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Find all real eigenvalue and eigenfunctions to the following:
(xy')' + rx^(1)y = 0
y'(0)=0 ; y(e^pi)=0
This is problem #19 to Fundamentals of Differential Equations and Boundary Value Problems 6th Edition.
 one year ago
 one year ago
Find all real eigenvalue and eigenfunctions to the following: (xy')' + rx^(1)y = 0 y'(0)=0 ; y(e^pi)=0 This is problem #19 to Fundamentals of Differential Equations and Boundary Value Problems 6th Edition.
 one year ago
 one year ago

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mahmit2012Best ResponseYou've already chosen the best response.1
x2y"+xy'+ry=0 a2+r=0 r should be w2 so a=+j(or i) y_1=coshlnwx y_2=sinhlnwx
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
y'(1)=0 so y_1 is eigen function y(epi)=0 so cospiw=0 then w=n1/2 then r=(n1/2)^2 are eigenvalue. That's it !
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
and fix your question y"(0)=0 change to y'(1)=0 !
 one year ago

jason432Best ResponseYou've already chosen the best response.0
your answer matches what is in the back of my book but i am confused as to how you did it in such a short manner. my professor always does a 3 case approach to these problems and they take about a page to solve.
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
Yes. I solved it with my own concept. That is different with all and your teacher also. So you just follow my way if you interested in, and be sure that is the best solution you can understand. r=0 and r=w2 never work for eigenvalue problem.
 one year ago

jason432Best ResponseYou've already chosen the best response.0
your way seems really convenient and short but i would truly appreciate it if you could go about solving the problem in the conventional manner as well. this technique is a bit foreign to me and i would want to practice a technique that i can reproduce during a test! thank you for your time. youre really helpful
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
Don't forget that eigenvalue problem always have result sin cos J(bessel) and H(Hilbert) and P(Logendr) and like them. But for any exam teachers give a sin and cos function because you can not find eigenvalue for others. r or lamda are chosen to get sin or cos functions then you should put constrain value to get r or Lamda as a eigenvalue.
 one year ago

jason432Best ResponseYou've already chosen the best response.0
i believe i am starting to understand what you are saying. Do you think you would be able to work this problem out the long way and attach a photo as an image? i am still unclear on a few steps in the process. i am currently reattempting it
 one year ago

jason432Best ResponseYou've already chosen the best response.0
my question was just answered on yahoo answers, but it doesnt match the same one you provided. Could you take a look and see what is wrong with it?? http://answers.yahoo.com/question/index?qid=20121117195712AAT2Ni6
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.1
it is not correct! he or she add k I don't know why? but he also concerned as always for r=0 and r=a2(or w2) I always suppose that I know they are not work, so I deal with the right answer. And why it was not correct because x2y+xy'+w2y=0 has solution sinh(w*lnx) and cos also But he wrote sinwlnx/x coswlnx/x both are not right> '
 one year ago

jason432Best ResponseYou've already chosen the best response.0
i still really dont understand the formal way of going about this. i am trying to use paul's online notes but i just cant seem to get a firm grasp on this
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
@mahmit2012 is an excellent mathematician, but his approaches can be very hard to decipher. I would have to review this topic to help you understand the conventional method, which I will do if/when I have time.
 one year ago

jason432Best ResponseYou've already chosen the best response.0
thank you so much! his answer is spot on according to my book, but his method was just a bit different then what i was used to. i truly appreciate any help you can provide me!
 one year ago
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