Solve Higher Order DE: Factoring to Find General Solution

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To solve higher order differential equations like y^{(4)}-4y^{(3)}+6y^{(2)}-4y^{(1)}+y=0, factoring can be approached using the roots derived from the characteristic equation m^4-4m^3+6m^2-4m=0. The discussion highlights the relevance of Pascal's Triangle in identifying the coefficients for binomial expansions, which can aid in factoring the polynomial. The equation can be expressed in binomial form as (y-1)^4, revealing a pattern in the coefficients that correspond to the rows of Pascal's Triangle. Understanding this pattern is crucial for finding the general solution of the differential equation. Overall, recognizing the connection between the polynomial's structure and Pascal's Triangle facilitates the factoring process.
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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?
 
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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 Alot!
 
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