# Best Guess for Partial Solution for Diff EQ?

1. Nov 24, 2014

### kq6up

1. The problem statement, all variables and given/known data

Find the specific solution for: y''-2y'+y=xe^x+4, y(0)=1, y'(0)=1.

2. Relevant equations
N/A

3. The attempt at a solution

Since xe^x is already in the general solution of the homogeneous version of this diff eq (complementary solution), my first guess for a partial solution term would be x^2e^x (xe^x is already taken by the complementary sol). However, it no worky. The correct guess is x^3e^x. I am not sure why this term has a higher power of x then the order of the DiffEq. I would have never have guessed this. Is there a reason I should have guessed this, or in general do I have to keep increasing powers until I find one that works? I am trying to find ways to save time on an exam.

Thanks,
Chris

2. Nov 24, 2014

### LCKurtz

The best way is to eliminate the guesswork with the annihilator method. It is really pretty simple. See
http://www.utdallas.edu/dept/abp/PDF_Files/DE_Folder/Annihilator_Method.pdf [Broken]

Last edited by a moderator: May 7, 2017
3. Nov 24, 2014

### kq6up

Ah, I vaguely remember that method from 20 years ago. Thanks for the post. According to the text my prof photocopied we are using the Method of undetermined coefficients, and I am going to assume there is guessing involved with my question above. So there is no straight forward way of knowing ahead of time with my professor's method -- just trial and error?

She assigned some painful extra credit for the whole class to repair a midterm grade, and I am thinking she wants us to just slog through it her way -- she is a bit sadistic like that :D

Thanks,
Chris

Last edited by a moderator: May 7, 2017
4. Nov 24, 2014

### LCKurtz

Actually, the method of undetermined coefficients is the method of annihilators. That's where the "guess" comes from. On an actual problem it is pretty quick and quicker if you guess wrong for your trial solution. For a particular problem it is really quite quick. To illustrate, in your problem$$y''-2y'+y = xe^x + 4$$your annihilator on the left is $(D-1)^2$ and on the right is $D(D-1)^2$ for a combined annihilator of $D(D-1)^4$. So you can annihilate both sides with$$\color{red}{Ae^x+Bxe^x}+Cx^2e^x+Dx^3e^x+E$$I have highlighted in red the part of that which is in the homogeneous solution. The rest of it is the correct "guess" to use for undetermined coefficients.

5. Nov 24, 2014

Thanks,
Chris