# Solve using Undetermined Coefficients

1. Oct 27, 2011

### VitaX

$y''(t) - \frac{2}{t^2}y(t) = 3 - \frac{1}{t^2}$

In this problem I had to solve two ways: Variation of Parameters and Undetermined Coefficients. I solved it using Variation of Parameters and got the correct answer for the particular solution in the back of the book being $y_p(t) = t^2ln|t| + \frac{1}{2}$

I can't seem to get the same answer when I'm solving using Undetermined Coefficients. I let my assumption be $y_p(t) = At^2 + Bt + C$ but I only end up with $C=\frac{1}{2}$ while $A=0$ and $B=0$ when I go back and substitute into the original to find their values. What am I doing wrong here? Is my assumption incorrect? I somehow have to get a $ln|t|$ to match up with a $t^2$ as an A value then I should be able to get the same particular solution.

2. Oct 27, 2011

### LCKurtz

If you multiply that equation through by t2 you get
$$t^2y''(t) - y(t) = 3t^2 - 1$$

It isn't a constant coefficient equation for which you would use undetermined coefficients. One way of working it is to look for a solution in the form y = tr to get a characteristic equation. Alternatively you can change the variable from t to x by x = ln(t) to make a constant coefficient DE out of it. This type of equation is called an Euler equation. Surely your text has something about these.

3. Oct 28, 2011

### VitaX

Yeah it does, I think I just misheard the teacher in him saying use undetermined coefficients when I think he meant use 2 methods of variation of parameters to get the particular solution. At least I hope he said that as that's what I did.

4. Oct 28, 2011

### lurflurf

Undetermined coefficients works fine with variable coefficients, it is just harder to guess the form, and you guessed wrong. A better guess would be

$y_p(t) = At^2 \log(t) + Bt + C$

If it is hard to guess the form undetermined coefficients becomes a hard to use method.

5. Oct 28, 2011

### Fusiontron

Treating it as an Euler equation would be better. Or maybe try a series solution.