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 $y_2$, $=y_1y'_2-y_1'y_2$