Solution for $-\frac{d^2u}{dx^2} + \frac{du}{dx} = 1$ In a book the author says the solution $u(x)$ for differential equation
$-\frac{d^2u}{dx^2} + \frac{du}{dx} = 1$
is the particular solution $u = x$ plus any $A + Be^{x}$. The $A$ and $B$ can be calculated from boundary conditions. I checked this and it works. I am confused however on how this solution is reached. Did he guess? Or did he find solutions for $\frac{du}{dx}$ and $-\frac{d^2u}{dx^2}$ seperately somehow add them up? 
Any ideas?
 A: The simplest thing to do here is to treat this as a first order equation in $y=\frac{du}{dx}$ first, and solve it. Note that the equation becomes
$$-y' + y = 1,$$
which is a first order linear separable equation. We can solve it the usual way:
if $y=1$ is constant, then the equation is satisfied; this is a particular/stable solution. Assume then that $y$ is not the constant function $1$. Then:
$$\begin{align*}
-\frac{dy}{dx} + y &= 1\\
y-1 &= \frac{dy}{dx}\\
dx &= \frac{dy}{y-1} &&\text{(this is where we use that }y\neq 1\text{)}\\
\int \,dx &= \int \frac{dy}{y-1}\\
x +C &= \ln|y-1| &&C\text{ arbitrary;}\\
e^{x+C} &= |y-1|\\
Ae^x &= |y-1| &&A\text{ a positive constant;}\\
Be^x &= y-1 &&B\text{ a nonzero constant;}\\
y &= 1+Be^x &&B\text{ a nonzero constant.}
\end{align*}$$
So $y(x) = 1+Be^x$ where $B$ is a nonzero constant. Since $B=0$ gives the stable solution, we have that the general solution is $y(x) = 1+Be^x$ with $B$ an arbitrary constant.
But $y(x) = \frac{du}{dx}$, and what we are really interested in is $u(x)$. But now we have the very easy differential equation
$$\frac{du}{dx} = 1 + Be^x,$$
which can be solved directly by integration. We get
$$\begin{align*}
\frac {du}{dx} &= 1 + Be^x\\
\int \frac{du(x)}{dx}\,dx &= \int(1+Be^x)\,dx\\
u(x) &= x + Be^x + C,
\end{align*}$$
where $B$ and $C$ are arbitrary.

If you don't realize you can use the trick of replacing $u'$ with $y$ and solving two first order equations instead of a single second order equation, you can still solve this very easily because this is a (nonhomogeneous) second order equation with constant coefficients.
First, we solve the associated homogeneous equation,
$$-u'' + u' = 0.$$
The associated equation is $-r^2 + r = 0$; the roots are $0$ and $1$, so this says that $e^{0x}=1$ and $e^{x}$ generate the space of solutions to the homogeneous equation.
So the general solution will be given by 
$$\mathbf{s} + A + Be^x$$
where $A$ and $B$ are arbitrary constants, and $\mathbf{s}$ is any particular solution to the equation $-u''+u'=1$. It is easy to find that $u(x)=x$ is a solution to this equation, so the general solution to the differential equation is then given by
$$u(x) = x + A + Be^x,$$
with $A$ and $B$ arbitrary constants. Same solution as above. 
