Factoring for Higher order ODE

In summary: As for the roots themselves, you could try using a calculator to approximate them or using quadratic equations to solve for them. 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. As for the roots themselves, you could try using a calculator to approximate them or using quadratic equations to solve for them.
  • #1
trap101
342
0
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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  • #2
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.
 
  • #3
In your characteristic polynomial, it is not clear why you have 1156 as the constant term rather than 564.
 
  • #4
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.
 
  • #5
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.
 

Related to Factoring for Higher order ODE

What is factoring for higher order ODE?

Factoring for higher order ODE (ordinary differential equation) is a method used to simplify and solve complex differential equations. It involves breaking down a higher order ODE into smaller, simpler equations that are easier to solve.

Why is factoring important in solving higher order ODEs?

Factoring is important because it allows us to reduce the complexity of a higher order ODE and make it solvable. It also helps us identify common factors and patterns in the equation, which can lead to faster and more efficient solutions.

What are the steps involved in factoring for higher order ODE?

The steps for factoring a higher order ODE may vary depending on the specific equation, but generally involve identifying common factors, using algebraic techniques to simplify the equation, and then solving for the remaining variables.

What are some common difficulties when factoring for higher order ODEs?

One common difficulty is identifying the correct factors to use, as there can be multiple potential factors for a given equation. Another challenge is knowing which algebraic techniques to use when simplifying the equation. Practice and familiarity with different types of equations can help overcome these difficulties.

How can factoring be used to solve real-world problems?

Factoring for higher order ODEs is a powerful tool for solving real-world problems in fields such as physics, engineering, and economics. It allows us to model and analyze complex systems and phenomena, and make predictions and decisions based on the solutions we obtain.

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