All Solution of Diff. Eq. Tend to Zero

1. Apr 13, 2012

TranscendArcu

1. The problem statement, all variables and given/known data

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)

3. 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?

2. Apr 13, 2012

Dick

Yeah, that works. It just seems a little overcomplicated. In the case where you have two real roots then you can factor your equation ar^2+br+c=a*(r-r1)*(r-r2). You don't really need the quadratic formula to say something about the signs of r1 and r2. You do need to think about the case of complex roots. If they are complex then r1 and r2 are complex conjugates. What can you say about the sign of the real part? And there is a third case you should mention. Where r1=r2. The solution looks a little different in that case.