Solve X'+2X'+(\lambda-\alpha)X=0 | Chris Struggling

[Chris_K]

I'm having trouble getting started on this problem... I just really don't understand what to do.


Solve
[tex]X'+2X'+(\lambda-\alpha)X=0, 0<x<1[/tex]
[tex]X(0)=0[/tex]
[tex]X'(1)=0[/tex]

a. Is [tex]\lambda=1+\alpha[/tex] an eigenvalue? What is the corresponding eigenfunction?
b. Find the equation that the other eigenvalues satisfy.


I appreciate any help you can give me!

Thanks,
Chris

 

[ReyChiquito]

are u sure that the edo is correct?

you have there a linear second order d.o with constant coeficients and initial values, so your solution will be some linear comb. of exp[rt] (you have to calculate r of course).

i don't really understand why would you end with an eigenvalue problem in this way but maybe I am not well informed.

 

[Chris_K]

Yeah, everything on there is correct. I'm not sure what you mean though...


nm... figured it out.

 

[ReyChiquito]

[tex]X''+2X'+(\lambda-\alpha)X=0[/tex]

let [itex]X=e^{rt}[/itex] implies

[tex]r^2+2r+(\lambda-\alpha)=0[/tex]

so

[tex]r=-1\pm\sqrt{1-(\lambda-\alpha)}[/tex]

[tex]X(t)=Ae^{(-1+\sqrt{1-(\lambda-\alpha)})t}+Be^{(-1-\sqrt{1-(\lambda-\alpha)})t}[/tex]

[tex]X(0)=A+B[/tex]

so [itex]A=-B[/tex]

[tex]X'(1)=A[r_{+}e^{r+}-r_{-}e^{r_{-}}]=0[/tex]

[itex]A=0[/itex] would lead to the trivial solution, so

[tex]r_{+}e^{\sqrt{1-(\lambda-\alpha)}}=r_{-}e^{-\sqrt{1-(\lambda-\alpha)}}[/tex]

[itex]\lambda=1+\alpha[/itex] is clearly an eigenvalue

and [itex]\lambda_{m}[/itex] satisfy the equation

[tex]\frac{-1+\sqrt{1-(\lambda_{m}-\alpha)}}{-1-\sqrt{1-(\lambda_{m}-\alpha)}}e^{2\sqrt{1-(\lambda_{m}-\alpha)}}=1[/tex]