Euler's rule or Non-homogenous method?

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Homework Help Overview

The problem involves a differential equation of the form x^2 y' + xy + 5x^5=0, which is being analyzed for solution methods, particularly focusing on Euler's rule and non-homogeneous methods.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the applicability of Euler's methods and non-homogeneous methods, questioning how to handle the presence of terms involving both x and y. There is mention of integrating factors and the potential for polynomial expansion or Taylor series.

Discussion Status

Some participants have suggested dividing by x and reformulating the equation, while others have pointed out the equivalence of different approaches. There is an ongoing exploration of integrating factors and the structure of the equation, with no explicit consensus reached.

Contextual Notes

The original poster notes the absence of starting conditions, which may influence the methods discussed. Assumptions about x being non-zero are also acknowledged in the conversation.

lost_math
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Homework Statement


x^2 y' + xy + 5x^5=0

Homework Equations


no starting conditions


The Attempt at a Solution


cannot figure out how to do this. Euler's equations methods have no pure x terms, and the non-homogenous methods have some kind of separable thing, where the x terms neatly land up on RHs and y terms on LHS. But what do i do with this? Is it the polynomial expansion or maybe some kind of taylor series?
 
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U can divide by "x" (assuming x different from 0) and then write the resulting eqn as

[tex](xy)'=-5x^{4}[/tex]

Integrate both terms and then see what you get.

Daniel.
 
I think dextercioby missed a term.

Start by putting in the form y' + y*(1/x) = -5x^3

Next, your integrating factor is p = e^[integral(1/x)dx] = e^(lnx) = x

Continue...
 
Stevecgz said:
I think dextercioby missed a term.

Start by putting in the form y' + y*(1/x) = -5x^3

Next, your integrating factor is p = e^[integral(1/x)dx] = e^(lnx) = x

Continue...

He didn't miss a term, your solution and his are identical, yours is just more detailed.
 
d_leet said:
He didn't miss a term, your solution and his are identical, yours is just more detailed.

Right :smile:
 
Thanks all, was really helpful.
 

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