# Solve using Undetermined Coefficients

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

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.

## Answers and Replies

LCKurtz
Science Advisor
Homework Helper
Gold Member
$y''(t) - \frac{2}{t^2}y(t) = 3 - \frac{1}{t^2}$

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.

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.

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.

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.

lurflurf
Homework Helper
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.

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