# Series Solutions of ODE's

1. Nov 10, 2011

### Angry Citizen

So.. I need.. 'help'. Meh. It seems I just don't understand what's going on with series solutions. I thought I did, but I just don't. It seems to be a fundamental misunderstanding of what the book and professor are trying to convey. Could someone please give me a rundown on what's going on and how to do them, both around ordinary points and singular points? I wish I could be more specific as far as questions, but I just don't know why I can't do them.

Thanks.

2. Nov 10, 2011

### jackmell

Know what, I use to play tennis. But my back-hand use to suck. So for a while I just hit it solely back-hand. Yeah, it went all over the place . . . for a little while. But then something happened. I started getting good at it and before long it didn't suck no more. Good cooks try again when they mess up a recipie and pretty soon they cookin' up something good. How they get so good? They're willing to try and fail and then try some more. So get out there, knock that ball all over the place and pretty soon you'll be cookin' up something nice too.

3. Nov 10, 2011

### Shane-O

What a statement.

4. Nov 10, 2011

### Angry Citizen

Problem is, I ran out of problems to practice with in my book - and I succeeded in solving precisely zero. I know practice makes perfect, but when you're failing miserably at every single one, you realize that no amount of problems are going to make any sense out of the material.

In a sense, it's like if you were trying to practice backhanding tennis balls ... if they were thrown from two states away.

5. Nov 10, 2011

### jackmell

A valuable lesson in the art of problem solving is when you run into problems with a problem, put it on the back-burner and work on something simpler and then start building back up to the original problem. So surely your book has some worked examples. Go over those and then change one only slightly. Say one in the book is:

$$y''+(2x+3)y'+4y=0$$

Ok, get that one straight then change it slightly like:

$$y''+(4x+3)y'+4y=0$$

Now do that on on your own. Bet you could if you got the first one straight.

Now, little more complex:

$$y''+(4x^2+3)y'+4y=0$$

Do that one. Same dif with the singular ones like:

$$(x^2+4x+2)y''-(2x+4)y'+3x^2y=0$$

First drop everything but just one x:

$$xy''+y'+y=0$$

Or if that one is too hard, move it over:

$$y''+y'+xy=0$$

Too hard?

$$y''+xy=0$$

Keep dropping it down until you get it, then start building it back up.