# All Solution of Diff. Eq. Tend to Zero

## Homework Statement

Assume that a,b,c are all positive constants. Show that all the solutions of:

ay'' + by' + cy = 0

tend to zero as x goes to infinity. (I presume that y is implicitly assumed to be a function of x)

## The Attempt at a Solution

This is what I have to far:

We write the characteristic polynomial: $ar^2 + br + c = 0$ and solve using the quadratic formula. Thus, we find two roots for the equation, call them $r_1 , r_2$. We can write solutions if we presume there exist solutions of the form y = C * exp(r*t). Thus, we have:

$y_1 = c_1 e^{\frac{-b + \sqrt{b^2 - 4(a)(c)}}{2a}}$
$y_2 = c_2 e^{\frac{-b - \sqrt{b^2 - 4(a)(c)}}{2a}}$

To show that a linear combination of these two solutions always tends to zero, we'll need that the exponents of e are negative. We must show that $|\frac{-b}{2a}| > |\frac{± \sqrt{b^2 -4(a)(c)}}{2a}|$ which implies $|-b| > |± \sqrt{b^2 -4(a)(c)}|$. We know that $b = \sqrt{b^2 - X}$ iff X = 0. By assumption, however, a,c are nonzero, so we must have some number inside the radical, call it $Q = b^2 - X$ such that |-b| > sqrt(Q). Thus, we have that all exponents are negative.

I'm not sure if this is right, or even if this is true for all cases. Do I need to do cases for instances where the exponents have complex elements?