# Stuck on equating co-efficents! DiffEQ

1. Feb 24, 2006

### mr_coffee

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 [Broken]

Last edited by a moderator: May 2, 2017
2. Feb 24, 2006

### quasar987

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

$$-At^2 + (2A-B)t + (12A+B-C) = t^2 + t$$

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!

3. Feb 24, 2006

### mr_coffee

THanks man worked great!!

4. Feb 25, 2006

### benorin

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.

5. Feb 25, 2006

### mr_coffee

shweet.
thanks for the tip