# Solve ODE

1. Oct 5, 2006

### ISU20CpreE

Hi i need to use Variation of Parameters to solve this ODE

$$x^2 y'' - 2xy' + 2y = x^(9/2)$$

So far I was thinking to use Euler's Equation and I really don't know if it will work please help me out with a hint. THanks.

Last edited: Oct 5, 2006
2. Oct 5, 2006

### arildno

Well, you will have two distinct contributions:
1. The general solution to the associated homogenous problem (i.e, with the right-hand side equal to zero)

2. A particular solution of your diff.eq. Here, try a simple power solution as your trial function, i.e $$y_{p}=Ax^{9}$$

3. Oct 5, 2006

### ISU20CpreE

Ok. so this is what I did...

$$y_p = Ax^9 y'_p = 9Ax^8 y''_p = 72Ax^7$$

Then i plug in into the original equation

$$72Ax^9 - 18Ax^9 + 2Ax^9 = x^9$$

after that I get $$1/52$$ as an answer.

Sorry i dont get your point.

4. Oct 5, 2006

### arildno

That would give you the particular solution, yes (A=1/52)
What have you done about the other contribution?

5. Oct 5, 2006

### arildno

If you don't get my point, then you have no business doing math, since you evidently haven't bothered to read your textbook or go to class. Goodbye.

6. Oct 5, 2006

### ISU20CpreE

$$y_p = Ax^(9/2) y'_p = 9/2 Ax^(7/2) y''_p = 63/4Ax^(95/2)$$

then i get $$A = 34/5$$

7. Oct 5, 2006

### ISU20CpreE

im sorry i haved worked on these problems for too long and my english is not that good maybe I word my sentence wrong. Would you reconsider.

8. Oct 5, 2006

### FunkyDwarf

You need a general solution and a particular solution. The sum of the two is the general solution you are after.

9. Oct 5, 2006

### arildno

Whatever nonsense is this???

You are to solve the homogenous equation:
$$x^{2}y_{h}''-2xy_{h}'+2y_{h}=0$$

Use the Euler trial solution: $$y_{h}=Cx^{n}$$
in order to determine allowable values for n.

10. Oct 5, 2006

### ISU20CpreE

I know but whats going on is that I need to use Variation of parameters and for that I need y_1 and y_2 so im stuck. One of my friends told me they have used Euler's method to find y_1 and y_2

11. Oct 5, 2006

### arildno

Well, why didn't you say at the beginning you were required to do this with variation of parameters, then?

12. Oct 6, 2006

### ISU20CpreE

I did. Oh and i was trying to solve the homogeneous eq and I dont know if I should take the x's into my quadratic formula or just the constants.??

13. Oct 6, 2006

### arildno

Asking that shows you haven't really tried it:
Let's see:
$$y_{h}''=n*(n-1)Cx^{n-2}, y_{h}=nCx^{(n-1}}$$
Inserting this into the homog. equation yields:
$$x^{2}n*(n-1)Cx^{n-2}-2xnCx^{n-1}+2Cx^{n}=0\to{n(n-1)Cx^{n}-2nCx^{n}+2Cx^{n}=0\to(n(n-1)-2n+2)Cx^{n}=0$$
Now, what must you require so that the last left hand side is zero for all choices of x, but so that in general, [itex]y_{h}[/tex] is distinct from the zero function?

14. Oct 6, 2006

### arildno

This would be correct particular solution for your EDITED ODE, something that occurred AFTER my first replies.

15. Oct 6, 2006

### ISU20CpreE

Sorry to bug you too much but I am confused on the post before the last one.

16. Oct 6, 2006

### arildno

What's your problem with it? You INSERT a trial solution into your diff.eq and see what happens!