Homework 3, due September 22, 2006.

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Homework 3, due September 22, 2006.
From pages 87 and 88 in the course text, do problems 2, 3, 5, 6 and 10.
Bill Goodwine, 376 Fitzpatrick
Problem 6  Boundary value issue
Hi there!
I heard there were some issues with Problem 6, and in particular with the initial condition y(0)=0 that would end up in a division by zero error.
What I did to overcome this issue was the following: I got to the point where I have a differential equation in u, with a constant of integration C flying around. We cannot solve for C because we just have two boundary values for y, and no values at the derivative y' (aka u). In order to solve for y, I studied three cases (before carrying on the second integration), when C=0, >0 and less than 0.
In the first case, C=0, I ended up with a singularity/division by zero when setting y(0)=0 to find C2 ... so I concluded that there is no solution for C=O. For the other two cases the singularity is now gone, and you can "easily" solve for C using the 2 given boundary values. However, for the C>0 case I ended up with a nasty equation to solve for C ... it's not singular, but I could not solve it explicitly for C. If you can ... I would like if you could share your finding! (by the way ... Dr. Goodwine suggested me to discuss this problem here on line with all of you ... so feel free to do the same!) For the case C<0 the diff eq is actually easy ... and I ended up with an inverse trig function.
I might be wrong, so don't take my way as The way ... but just wanted to share it with you!
Good luck!
~Gian
P.S. Refer to example 2.8 in the text for a similar procedure!
I heard there were some issues with Problem 6, and in particular with the initial condition y(0)=0 that would end up in a division by zero error.
What I did to overcome this issue was the following: I got to the point where I have a differential equation in u, with a constant of integration C flying around. We cannot solve for C because we just have two boundary values for y, and no values at the derivative y' (aka u). In order to solve for y, I studied three cases (before carrying on the second integration), when C=0, >0 and less than 0.
In the first case, C=0, I ended up with a singularity/division by zero when setting y(0)=0 to find C2 ... so I concluded that there is no solution for C=O. For the other two cases the singularity is now gone, and you can "easily" solve for C using the 2 given boundary values. However, for the C>0 case I ended up with a nasty equation to solve for C ... it's not singular, but I could not solve it explicitly for C. If you can ... I would like if you could share your finding! (by the way ... Dr. Goodwine suggested me to discuss this problem here on line with all of you ... so feel free to do the same!) For the case C<0 the diff eq is actually easy ... and I ended up with an inverse trig function.
I might be wrong, so don't take my way as The way ... but just wanted to share it with you!
Good luck!
~Gian
P.S. Refer to example 2.8 in the text for a similar procedure!

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The main justification is that for an nth order equation, you are looking for n unknown functions. If you just substitute the assumed form of the solution into the differential equation, that gives you only one equation for n unknowns. Assuming that "condition" does two things: first it will result in n equations for the n unknown functions and second, it greatly simplifies the form of the next "level" of equations when you compute the next higher order derivative of the assumed form of the solution.I have some questions about the variable of parameters technique that you
mentioned in class:
In the book on page 78 they set each u prime and u double prime term to zero.
Why is that?
I don't really see any justification.
Bill Goodwine, 376 Fitzpatrick

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problem 6
The "x missing" approach to reduce it to a first order equation with y as a dependent variable does, in fact work. Check above for a hint on dealing with a complication that arises after you do that.I am having some trouble with problem 6. Could you give me a hint on setting up the problem? I tried an x absent method but it doesn't look like that is going to work.
Bill Goodwine, 376 Fitzpatrick

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Sometimes VoP may give extra terms that are actually homogeneous solutions (i.e., complementary functions). In such a case, you have still found a particular solution since it still satisfies the differential equation. Also, the general solution will still be correct since, when you add a linear combination of the complementary functions to it, the only effect will be to alter the values of the constants necessary to satisfy the initial conditions.I was wondering about my answer to problem five of the homework. I
solved the problem using both variation of parameters and undetermined
coefficients  however, the two methods results in different particular
solutions (to be exact, the method of variation of parameters produced
two terms in particular solution whereas undetermined coefficients only
produces one term  albeit the first term of the variation of
parameters). I was just wondering if you could talk about this in
class for a second tomorrow.
Bill Goodwine, 376 Fitzpatrick