Limit of y(x) in Second Order Differential Equation So, even though I know how to solve ODEs, I don't know how I should proceed about this question:
Let $a$, $b$, and $c$ positive constants. If $y = y(x)$ is solution to the differential equation $ay'' + by' + cy = 0$, then $\lim_{x\to\infty}$ $y(x)$:
(a) doesn't exist and tends to $+\infty$. 
(b) exists and is $0$. 
(c) doesn't exist and tends to $-\infty$. 
(d) exists and is $\pi$. 
(e) exists and is $e$.
I tried to take the limit of the possible solutions but even if $a$, $b$, and $c$ are positive-only numbers, there are many possibilities so I couldn't achieve anything.
Thanks for the reading!
 A: Since 
$a, b, c > 0, \tag 1$
we can set
$\alpha = \dfrac{b}{a}, \; \beta = \dfrac{c}{a}, \tag{2}$ 
and obtain an equivalent equation
$y'' + \alpha y' + \beta y = 0, \; \alpha, \beta > 0; \tag{3}$
then setting
$\mu_\pm = \dfrac{1}{2}(-\alpha \pm \sqrt{\alpha^2 - 4 \beta}), \tag 4$
we may distinguish two cases:  first
$\alpha^2 \ne 4 \beta, \tag 5$
whence
$\sqrt{\alpha^2 - 4\beta} \ne 0, \tag 6$
and the roots $\mu_\pm$ are thus distinct; in this case the general solution is of the form
$y(t) = c_+ e^{\mu_+ t} + c_- e^{\mu_- t}; \tag 7$
since from (4) each of $\mu_\pm$ has negative real part, 
$\displaystyle \lim_{t \to \infty} y(t) = 0; \tag 8$
if, on the other hand, 
$\alpha^2 = 4\beta, \tag 9$
then we may write
$\mu_+ = \mu_- = \mu = -\dfrac{\alpha}{2} < 0, \tag{10}$
and the solution now takes the form
$y(t) = e^{\mu t}(c_1 + c_2 t) \to 0 \; \text{as} \; t \to \infty, \tag{11}$
since the exponential term dominates $c_1 + c_2 t$ for $t$ sufficiently large.
In any event, whether $\mu_+ = \mu_-$ or not, we have $\lim_{t \to \infty} y(t) = 0$; thus (b) is the correct result.
Just to verify that $t e^{\mu t}$ solves (3): if
$y(t) = te^{\mu t}, \tag{12}$
then
$y'(t) = e^{\mu t} + \mu t e^{\mu t}; \tag{13}$
$y''(t) = \mu e^{\mu t} + \mu e^{\mu t} + \mu^2 t e^{\mu t} = \mu^2 t e^{\mu t} + 2 \mu e^{\mu t}; \tag{14}$
$y''(t) + \alpha y'(t) + \beta y(t) = \mu^2 t e^{\mu t} + 2 \mu e^{\mu t} + \alpha \mu t e^{\mu t} + \alpha e^{\mu t} + \beta t e^{\mu t}$
$= (\mu^2 + \alpha \mu + \beta) t e^{\mu t} + (\alpha + 2 \mu) e^{\mu t} = 0 \tag{15}$
since
$\mu^2 + \alpha \mu + \beta = (\mu + \dfrac{\alpha}{2})^2 = 0 = 2\alpha + \mu, \tag{16}$
which follows from (4) when (9) binds.
A: There are indeed several possibilities, but I wouldn't necessarily call that "many". And the key is that all three coefficients $a$, $b$, and $c$ are positive real numbers.
The roots of the characteristic equation are
$$\lambda_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=-\frac{b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}.$$
Jumping ahead, the $\displaystyle \color{red}{-\frac{b}{2a}}$ part, which is obviously a negative number here, is the key! Anyways, now we only have three significantly different cases based on the discriminant.


*

*If $b^2-4ac>0$, then we have two distinct real roots, and it's easy to see that both $\lambda_{1,2}$ are negative.

*If $b^2-4ac=0$, then we have a repeated real root $\displaystyle\lambda_{1,2}=-\frac{b}{2a}$, which is again a negative number.

*If $b^2-4ac<0$, then we have two complex conjugate roots, whose real part is the same negative value $\displaystyle -\frac{b}{2a}$.
Setting up the corresponding solutions $y(x)$ in all three cases leads to functions that have the same limit as $x\to+\infty$. In the first case the solution is a linear combination of $e^{\lambda_1}$ and $e^{\lambda_2}$, where both $\lambda_{1,2}<0$; and in the second and third cases the key is that the solution has a factor of $e^{-b/(2a)}$ (times something that doesn't grow fast enough or doesn't even grow at all).
A: Note that your differential equation is linear. If $y$ is a solution to your problem, then $\lambda y$ is also a solution to your equation. No initial conditions are given in your problem.
Consequently, if $\lim_{x\rightarrow\infty} y(x)$ has value $A$ (where $A$ is possibly $\pm\infty$), then $\lim_{x\rightarrow\infty} \lambda y(x)$ has value $\lambda A$.
Given that one of the answers a to e is correct, so knowing that $A$ exists, the only way it can be uniquely defined if is $A=\lambda A$ for all $\lambda$, or in other words $A=0$, and answer b is correct.
A: Always check for the trivial solution.  $y \equiv 0$ is a solution of the given equation.  This makes answer (b) correct.  (The other correct answers here arrive at the same conclusion by other means.)
A: According to the Laplace final value theorem for a stable system under initial conditions,
$$
\lim_{t\to\infty}y(t) = \lim_{s\to 0}sY(s)
$$
and here 
$$
Y(s) = \frac{s \dot y(0)+y(0)}{a s^2+bs+c}
$$
which is the Laplace transform for $y(t)$. Here the poles are
$$
s^* =\frac{-b\pm\sqrt{b^2-4ac}}{2a}
$$
then as $\left(\frac{-b}{2a}\right) < 0$ the system response is stable and under those conditions we have
$$
\lim_{t\to\infty}y(t) = \lim_{s\to 0}sY(s) = 0
$$
