Homework 6, due October 16, 2009.
Posted: Mon Oct 12, 2009 9:43 am
From Chapter 5 in the course text, problems 5.1-5.4.
5.5 is extra credit.
5.5 is extra credit.
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Yes, that's fine. lambda=1 is an ok choice.Apuente wrote:For problem 5.1, should we just use some arbitrary value for gamma when plotting?
I don't remember the details of this problem specifically, but there isn't necessarily anything wrong with this. a0 and a1 would give a2; a1 and a2 would give a3, etc.Flygirl89 wrote:1. When I am solve for the coefficients I am finding them in terms of multiple a's. For example
a(n+2) = something*a(n+1) +something*a(n-2)
Does this indicate that I am doing something incorrectly or is that typical? Specifically I was confused when attempting to find the range of convergence for problem 5.3.
You should be able to compute all the coefficients. Why the problem says a0 and a1 is because, I think, there are not initial conditions specified.2. In problem 5.4 the problem indicates that the general solution should be in terms of a0 and a1, does that mean exclusively in terms of a0 and a1?
Yes.3. For problem 5.5 do you have to convert the equation into three equations in order to calculate the numerical solution?
You need to expand those functions in a Taylor series too so you can set the coefficients of the powers of t to zero,mkiener wrote:For problem 5.4:
I put the sint and e^t terms into the sums, which would yield coefficients that were functions of t. I'm unsure about this approach since I'm confused if you can set the coefficients of the powers of t equal to zero since there are functions of t in there now, which might mess with the linear independence of the powers of t.
Pretty much everything applies to any order for this stuff.mkiener wrote:For Problem 5.5:
does the "Huge Theorem" that was given on Monday apply to third order DEs, since p and q are not analytic at t=0?
I'm not sure why you divided by t^2. I'm pretty sure you don't want to do that. In general, thoughk there isn't any problem with a negative starting index as long as it doesn't result in any negative powers of t or any coefficient less than a0.For problem 5.2 , I am getting m=-2 or m=-1 as the starting number for the summation term when I shift the index for non-t^n terms. I divided through by t^2 to get x by itself, so when I shifted the xdd term, sum(n(n-1)*a_n*t^(n-4)) from 1 to infinity, letting m=n-4, I get sum((m+4)*(m+3)*a_(m+4)*t^m) from -2 to infinity. How do I compute the summation that starts at a negative number if the series solution is supposed to start at a_0? Or am I just not approaching the problem in the right way?
You should not end up with negative powers of t or any a_n with an n<0. I don't see how you can get those since t^2 is multiplying x, and x starts at t^0 and goes up to higher powers, so t^2 x will then start with t^2. Maybe you are doing something wrong when shifting the summation indices. A way to check that you are doing it right is to write out the first few terms before and after you do it -- they should be exactly the same because what you are doing is just manipulating the index, but not changing the series.Flygirl89 wrote:In your response to the question about 5.2 you said
"I'm not sure why you divided by t^2. I'm pretty sure you don't want to do that. In general, though there isn't any problem with a negative starting index as long as it doesn't result in any negative powers of t or any coefficient less than a0."
If you do not divide by t^2 and compute the series with the method presented, you do end up with negative powers of t if you allow it to start with a negative starting index. Is there another method to calculate the coefficients?