Y''+xy'-y=0 differential equation

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Homework Help Overview

The discussion revolves around solving the differential equation y'' + xy' - y = 0, given that y_1 = x is a known solution. Participants are exploring methods to find a second solution using variation of parameters.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to find a second solution by varying a parameter A in the form y_2 = Ax, leading to a new differential equation. Some participants suggest reducing the order of this new equation through substitution.

Discussion Status

Participants are actively discussing the steps taken to reduce the order of the differential equation and are sharing insights about the challenges faced in integrating the resulting expressions. There is mention of using Mathematica to verify solutions, indicating a collaborative exploration of the problem.

Contextual Notes

There are mentions of issues with TeX tags and formatting in the forum, which may affect the clarity of mathematical expressions shared among participants. Additionally, there is uncertainty regarding the correctness of derived solutions and the integration process involved.

TheFerruccio
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Homework Statement


Given y_1=x is a solution, solve the differential equation


Homework Equations



y''+xy'-y=0

The Attempt at a Solution



Since I am given y_1=x (is there a hotkey for adding TeX tags so I don't have to manually type these tags over and over? So tedious.) then I wish to vary a parameter A that y is multiplied by, A(x)...

y_2=Ax
y_2'=A'x+A
y_2''=A''x+2A'
Plugging into the original equation, I end up with

A''x+(x^2+2)A'=0

This is another differential equation, and I do not know where to go from here.
 
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You could reduce the order of your new DE with a simple substitution.
 
Code:
 Use ##tex here## instead of [itex] tags for inline and
$$tex here$$ instead of [tex] tags.
 
FeDeX_LaTeX said:
You could reduce the order of your new DE with a simple substitution.

Regarding the # # tags, I was using those for a while, but, a few posts ago, they completely stopped working for me. Maybe it was a script error in the browser. I've noticed that I have to refresh quite a few times to get the equations to show, no matter which browser or computer I'm using.

Anyway, I reduced the order of the equation, by substituting P = A', then I get a separable DE.

I get a solution of the form:##\text{Ce}^{-\frac{x^3}{3}-2 x}##

I do not think this is right. I simply replaced every instance of A' with P. I do not know how to integrate from this point, because I have a nasty integral that evades substitution methods. This just feels wrong.
 
(can't delete this)
 
TheFerruccio said:
Regarding the # tags, I was using those for a while, but, a few posts ago, they completely stopped working for me. Maybe it was a script error in the browser. I've noticed that I have to refresh quite a few times to get the equations to show, no matter which browser or computer I'm using.

Anyway, I reduced the order of the equation, by substituting P = A', then I get a separable DE.

I get a solution of the form:


##\text{Ce}^{-\frac{x^3}{3}-2 x}##

I do not think this is right. I simply replaced every instance of A' with P. I do not know how to integrate from this point, because I have a nasty integral that evades substitution methods. This just feels wrong.
Using Mathematica, I find the solution is ##P = Ce^{-x^2/2}/x^2##. Not sure that makes it any easier for you though.
 
vela said:
Using Mathematica, I find the solution is ##P = Ce^{-x^2/2}/x^2##. Not sure that makes it any easier for you though.

I've been using Mathematica, as well, which is why I figured that my answer was wrong.
 
TheFerruccio said:
I've been using Mathematica, as well, which is why I figured that my answer was wrong.

Where are you getting the ##\frac{x^3}{3}## term from?
 
  • #10
You seem to have forgotten the factor of ##x## multiplying ##P' = A''##.
 

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