t,s)ds\tag{5}$$ 
(and this formula can be derived using the chain rule + fundamental theorem of calculus).

Going from $(1)$ to $(2)$ with by-parts integration: Alright, first let's look at
$$\int_0^v\int_0^u e(s)dsdu. \tag{6}$$
Use by-parts ($X=\int_0^u e(s)ds$ and $Y=u$) to get
$$\int_0^v XdY=[XY]_0^v-\int_0^v YdX=v\int_0^v e(s)ds-\int_0^v ue(u)du=\int_0^v(v-s)e(s)ds. \tag{7}$$
Thus (using by-parts again)
$$\int_0^t\int_0^v\int_0^ue(s)dsdudv=\int_0^t\int_0^v(v-s)e(s)dsdv \tag{8}$$
$$=\int_0^tv\int_0^ve(s)dsdv-\int_0^t\int_0^vse(s)dsdv \tag{9}$$
($dY=vdv$ and $X=\int_0^ve(s)ds$ in the first integral, same by-parts as $(6)$-$(7)$ in second integral)
$$=\left[\frac{t^2}{2}\int_0^te(s)ds-\int_0^t\frac{v^2}{2}e(v)dv\right]-\left[\int_0^t (t-s)se(s)ds\right] \tag{10}$$
$$=\int_0^t\frac{t^2-s^2-2(t-s)s}{2}e(s)ds=\int_0^t\frac{(t-s)^2}{2}e(s)ds. \tag{11}$$

Going from $(1)$ to $(2)$ with reparametrization: The region of integration in ${\bf R}^3$ is
$$D=\{(s,u,v):0\le s\le u\le v\le t\}. \tag{12}$$
Therefore
$$\int_0^t\int_0^v\int_0^ue(s)dsdudv=\iiint_D e(s)dV=\int_0^t\int_s^t\int_u^te(s)dvduds \tag{13}$$
$$=\int_0^t e(s) \left(\int_s^t\int_u^t 1dvdu\right)ds=\int_0^t\frac{(t-s)^2}{2}e(s)ds. \tag{14}$$

Simply checking that the integral expression is a particular solution: differentiating once,
$$\frac{d}{dt}\int_0^t\frac{(t-s)^2}{2}e(s)ds=\frac{(t-t)^2}{2}e(t)+\int_0^t(t-s)e(s)ds. \tag{15}$$
Differentiating a second time,
$$\frac{d}{dt}\int_0^t(t-s)e(s)ds=(t-t)e(t)+\int_0^te(s)ds. \tag{16}$$
Differentiating a third time,
$$\frac{d}{dt}\int_0^te(s)ds=e(t). \tag{17}$$
Hence $u(t)=\int_0^t\frac{(t-s)^2}{2}e(s)ds$ satisfies $u'''(t)=e(t)$.

Proof of $(5)$: let $G(u,t)$ be such that $\frac{dG}{dt}(u,t)=f(u,t)$. Then
$$\frac{d}{dt}\int_0^tf(t,s)ds=\frac{d}{dt}G(t,t)=\frac{dG}{d\,{\small\rm 1st\,coord}}(t,t)+\frac{dG}{d\,{\small\rm 2nd\,coord}}(t,t) \tag{18}$$
$$=\int_0^tf_u(u,s)|_{u=t}ds+f(u,t)|_{u=t}=\int_0^tf_t(t,s)ds+f(t,t). \tag{19}$$
A: The particular solution 
$$u(t)=\frac{1}{2}\int^{1}_{0}(t-s)^2e(s)ds $$
is the Green's function solution $G(t,s)$ for this problem. 
