Solve Higher Order DE: Factoring to Find General Solution

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Homework Help Overview

The discussion revolves around solving a higher order differential equation, specifically the equation y^{(4)}-4y^{(3)}+6y^{(2)}-4y^{(1)}+y=0. The original poster is seeking guidance on how to factor the characteristic polynomial associated with this equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to rewrite the differential equation in terms of its characteristic polynomial and seeks to understand the factoring process. Some participants question the clarity of the original equation and provide references to Pascal's Triangle as a potential tool for understanding the coefficients.

Discussion Status

The discussion is ongoing, with participants exploring the connection between the equation and Pascal's Triangle. Some guidance has been offered regarding the pattern in the coefficients, but no consensus or resolution has been reached yet.

Contextual Notes

There appears to be some uncertainty regarding the original equation's formulation, as one participant suggests a possible typographical error. Additionally, the original poster expresses a lack of familiarity with Pascal's Triangle, which may impact their understanding of the discussion.

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When trying to solve a higher order differential equation, how can I factor it to find the general solution. A question like :y^{(4)}-4y^{(3)}+6y^{(2)}-4y^{(1)}+y=0.
So I can write this as:m^4-4m^3+6m^2-4y^1=0. How can I factor m? The prof mentioned something about Pascal's Triangle, but I'm not sure what to do, I haven't even looked at Pascals Triangle in I don't even know how many years.
 
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Did you mistype something, please check it again.

Here is Pascal's Triangle.

http://en.wikipedia.org/wiki/Pascal's_triangle

The first equation you gave, in binomial form is: (y-1)^4=y^4(-1)^0+4y^3(-1)^1+6y^2(-1)^2+4y^1(-1)^3+y^0(-1)^4=y^4-4y^3+6y^2-4y+1

Do you see the pattern?
 
Last edited:
Yeessss, I suppose
 
(y+1)^0=1=1
(y+1)^1=y+1=1 1
(y+1)^2=y+2y^2+1=1 2 1
(y+1)^3=y^3+3y^2+3y+1=1 3 3 1

etc
 
Thanks a lot!
 

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