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 [quote="kevinatilusa"]Right, you start by assuming you have a solution looking like f=acos(x)+bsin(x). Differentiating twice, f'=a'cos(x)-asin(x)+b'sin(x)+bcos(x) f''=a''cos(x)-2a'sin(x)-acos(x)+b''sin(x)+2b'cos(x)-bsin(x) f+f''=a''cos(x)+b''sin(x)-2a'sin(x)+2b'cos(x) I got some cancellation already because I picked cos and sin to satisfy g''+g=0. We can still get a little more, however. Right now we have two unknown functions (a and b), but we're only looking for one solution. This means we can stick in an extra equation, require our functions to satisfy BOTH equations, and still have a solution. The second equation I'm going to pick here is that a'cos(x)+b'sin(x)=0. I'm choosing this so as to get cancellation early in my argument, but for a different problem I could always replace cos and sin by the solutions to the homogenous equation. Assuming this second equation is satisfied, we now have f'=a'cos(x)-asin(x)+b'sin(x)+bcos(x)=bcos(x)-asin(x) f''=-bsin(x)+b'cos(x)-acos(x)-a'sin(x), and adding f in gives f+f''=[b]b'cos(x)-a'sin(x)=1/cos(x)[/b] Which must be solved simultaneously with [b]a'cos(x)+b'sin(x)=0[/b] Doing so gives me that b'=1, a'=-tan(x) I can now integrate and get [b]b=x, a=ln|Cos(x)|[/b]. The same method works for any 2nd order equation once you've found the homogenous solutions. The only hangup sometimes is actually doing the integrals in the last step.[/quote]
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Antrax
Posted: Thu Jul 15, 2004 9:08 am    Post subject: 1

ah, so that's what I was missing. Instead of just adding an equation, I saw I had a''cos(x), and knew if I picked a to be 1/cos 2 (x) then a' is tg and maybe I can work from there. That's why I referred to it as guessing
Antrax
kevinatilusa
Posted: Thu Jul 15, 2004 7:31 am    Post subject: 0

Right, you start by assuming you have a solution looking like

f=acos(x)+bsin(x).

Differentiating twice,

f'=a'cos(x)-asin(x)+b'sin(x)+bcos(x)
f''=a''cos(x)-2a'sin(x)-acos(x)+b''sin(x)+2b'cos(x)-bsin(x)

f+f''=a''cos(x)+b''sin(x)-2a'sin(x)+2b'cos(x)

I got some cancellation already because I picked cos and sin to satisfy g''+g=0. We can still get a little more, however. Right now we have two unknown functions (a and b), but we're only looking for one solution. This means we can stick in an extra equation, require our functions to satisfy BOTH equations, and still have a solution.

The second equation I'm going to pick here is that a'cos(x)+b'sin(x)=0. I'm choosing this so as to get cancellation early in my argument, but for a different problem I could always replace cos and sin by the solutions to the homogenous equation. Assuming this second equation is satisfied, we now have

f'=a'cos(x)-asin(x)+b'sin(x)+bcos(x)=bcos(x)-asin(x)
f''=-bsin(x)+b'cos(x)-acos(x)-a'sin(x), and adding f in gives

f+f''=b'cos(x)-a'sin(x)=1/cos(x) Which must be solved simultaneously with
a'cos(x)+b'sin(x)=0

Doing so gives me that b'=1, a'=-tan(x)

I can now integrate and get b=x, a=ln|Cos(x)|.

The same method works for any 2nd order equation once you've found the homogenous solutions. The only hangup sometimes is actually doing the integrals in the last step.
Antrax
Posted: Wed Jul 14, 2004 5:44 am    Post subject: -1

kevin,

Basically we were taught to find a solution to the homogenous equation, then add, uh, functions? to solve the non-homogenous. So you get:

y h =c 1 cos(x)+c 2 sin(x)
Here I tried thinking what kind of function I'd add to that so I'll get sec(x), but came up completely blank -- in the exercises I did solve, on the right we had things like x 2 e x , or other easily guessable things (where the most difficult problem was if you got something that also corresponds to a solution of the homogenous, then you pad it with x or something).
So, after being stumped on that, I used paramter variation, and looked for
y p =c 1 (x)cos(x)+c 2 (x)sin(x)
After deriving, I got a fairly nasty equation, and from there I just sorta guessed my way -- "hey, if c 1 ' would be 1/cos 2 (x) then I'll get my sec(x) and then I could say b is ln..." stuff. I was wondering if there's a more analytic approach.
kevinatilusa
Posted: Wed Jul 14, 2004 2:44 am    Post subject: -2

I think the right way depends on what tools you have/are supposed to use.

Normally what I would do with an equation like this is find all solutions to y''+y=0, and add on ONE solution to y''+y=sec(x).

Unfortunately, I don't know of an intuitive way to just "guess" any one solution. There is a method called "variation of parameters" which works though, and using it I get as my particular solution cos(x)Ln|cos(x)|+xsin(x), so general solutions look like cos(x)Ln|cos(x)|+xsin(x)+A*cos(x)+B*sin(x)
Huey
Posted: Tue Jul 13, 2004 11:44 pm    Post subject: -3

err

- y on both side.

Then take the intergrals of both sides.

And look up what the intergral of (1/cos x ) - y is on the cheat sheet you brought along with you.
Antrax
Posted: Tue Jul 13, 2004 10:39 pm    Post subject: -4

From my final. I solved it using a lot of intuition, and I wonder what the right way was.
Solve:
y''+y=1/cos(x)

Antrax