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smartdude1212
09-21-2011, 08:26 PM
Okay so in my Differential Equations class we have been learning the basics when it comes to solving homogeneous second order linear differential equations. We looked at Reduction of Order, and right now we are studying the types with constant coefficients:

ay'' + by' + cy = 0, where a, b, and c are constant coefficients.

Now, this is solved using the characteristic equation ar^2 + br + c = 0 and solving for r given these three cases:

1. two real roots, r1 =! r2 ; y1 = e^(r1)x, y2 = e^(r2)x
2. double root, r1 = r2 = r; y1 = e^rx, y2 = x*e^rx
3. two complex conjugate roots where r = A +/- Bi; y1 = e^Ax*cos(Bx), y2 = e^Ax*sin(Bx)

Now, my professor said that with these kinds of differential equations, the same ideas extend to higher orders. He gave us some straightforward examples, but the one that confused me was y''' + y = 0.

The characteristic equation for this would be:
r^3 + 1 = 0
r^3 = -1

And from this point he simply stated that:
r = -1
y1 = e^-x
y2 = x*e^-x
y3 = x^2*e^-x

Now in my first university calculus class (with a freaking amazing professor), we were learning some sort of concept that involved factoring and finding the roots of an equation (I can't even remember what) and the professor always said that even if we got an equation like r^3 + 1 = 0, we should always factor it completely and show that there are three answers, two of which are not real. I probably recalled her saying this because she's also recognized for being an awesome differential equations professor (wish I could have her for this class) and at the time it seemed pointless to me to solve r^3 + 1 = 0 by factoring it just to show that it only has one real root.

Anyhow, keeping her advice in mind, this is how I solved y''' + y = 0:

r^3 + 1 = 0
(r + 1)(r^2 - r + 1) = 0

Two cases:
r = -1
So, y1 = e^-x

r = -1/2 +/- (i sqrt3)/2
So, y2 = e^-(x/2)*cos[(sqrt(3)x)/2], y3 = e^-(x/2)*sin[(sqrt(3)x)/2]

And then the general solution would be y = c1y1 + c2y2 + c3y3 where c1/c2/c3 are constants.

It seems absurd to me that:

x*e^-x and e^-(x/2)*cos[(sqrt(3)x)/2]
or
x^2*e^-x and e^-(x/2)*sin[(sqrt(3)x)/2]
would be equivalent.

So, which method is correct?

kaiten123
09-22-2011, 08:19 PM
before i even attempt to give you a definite answer or solve it myself or w/e i'll tell you this: Many diff eqs have multiple solutions, why don't you try plugging your different answers into the original to easily see if they work?

MaiChanDesu
09-22-2011, 08:30 PM
The second method is correct because the characteristic polynomial implies that there are three values, (essentially they are eigenvalues). Clearly two of them are complex and one of them is real.

The characteristic equation for this would be:
r^3 + 1 = 0
r^3 = -1

And from this point he simply stated that:
r = -1
y1 = e^-x
y2 = x*e^-x
y3 = x^2*e^-x

You would only use this if you have repeated eigenvalues. Such examples of this case would be p(r) = r^2(r+1).

This would imply that this characteristic polynomial have eigenvalues 0 (with algebraic multiplicity 2) and 1 (with algebraic multiplicity 1). When you have repeated eigenvalues, the first method you mentioned is the way to go when searching for the general solution of the differential equation!

Let me know if you don't understand this!

Sincerely,

Broddas =w=