Question

Trouble with eigen value problems. I don't get how to do these when the given conditions are on different derivatives, i.e. y(0) and y'(pi).

Find all real eigen values and eigen functions for the following problem:

y'' + (r-2)y = 0
y(0)=y'(pi)=0

Answer 1

r is a constant?

Answer 2

\[y''=-k^2y\]\[y=(A+iB)e^{ikt}\]\[y=(Acos(kt)-Bsin(kt))\]\[y=(Acos(\sqrt{2-r}t)-Bsin(\sqrt{2-r}t))\]

\[y(0)=0\]
\[ \Rightarrow (Acos(\sqrt{2-r}t0)-Bsin(\sqrt{2-r}0))=A=0\]\[\Rightarrow A=0\]

Answer 3

\[y'=(-A\sqrt{2-r}sin(\sqrt{2-r}t)-B\sqrt{2-r} cos(\sqrt{2-r}t))\]
\[y'(0)=0\]
\[ \Rightarrow (-A\sqrt{2-r}sin(\sqrt{2-r}0)-B\sqrt{2-r} cos(\sqrt{2-r}0))=-B\sqrt{2-r}=0\]\[\Rightarrow B\sqrt{2-r}=0\]

Answer 4

in the problem the r was actually a lambda. i rewrote it as r for ease in typing! thanks for the continued help =]

Answer 5

Is that the format of the answer you were looking for?

Answer 6

in the examples that my professor has done in class, there has always been a 3 case approach to solving these problems: when r=0 ; r >0 ; and r<0. these cases will either produce trivial solutions or eigenvalues.

Answer 7

it seems you only performed one case. did you just not show the others because they were trivial?

Answer 8

I think all cases are contained in \[B \sqrt{2-r}=0\]
Either r=2 and B=anything or B=0 and r=anything

r cannot be >2 as you asked for real solutions

Answer 9

I'm not sure which counts as trivial (probably B=0, as then y=0)

Answer 10

is there any reason you did not use the second given condition, y'(pi)=0?

Answer 11

Sorry, I misread that as y'(0)=0

Answer 12

It doesn't affect it though

Answer 13

as sin(pi)=sin(0)=0 and cos (pi)=-cos(0)

Answer 14

that makes sense to me now. i am having a bit of trouble understanding how you got to your first step tho. would you mind if i uploaded a photo of how i usually approach these problems?

Answer 15

I defined it like that to make the DE easier

Answer 16

as you can see, its usually a pretty lengthy process. im just not sure how you were able to solve it so fast

Answer 17

no i did not. i believe in this problem the variable r could be used to represent x if wanted. when simplified, the problem comes to (xy')' + ry - 2y = 0

Answer 18

So is r not a constant?

Answer 19

i dont believe so. the questions does not explicitely state it as such, so i would assume it to be a variable

Answer 20

Sorry, I assumed this

Answer 21

oh dear...

Answer 22

no problem! do you know how to go about it with it being a variable?