ODE theory. Need someone to jog my memory I have two questions regarding the same thing.


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*Let's say I have a homogeneous ODE (is that what they are called?) $ay'' + by' + cy = 0$


The trick in this problem is to multiply both sides by $e^{rt}$ and do some trick to get the homogeneous solution which usually includes strictly exponential or sometimes a mixed with trigonometric functions. This is what most book does, but now that I had forgotten most of this stuff, could someone remind me why we ignore the trivial solution $y = 0$?


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*Let's say the ODE $ay'' + by' + cy = 0$ has the homogeneous solution $y = Ae^{-r_1t} + Be^{r_2t}$ (Let $r_1, r_2 > 0$ and are roots of the characteristic equation). If it happens that $\lim_{x\to \infty} y = 1$. Why can't there be a solution? Is there a theorem that says this? I can't remember it out for the life of me.


Thank you for reading
 A: Yes, they are called homogeneous ODEs.
1 Simply speaking, We don't ignore it. Set $A=B=0$ in your solution and it comes right out. It doesn't have to be treated separately this time. There are existence and uniqueness theorems that cover when this is the case.
2 In your ODE, take $a=1,b=1,c=0$. So your ODE is $y''+y'=0$ and your characteristic equation is $r^2+r=0$. This has roots at $r=0,-1$. Subbing this in we get a general solution as $Ae^{-t}+Be^{0t}=Ae^{-t}+B$. Setting $B=1$ will give you an equation that has $\lim_{x\rightarrow\infty}=1$. What made you think there wasn't such a solution? Did the book specifically say that $a,b,c$ in the ODE are non-zero?
^leaving that up there just for future reference, but to the modified question:
2 A limit of sums is the sum of the limits, so you can treat each term separately. $r_1>0$, so then $\lim_{x\rightarrow\infty}Ae^{-r_1t}=0$, since it's a decaying exponential. Since $r_2>0$, $\lim_{x\rightarrow\infty}Be^{r_2t}=\pm\infty$, depending on the sign of $B$. $0\pm\infty=\pm\infty$, so the limit can't possibly be anything else with those restrictions.
Basically, you're adding one decaying exponential (which goes to zero) and one that blows up. They can't possibly add to anything other than a different equation that still blows up, essentially because the first one becomes smaller and smaller as $x\rightarrow\infty$ and so affects the term that already diverges less and less.
A: Consider the homogeneous ODE of second order, 

$$ \tag 1 y''+ay'+by=0$$

Where $a$ and $b$ are constant.
The trivial solution is ignored simply because it is trivial. It's not significant in the study of the ODE. We're interested in non-trivial solutions. 
What can be done is assume a solution of the form $y=e^{rx}$, $r$ not necessarily real. Then
$$r^2e^{rx}+ar e^{rx}+b e^{rx}=0 $$
Since $e^{rx}$ is never zero,

$$\tag 2 r^2+ar +b =0 $$

Thus, let $r_1$ and $r_2$ be the roots of this equation, we have two solutions:

$$y_1= e^{r_1 x}$$
  $$y_2= e^{r_2 x}$$

If the roots aren't identical, then the solutions are linearly independent, so that the general solution of the hom. ODE is given by

$$y_h = c_1 y_1 +c_2y_2$$

where the $c$s are arbitrary constants. 
In genera, we call $(2)$ the characteristic equation of $(1)$, $r_1$, $r_2$ the roots of the equation, and each $y_1$ ,$y_2$ the partical solution.
Depending on the nature of the roots (complex or real), we'll get solutions of the form:
$(1)$ Real roots, $r_1$,r_2$
$$y = c_1 e^{r_1 x} +c_2 e^{r_2 x}$$ 
$(2)$ Conjugate complex roots, $r_1=a+bi$,$r_2=a-bi$
$$ y = {e^{ax}}\left( {{C_1}\sin bx + {C_2}\cos bx} \right) $$
I guess you can now see it is not possible that the limit of such functions is $1$.
Thus, no solution can exists fullfilling that initial condition.
Suppose now we have the following non-homogeneous ODE,

$$ \tag 3 y''+ay'+by=\rm F$$ 

where $\mathrm F$ is a function of $x$.
There's a theorem that states that if $y_p$ is a particular solution of $(3)$ and  $y_h$ is the solution to the related hom. ODE, then all the solutions are given by

$$ \tag 4 Y=y_h+y_p$$

We can get particular solutions by various methods, and they will depend on $F$. 
In particular, one method gives the general solution, and thus the particular solution: 
If $y_h$ is $y_h = c_1 y_1 +c_2 y_2$ then the particular solution is

$$ y_p=y_2 \int \frac{{\rm F} \cdot y_1}{{\rm W}(y_1,y_2)}dx-y_1\int \frac{{\rm F}  \cdot y_2}{{\rm W}(y_1,y_2)}dx$$ 

where ${\rm W} (y_1,y_2)$ is the wronskian determinant of $y_1$ and $y_2$, $=y_1y'_2-y_1'y_2$
