- #1
BlkDaemon
- 15
- 0
I'm creeping my way through DiffEq, and recently started reading Paul Dawkins' PDF, which is actually pretty helpful. He does, however, tend to assume that his readers know how to approach what he calls simple problems.
Well, one of 'em has me stumped.
2t*y' + 4y = 3
I need to find the general solution. Here's what I tried:
1) I tried isolating y. I get y=(3 - 2t*y')/4. That doesn't work because I've got y' as part of the solution. Eeek.
2) I tried isolating y'. That didn't work either, for the same reason. I got
y' = (3-4y)/2t
But if I take the derivative of that, I get weirdness. Now, it's really possible that I'm just calculating the derivative incorrectly.
I also can't figure out how to ditch the "t", but when I look at the answer, there's t, as well as C, so I know that (because of the arbitrary constant) there's a derivative involved, and that means the "t" isn't going anywhere.
Anyone have any suggestions on how to work this? I don't need the answer. I need to know how to approach the problem.
Thanks in advance.
Well, one of 'em has me stumped.
2t*y' + 4y = 3
I need to find the general solution. Here's what I tried:
1) I tried isolating y. I get y=(3 - 2t*y')/4. That doesn't work because I've got y' as part of the solution. Eeek.
2) I tried isolating y'. That didn't work either, for the same reason. I got
y' = (3-4y)/2t
But if I take the derivative of that, I get weirdness. Now, it's really possible that I'm just calculating the derivative incorrectly.
I also can't figure out how to ditch the "t", but when I look at the answer, there's t, as well as C, so I know that (because of the arbitrary constant) there's a derivative involved, and that means the "t" isn't going anywhere.
Anyone have any suggestions on how to work this? I don't need the answer. I need to know how to approach the problem.
Thanks in advance.