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

[itex]y_1 = c_1 e^{\frac{-b + \sqrt{b^2 - 4(a)(c)}}{2a}}[/itex]

[itex]y_2 = c_2 e^{\frac{-b - \sqrt{b^2 - 4(a)(c)}}{2a}}[/itex]

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 [itex]|\frac{-b}{2a}| > |\frac{± \sqrt{b^2 -4(a)(c)}}{2a}|[/itex] which implies [itex]|-b| > |± \sqrt{b^2 -4(a)(c)}| [/itex]. We know that [itex]b = \sqrt{b^2 - X}[/itex] iff X = 0. By assumption, however, a,c are nonzero, so we must have some number inside the radical, call it [itex]Q = b^2 - X[/itex] 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?