How can Euler differential equations be solved using a substitution?

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Discussion Overview

The discussion revolves around solving a second-order differential equation of the form 4x²y'' + y = 0, specifically focusing on the use of substitutions to simplify the equation. Participants explore various substitution methods, including power series and transformations, to find solutions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes the substitution y' = p and y'' = p' to transform the equation into 4x²p' + p = 0, suggesting that it can be solved using separation of variables.
  • Another participant challenges the initial substitution, stating that it cannot be applied since the equation does not contain y'. They recommend using the substitution y(x) = √x p(x) to arrive at a different form of the equation.
  • A participant questions the rationale behind choosing specific substitutions, asking if there is a theorem that guides such choices.
  • Another participant suggests substituting y = x^n and solving for n, arguing that this method might be more straightforward since it preserves the powers of x.
  • Some participants discuss the Power Series Method, proposing that it could be used to express the solution as a series, although one participant notes they have not yet covered this method in their studies.
  • A later reply introduces the idea of using the substitution z = ln(x) to convert the equation into one with constant coefficients.

Areas of Agreement / Disagreement

Participants express differing opinions on the appropriate substitutions to use, with no consensus reached on a single method. Multiple competing views remain regarding the best approach to solve the differential equation.

Contextual Notes

Some participants express uncertainty about the effectiveness of their proposed methods, and there are mentions of limitations based on the current curriculum and familiarity with certain techniques, such as the Power Series Method.

sutupidmath
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Second order diff.eq. help??

well i am trying to find a solution to this diff. eq, but i get stuck somewhere.
[tex]\ 4x^{2}y''+y=0[/tex]

I first took this substitution

y'=p, y"=p' so the diff. eq becomes of this form

[tex]\ 4x^{2}p'+p=0[/tex]
i think this can be done with the separable of variables thing. so
[tex]\ 4x^{2}\frac{dp}{dx}=-p[/tex]

[tex]\frac{dp}{p}=-\frac{dx}{4x^{2}}[/tex] now i integrate bot sides
[tex]\int\frac{dp}{p}[/tex] = [tex]-\int\frac{dx}{4x^{2}}[/tex], after some calculations i get

[tex]p=A e^{\frac{1}{4x}[/tex], where A is a constant that we get because of integrating bot siedes. NOw i go back to p=y'
so

[tex]y'= A e^{\frac{1}{4x}[/tex], now i think all i need to do is integrate, but i get stuck integrating the right part of this eq. I do not think it has any closed form, right?


but i know that the answer to this equation is in a closed form, that is, it is an elementary function, so where am i going wrong?
I also tried to consider the eq [tex]\ 4x^{2}p'+p=0[/tex] as a linear one, but also when i tried to derive an integrating factor i ended up with a similar expression as [tex]y'= A e^{\frac{1}{4x}[/tex],
so, any hints on how to solve this?
 
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You can not apply the substitution [itex]y'(x)=p(x),\,y''=p'(x)[/itex] in your equation since it does not contains [itex]y'(x)[/itex]
Try the substitution [itex]y(x)=\sqrt{x}\,p(x)[/itex] to arrive to [itex]x\,p''(x)+p'(x)=0[/itex] and then try your substitution. i.e. [itex]p'(x)=w(x),\,p''(x)=w'(x)[/itex]
 
Rainbow Child said:
Try the substitution [itex]y(x)=\sqrt{x}\,p(x)[/itex] to arrive to [itex]x\,p''(x)+p'(x)=0[/itex]
How does one know that such a substitution will work?? Is there a theorem that provides us with these kind of substitutions or?
 
I'm not sure how the above works. Why not substitute into the equation y=x^n and then solve for n. I say this because if you differentiate twice you lose two powers of x.
 
Why can't you use the Power Series Method where you let the solution y=[tex]\sum_{0}a_n{}x^{n}[/tex] where a is from to infinite. The you will replace y along with its derivatives in that equation
 
kthouz said:
Why can't you use the Power Series Method where you let the solution y=[tex]\sum_{0}a_n{}x^{n}[/tex] where a is from to infinite. The you will replace y along with its derivatives in that equation

I guess i have to wait for another month then, until we get to that chapter! We haven't yet done this method of inifinite taylor series, or whatever!.
 
sutupidmath said:
well i am trying to find a solution to this diff. eq, but i get stuck somewhere.
[tex]\ 4x^{2}y''+y=0[/tex]

DEs with this form are called Euler differential equations. A substuition of the form:

z = ln(x)

with [tex]\frac{dy}{dx} = \frac{dy}{dz}\frac{dz}{dx}[/tex], can be used to convert this to a constant coefficient second order equation.
 

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