Values of Constants for an Exponentially Decaying General Solution

Question

I have a doubt about the following question:

For the differential equation with constant coefficients $$y'' + by' + cy = 0 $$

the general solution $y(x)$ approaches zero as $x$ approaches $\infty$ if

(a) $b \gt 0, c \gt 0 $

(b) $b \gt 0, c \lt 0 $

(c) $b \lt 0, c \gt 0 $

(d) $b \lt 0, c \lt 0 $

Doubt: On solving the differential equation by forming the auxiliary equation, I get the general solution (assuming $\sqrt{b^2-4c} \gt 0$) as: $$y=c_{1}e^{x{\frac{-b + \sqrt{b^2 - 4c}}{2}}} + c_{2}e^{x{\frac{-b - \sqrt{b^2 - 4c}}{2}}}$$

I tried doing $\lim\limits_{x \to \infty} y$, which equated to zero (I might be stupid by doing so). But I have no idea on answering the actual question. I would appreciate if someone could help me and let me know if I am missing anything out.

Edit 1: The solution according to the answer key is (a).

Answer 1

The general solution to your differential equation is $$y= C_1e^{\lambda_1x}+ C_2e^{\lambda_2x}$$ where $\lambda _1$ and $\lambda_2$ are roots of $$\lambda ^2 +b\lambda +c =0$$

Note that the sum of your eigenvalues is $\lambda _1+ \lambda _2=-b$ and the product is $\lambda _1\lambda _2=c$

For both eigenvalues to be negative you need $b>0$ and $c>0$ that is case (a)

Answer 2

According to your solution,

$$ C_1e^{\frac x2(-b+\sqrt{b^2-4c})}+C_2e^{\frac x2(-b-\sqrt{b^2-4c})} $$

If

a) we have $-b < 0\;$ and $\sqrt{b^2-4c} < b\;$ with $y$ exponentially decaying. For $c \le \frac{b^2}{4}\;$ and if $c > \frac{b^2}{4}\;$ then we will have $$ y = e^{-\frac{bx}{2}}\left(C_1\cos\omega t + C_2\sin\omega t\right) $$ hence $b > 0, c > 0\;\Rightarrow \lim_{x\to\infty} = 0$

b) If $b > 0, c < 0\Rightarrow \sqrt{b^2-4c} > b$ and then the exponential $e^{(-\frac{b}{2}+\frac{\sqrt{b^2-4c}}{2})x}$ grows positively hence $b > 0, c < 0\;\Rightarrow \lim_{x\to\infty} = \pm\infty$

c) If $b < 0$ then independent of the $c$ value, we will have one of the exponential positive hence $b < 0, c > 0\;\Rightarrow \lim_{x\to\infty} = \pm\infty$

d) This item is treated in c) hence $b < 0, c < 0\;\Rightarrow \lim_{x\to\infty} = \pm\infty$