# Proving that $\lim_{x\to\infty}y(x) = 0$ if $y(x)$ is the solution to a homogeneous second-order linear equation with positive constant functions.

As the title says:

Suppose $a,b,c > 0$ are constants. Consider $ay'' + by' + cy = 0$. Suppose that $y(x)$ is a solution. Prove that $\lim_{x\to\infty}y(x) = 0$. This is problem 34 from p.1160 in section 17.1 of Stewart Single Variable Calculus, 8th edition (2015).

The characteristic equation is of course $ar^2 + br + c = 0$, for which $r = \frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$.

1. For $b^2 - 4ac > 0$ and $r_1 \neq r_2$ then $y(x) = C_1e^{r_1x} + C_2e^{r_2x}$ for some constants $C_1, C_2$.
2. For $b^2 - 4ac = 0$ and $r = r_1 = r_2$ then $y(x) = C_1e^{rx} + C_2xe^{rx}$ for some constants $C_1, C_2$.
3. For $b^2 - 4ac < 0$ and $r = \alpha + \beta i$ then $y(x) = e^{\alpha x}(C_1cos(\beta x) + C_2isin(\beta x))$ for some constants $C_1, C_2$.

I can’t see any obvious reason that the limit of any of these is $0$. How do I approach this?

• For cases 1 and 2 obviously $r_1$ and $r_2$ should be negative from which follows the fact. – pointer Apr 16 '16 at 8:45

When will the solution go to zero. Analyse each case separately. For the first case both roots should be negative.hence the sum should be negative.What is the sum of roots of a quadraric equation.? Same is the case for the second For the third you have to see that the real part of the root is negative because the imaginary part is bounded. Do you see any relation between $a,b,c$ that is given to you?. $-\frac {b}{a}<0$