Solving ODE with Variation of Parameters

[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.

 

[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 $$y_{p}=Ax^{9}$$

 

[ISU20CpreE]

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 $$y_{p}=Ax^{9}$$

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 don't get your point.

 

[arildno]

Ok. so this is what I did...

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

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

 

[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.

 

[ISU20CpreE]

That would give you the particular solution, yes.
What have you done about the other contribution?

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

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

 

[ISU20CpreE]

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.

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.

 

[FunkyDwarf]

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

 

[arildno]

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

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

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.

 

[ISU20CpreE]

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

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

 

[arildno]

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

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

 

[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.

I did. Oh and i was trying to solve the homogeneous eq and I don't know if I should take the x's into my quadratic formula or just the constants.??

 

[arildno]

I did. Oh and i was trying to solve the homogeneous eq and I don't know if I should take the x's into my quadratic formula or just the constants.??

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?[/itex]

 

[arildno]

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

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

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

 

[ISU20CpreE]

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

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

 

[arildno]

What's your problem with it? You INSERT a trial solution into your diff.eq and see what happens!
Please say EXACTLY where your problem with that lies!

 

[ISU20CpreE]

What's your problem with it? You INSERT a trial solution into your diff.eq and see what happens!
Please say EXACTLY where your problem with that lies!

I got it! LOL thank you so much for your time and I am sorry to bother you Ill be prepared next time.