Solving a differential equation using Cauchy-Euler Method

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Homework Help Overview

The discussion revolves around solving a differential equation using the Cauchy-Euler method, specifically the equation xy⁴ + 6y^m = 0, where the notation for derivatives is a point of confusion. The original poster expresses difficulty in understanding how to handle the term involving 6y^m.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to substitute y = xm into the derivatives and has calculated the first few derivatives. Some participants question the notation used for derivatives, clarifying that y⁴ refers to the fourth derivative of y. Others suggest looking for patterns in the derivatives and consider alternative notations for clarity.

Discussion Status

The discussion is ongoing, with participants clarifying notation and exploring the implications of the original equation. There is no explicit consensus, but some guidance has been offered regarding the interpretation of the notation and the relationships between the derivatives.

Contextual Notes

There is confusion regarding the notation used in the textbook, particularly the representation of derivatives, which has led to misunderstandings about the problem setup. The original poster also mentions struggling with the problem for an extended period, indicating a potential need for clearer communication of mathematical notation.

madcattle
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Homework Statement


Hey all, this is my first time posting here. I've used your help before, but have not actually ever had to post a question. Thanks for any help you can give, I am excited to finally join the community.

I am doing calculus homework and I am having trouble solving this problem using the Cauchy-Euler method that we're supposed to solve it with.

Solve: xy4 + 6ym = 0

It is the part about the 6ym that I am not sure what to do with.


Homework Equations



andny / dxn ...a0y = 0

The Attempt at a Solution



My book toldme to try and substitute y=xm into the derivative parts, and all of the other problems I've done work with that method. This is what I've done so far.

y=xm
y1=mxm-1
y2=m(m-1)xm-2
y3=m(m-1)(m-2)xm-3
y4=m(m-1)(m-2)(m-3)xm-4

I know how to treat the 4th derivative term; again, I just don't know what or how to think about the y^m term. I know the solution to the problem, as it is in the back of my book, I just don't know how to get there. Can anyone give me a prod in the right direction?
Thanks
 
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I don't see any derivative in the original equation!
 
I'm sorry, our book uses notation so that y^4 is supposed to be the fourth derivative of y.
So, for example, y1=y´ and y^2 = y´´
 
madcattle said:
I'm sorry, our book uses notation so that y^4 is supposed to be the fourth derivative of y.
So, for example, y1=y´ and y^2 = y´´

The usual notation is y^(2) (so we can tell the difference between that and the square of y). Anyway, you have found the first few y^(m) for y = x^n. If you don't see the pattern, I suggest you look instead at y^(m)/m! and see if you recognize the coefficients there.

RGV
 
I see that there is a relationship in that
y^(m)/m = (my^(m-1) + m(m-1)y^(m-2)...)/(m(m-1)(m-2)...) but I don't actually understand how to apply that relationship at all yet
 
Nevermind, I'm really sorry about this. I mistook the book for having written y^m when they really wrote y^´´´. I guess my new problem is poor eyesight. Thanks a lot for the help, I struggled trying to solve this for a solid 40 minutes.
 

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