# Homework Help: Y''+xy'-y=0 differential equation

1. Dec 2, 2013

### TheFerruccio

1. The problem statement, all variables and given/known data
Given $y_1=x$ is a solution, solve the differential equation

2. Relevant equations

y''+xy'-y=0

3. 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.

2. Dec 2, 2013

### FeDeX_LaTeX

You could reduce the order of your new DE with a simple substitution.

3. Dec 2, 2013

### vela

Staff Emeritus
4. Dec 2, 2013

### LCKurtz

Code (Text):

Use $tex here$ instead of [itex] tags for inline and
$$tex here$$ instead of [tex] tags.

5. Dec 2, 2013

### TheFerruccio

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.

6. Dec 2, 2013

### TheFerruccio

(can't delete this)

7. Dec 2, 2013

### vela

Staff Emeritus
Using Mathematica, I find the solution is $P = Ce^{-x^2/2}/x^2$. Not sure that makes it any easier for you though.

8. Dec 2, 2013

### TheFerruccio

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

9. Dec 2, 2013

### FeDeX_LaTeX

Where are you getting the $\frac{x^3}{3}$ term from?

10. Dec 2, 2013

### vela

Staff Emeritus
You seem to have forgotten the factor of $x$ multiplying $P' = A''$.