Stuck on equating co-efficents DiffEQ

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Homework Help Overview

The discussion revolves around the process of equating coefficients in the context of solving differential equations, specifically when finding a particular solution involving polynomials and exponentials.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to find a particular solution by separating polynomial and exponential components and expresses confusion about the process of adding coefficients. Some participants suggest using the fact that two polynomials are equal if their coefficients match, proposing a method to set up a system of equations based on this principle.

Discussion Status

Participants are exploring different methods for equating coefficients, with one suggesting a linear system approach and another proposing a substitution method using specific values of t. There appears to be a productive exchange of ideas, but no explicit consensus has been reached.

Contextual Notes

The original poster's approach involves separating components of the solution, which may reflect specific constraints or preferences in their problem-solving strategy. The discussion also hints at the potential complexity of combining solutions.

mr_coffee
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I'm confused on how to equate coniffecents. What I'm doing is, finding a particular solution for just the polynomial then I'm going to find it just for the exponential and add them together rather then putting it all together and making it a mess, but I'm stuck when i try to add the Co-efficents, can someone explain that process to me? Thanks!

Here is my work!
http://suprfile.com/src/1/1ta5k1/lastscan.jpg
 
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I kinda see.

Fact: Two polynomials are equal iff their respective coefficients are the same. (You might want to try proving this (hint set x=0. u get 1 relation, differentiate, set x=0, u get a second equality, differentiate, set x=0, etc))

This is what you have here. So you want to rewrite the LHS as

[tex]-At^2 + (2A-B)t + (12A+B-C) = t^2 + t[/tex]

Now what the theorem/fact stated above is telling you is that

-A=1
2A-B=1
12A+B-C=0

It's a linear system of equation. Start row-reducin' friend!
 
THanks man worked great!
 
There is another method: just plug-in three different values of t (like, say t=0,1,2,) into the given equation to generate three equations in A,B, and C.
 
shweet.
thanks for the tip
 

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