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

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.
 
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$$tex here$$ 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.
 
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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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