Can anyone help with eigenfunction?

[samleemc]

if y''-2y'+y=ky k=eigenvalue, y(0,pi)=0, 0<x<pi
find corresponding eigenvalues and eigenfunctions.

thx a lot!

 

[elibj123]

How would you solve the ode

y''-2y'+(1-k)y=0

Regarding k as a parameter for the moment

 

[samleemc]

I got this

sqrt{k} = m
A(1-m)² -2A(1-m)+A(1-k)=0
B(1+m)² -2B(1+m)+B(1-k)=0

and what then !??

Please help! thank you very much!

 

[samleemc]

can anyone help!?

 

[Jerbearrrrrr]

It's a 2nd order linear ODE with constant coefficients.
Trying the solution y=e^mx (as I think you've done) gets your solutions (unless...etc, look it up).
Solve for m, then you have one function that's an eigenfunction, with eigenvalue corresponding to m.

 

[HallsofIvy]

if y''-2y'+y=ky k=eigenvalue, y(0,pi)=0, 0<x<pi
find corresponding eigenvalues and eigenfunctions.

thx a lot!

The characteristic equation for y"- 2y'+ (1- k)y= 0 is $r^2- 2r+ 1-k= 0$ which is the same as $r^2- 2r+ 1= (r- 1)^2= k$ and has roots $r= 1\pm \sqrt{k}= 1\pm m$ with your choice of m as $\sqrt{k}$.

The general solution is $y= Ae^{(1+m)t}+ Be^{(1-m)t}$
Setting that equal to 0 at x= 0 and $\pi$, we find that as long as 1+ m and 1- m are real, A and B must be 0.

In order for k to be an eigenvalue, k will have to be negative so that m is imaginary. Given that, and writing m= ni, $y= Ae^{(1+ni)t}+ Be^{(1-nit)}= e^t(Ae^{nit}+ Be^{-nit})$. We can write that as $y= e^t(C cos(nt)+ D sin(nt))$.

Now we have $y(0)= e^0(Ccos(0)+ D sin(0))= C= 0$ and $y(\pi)= e^{\pi}(Ccos(n\pi)+ Bsin(n\pi)= Be^{\pi}sin(n\pi)$ (because C= 0) and that must be equal to 0. That will be true either for B= 0 or for $sin(n\pi)= 0$ which will be the case as long as n is an integer.

Can you find the eigenvalues and eigenvectors from there?

 

[samleemc]

The characteristic equation for y"- 2y'+ (1- k)y= 0 is $r^2- 2r+ 1-k= 0$ which is the same as $r^2- 2r+ 1= (r- 1)^2= k$ and has roots $r= 1\pm \sqrt{k}= 1\pm m$ with your choice of m as $\sqrt{k}$.

The general solution is $y= Ae^{(1+m)t}+ Be^{(1-m)t}$
Setting that equal to 0 at x= 0 and $\pi$, we find that as long as 1+ m and 1- m are real, A and B must be 0.

In order for k to be an eigenvalue, k will have to be negative so that m is imaginary. Given that, and writing m= ni, $y= Ae^{(1+ni)t}+ Be^{(1-nit)}= e^t(Ae^{nit}+ Be^{-nit})$. We can write that as $y= e^t(C cos(nt)+ D sin(nt))$.

Now we have $y(0)= e^0(Ccos(0)+ D sin(0))= C= 0$ and $y(\pi)= e^{\pi}(Ccos(n\pi)+ Bsin(n\pi)= Be^{\pi}sin(n\pi)$ (because C= 0) and that must be equal to 0. That will be true either for B= 0 or for $sin(n\pi)= 0$ which will be the case as long as n is an integer.

Can you find the eigenvalues and eigenvectors from there?

k have to be negative and n have to be integer, do u mean k=0 ?!
Please answer ! Thanks !

 

[elibj123]

k have to be negative and n have to be integer, do u mean k=0 ?!
Please answer ! Thanks !

k=m²=-n²

Therefore, any k that is a square of an integer (with a minus in front) is an eigenvalue to this problem.