Solving a differential equation using Cauchy-Euler Method

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.
 
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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