Variation of Parameters - Higher order DE

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Homework Help Overview

The problem involves finding a particular solution to a higher-order differential equation, specifically one that is expressed in terms of variation of parameters. The homogeneous solutions provided are x, x^2, and 1/x, and the equation is given as x^3y''' + x^2y'' - 2xy' + 2y = 2x^4 for x > 0.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss using simultaneous equations to apply the variation of parameters method, but some express confusion over the results they are obtaining. There is also mention of guessing a solution based on the form of the non-homogeneous term.

Discussion Status

Some participants have provided guidance on alternative approaches, including the suggestion to guess a solution. There is a recognition of the potential for different methods to yield valid results, although concerns about adhering to specific homework requirements are noted.

Contextual Notes

Participants question the necessity of using variation of parameters specifically, given that the problem does not explicitly require it. There is also mention of a potential error in the application of the variation of parameters technique related to the coefficient of the highest derivative.

Wellesley
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Homework Statement


Given that x, x2 and 1/x are solutions of the homogeneous equation corresponding to:


[tex] x^3y''' + x^2y''-2xy'+2y=2x^4 [/tex]

x>0
determine a particular solution.


Homework Equations





The Attempt at a Solution


I'm trying to solve this problem using three simultaneous equations:

[tex] u_1'x + u_2'x^2 + u_3'*(1/x) = 0[/tex]

[tex]u_1' + 2u_2'x - u_3'x^{-2}= 0[/tex]

[tex]2u_2' + 2u_3'x^{-3}= 2x^4[/tex]

Then, when I try and solve for u1, u2 and u3, I seem to doing it incorrectly, the wrong answer. I've come back to this problem several times over the last three days, and all that has changed were the answers I'm getting! Should I be trying a different method in solving this problem?

Thanks.
 
Last edited:
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Variation of parameters is the hard way. The easy way is to try and guess a solution. I'm guessing that there is a solution of the form a*x^4. Do you see why?
 
Thanks for the response.

I do see what you are saying about the guess, and it works. ax4 comes from 2x4. Plugging in the derivatives of ax4 into the original equation, you can solve for a...a=1/15 so the answer is x4/15 which is the right answer.

Will this process be okay for a homework problem in the variation of parameters chapter? I just don't want to get marked down for not using the 'correct' method. When would you use the variation of parameters?
 
Last edited:
Wellesley said:
Thanks for the response.

I do see what you are saying about the guess, and it works. ax4 comes from 2x4. Plugging in the derivatives of ax4 into the original equation, you can solve for a. a=1/15 so the answer is x4/15 which is the right answer.

Will this process be okay for a homework problem in the variation of parameters chapter? I just don't want to get marked down for not using the 'correct' method.

Hard to say. But if they say 'use variation of parameters' then I guess you had better use variation of parameters.
 
Dick said:
Hard to say. But if they say 'use variation of parameters' then I guess you had better use variation of parameters.

Well, the problem just says to determine a particular solution. They didn't say which method to use, like they did on other problems so I think I'll be fine. Thanks for the help!
 
BTW the your variation of parameters was going wrong because in the technique the coefficient of the highest derivative should be 1. You should have divided both sides of the ODE by x^3 before you started. So the 2*x^4 in your last equation should be 2x. Then it will work.
 
Dick said:
BTW the your variation of parameters was going wrong because in the technique the coefficient of the highest derivative should be 1. You should have divided both sides of the ODE by x^3 before you started. So the 2*x^4 in your last equation should be 2x. Then it will work.

Thanks! I knew something was going wrong but I didn't know where.
 

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