Limit of y(x) in Second Order Differential Equation

Question

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!

Answer 1

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

$$ \lim_{t\to\infty}y(t) = \lim_{s\to 0}sY(s) = 0 $$

Answer 2

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).

Answer 3

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.

Answer 4

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.

Answer 5

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.)