Factoring for Higher order ODE

trap101
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Solve the differential equation:

y(5)+12y(4)+104y(3)+408y''+564y'=0

where the (n) is the nth derivative.

So it's a 5th order DE. Now I'm trying to find the roots:

One of the roots is r = 0, which I obtain by factoring the equation into this form:

r(r4+12r3+104r2+408r+1156) = 0

No problem there. Now the other solutions are complex, my issue is how can I find those solutions from this 4th degree polynomial? I can't synthetically divide like it was just real numbers, so what do I do? The solution get's it into the form:

(r2+6r+34)2 from here I see how to get the complex, but how do I factor my above equation even to get this equation?

Besides that factoring issue I understand the problem.

Thanks
 
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You can look up "roots of quartic" in wikipedia and use the general formula they supply or you just guess. (Or use a math program.)

Since it is a 4th order polynomial, not obviously a quadratic in r^2, you would guess something like (ar^2+br+c)^2 - expand it out and compare the coefficients.
 
In your characteristic polynomial, it is not clear why you have 1156 as the constant term rather than 564.
 
Simon Bridge said:
You can look up "roots of quartic" in wikipedia and use the general formula they supply or you just guess. (Or use a math program.)

Since it is a 4th order polynomial, not obviously a quadratic in r^2, you would guess something like (ar^2+br+c)^2 - expand it out and compare the coefficients.



LOL, using that formula is just a cruel joke. If that's the only way of being able to solve these sorts of problems. How would they ask them on a quiz? I ask because I have a quiz tomorrow and it is mainly on higher order DE's if this is the general format to get them, how in tarnation are they going to ask me to solve any higher order DE's beyond one's with real solutions? That would be a nightmare to solve by hand.
 
trap101 said:
LOL, using that formula is just a cruel joke. If that's the only way of being able to solve these sorts of problems. How would they ask them on a quiz? I ask because I have a quiz tomorrow and it is mainly on higher order DE's if this is the general format to get them, how in tarnation are they going to ask me to solve any higher order DE's beyond one's with real solutions? That would be a nightmare to solve by hand.

There are almost no practical uses for that formula. If they give you a high order polynomial to solve they will have set it up to be easy enough so it will factor with some guessing.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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