Solving $-\epsilon u''(x) + \beta u'(x) = 1$ Consider the equation:
$$-\epsilon u''(x) + \beta u'(x) = 1, \;\; x \in (0,1)$$
$$u(0) = 0, \;\; u(1) = 0.$$
$\beta > 0, \;\; \epsilon > 0$.
Can someone please help me to solve this equation? 
(I don't know very well differential equations, so I need help).
Thank you!
 A: First, you need to find the general solution of the associated homogeneous equation
$$
-\epsilon u''(x) + \beta u'(x) = 0\ ,
$$
by setting $u'(x)=y(x)$. You have
$$
-\epsilon y^\prime(x)+\beta y(x)=0\ ,
$$
which is separable
$$
\frac{\epsilon}{\beta}\frac{dy}{y}=dx\ ,
$$
yielding
$$
y(x)=\kappa_1 e^{\beta x/\epsilon}\ ,
$$
where $\kappa_1$ is an arbitrary constant. Therefore
$$
\frac{du}{dx}=\kappa_1 e^{\beta x/\epsilon}\Rightarrow u(x)=\kappa_1 \frac{\epsilon}{\beta}e^{\beta x/\epsilon}+\kappa_2\ .
$$
Next, find a particular solution of the inhomogeneous equation
$$
-\epsilon u''(x) + \beta u'(x) = 1\ ,
$$
for example take $u(x)=A x$, where you fix $A$ by requiring that
$$
\beta A=1\Rightarrow A=1/\beta\ .
$$
Therefore the general solution of your original equation is
$$
u(x)=\kappa_1 \frac{\epsilon}{\beta}e^{\beta x/\epsilon}+\kappa_2+\frac{1}{\beta}x\ .
$$
Next, fix $\kappa_1$ and $\kappa_2$ by requiring that $u(0)=u(1)=0$ as
$$
\kappa_1\epsilon/\beta+\kappa_2=0 \quad\text{and}\quad \kappa_1(\epsilon/\beta)e^{\beta/\epsilon}+\kappa_2+1/\beta=0\ .
$$
A: Consider the homogenous equation $$-\varepsilon y'' + \beta y' =0 $$ its general solutions are of the form $$y=C_1 +C_2 e^{\frac{\varepsilon}{\beta} x}$$ thus a general solutions of your equation are $$u(x) =C_1 +\frac{x}{\beta} +C_2 e^{\frac{\varepsilon}{\beta} x}$$ since $u(0)=0 , u(1)=0.$ We get $C_1 +C_2 =0 \wedge C_1 + \frac{1}{\beta} + C_2 e^{\frac{\varepsilon}{\beta} } =0$ therefore $$C_1 =\frac{1}{\beta e^{\frac{\varepsilon}{\beta} } -\beta} , C_2 =\frac{1}{-\beta e^{\frac{\varepsilon}{\beta} } +\beta}$$ and the solution is equal to $$u(x)=\frac{1}{\beta e^{\frac{\varepsilon}{\beta} } -\beta} +\frac{x}{\beta} +\frac{1}{-\beta e^{\frac{\varepsilon}{\beta} } +\beta} e^{\frac{\varepsilon}{\beta} x}$$
A: This is a 2nd order inhomogeneous ODE with constant coefficients. Such an equation can always be solved with the method of variation of parameters. There 4 steps:


*

*Find the general solution to the homogeneous equation
$$ -\epsilon u'' + \beta u' = 0 $$
using the ansatz $u(x) = e^{kx}$. You will find two values $k_1,k_2$ for which $e^{kx}$ is a solution, so the general solution is then $u(x) = c_1 e^{k_1 x} + c_2 e^{k_2 x}$.

*Reduce the inhomogeneous equation to a 1st order equation as follows: assume that $u(x) = c_1(x) e^{k_1 x} + c_2(x) e^{k_2 x}$, subject to the constraint $$c_1'(x) e^{k_1 x} + c_2'(x) e^{k_2 x} = 0.$$
Insert this expression for $u(x)$ back into the inhomogeneous equation, and using the fact that $e^{k_1 x}, e^{k_2 x}$ are solutions to the homogeneous equation, together with the constraint above, you will see that many terms cancel, and in the end you are left with
$$ -\epsilon( k_1 c_1'(x) e^{k_1 x} + k_2 c_2'(x) e^{k_2 x}) = 1.  $$

*Solve this 1st order equation. From the constraint, we have that $c_2'(x) = -c_1'(x) e^{(k_1-k_2)x}$. Plug this into the 1st order equation from (2), and rearrange to solve for $c_1'(x)$. You will get an equation of the form
$$ c_1'(x) = F(x), $$
so the solution is $c_1(x) = \int F(x) dx + a_1$, where $a_1$ is an integration constant. Now that you know $c_1(x)$, you can similarly solve for $c_2(x)$ (you will get another integration constant).

*Use your boundary conditions $u(0) = 0, u(1) = 0$ to solve for the integration constants.
Since this smells of homework, I hesitate to give additional details. But it is easy to find many worked examples of variation of parameters, which should make the technique very easy to understand.
A: First of all make the substitution $t=u'$ so you'll have a first order equation: $$
-\epsilon t'+\beta t=1
$$ Then solve the homogenous equation: $$
-\epsilon t_0'+\beta t_0=0 \iff
t_0'-\frac{\beta}{\epsilon}t_0=0 \iff
\frac{dt_0}{dx}=\frac{\beta}{\epsilon}t_0 \iff
\frac{dt_0}{t_0}=\frac{\beta}{\epsilon}dx \iff
$$ $$
\iff \ln t_0=\frac{\beta}{\epsilon}x+k \iff t_0=ce^{\beta / \epsilon x}
$$ Now looking at the RHS of the equation we know that the solutiono for the inhomogeneus equation differs from the solution of the homogenus by a zeroth degree polynomial: $$
t=t_0+q$$
Substituting in the equation youll find that $q=\frac{1}{\beta}$ so: $$
t=ce^{\beta / \epsilon x}+\frac{1}{\beta}
$$ Now we can integrate $t$ to find $u$: $$
u=\int t dx \iff u=c\frac{\epsilon }{\beta}e^{\beta / \epsilon x}+\frac{x}{\beta}+d
$$ Now we can force the contour conditions to find $c$ and $d$.$$
\left\{\begin{align*}
u(0) &= c\frac{\epsilon }{\beta}+d &= 0\\ 
u(1) &= c\frac{\epsilon }{\beta}e^{\beta / \epsilon }+\frac{1}{\beta}+d &= 0
\end{align*}\right.
$$ then $$
c=\frac{1}{\epsilon}\frac{1}{1-e^{\beta / \epsilon }}
$$ $$
d=-\frac{1}{\beta}\frac{1}{1-e^{\beta / \epsilon }}
$$ and the final expression for $u$ is: $$
\frac{1}{\beta}\frac{1}{1-e^{\beta / \epsilon }}\left ( e^{\beta / \epsilon x}+(1-e^{\beta / \epsilon })x-1\right )
$$
