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
X'+2X'+(\lambda-\alpha)X=0, 0<x<1
X(0)=0
X'(1)=0

a. Is \lambda=1+\alpha 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]

X''+2X'+(\lambda-\alpha)X=0

let X=e^{rt} implies

r^2+2r+(\lambda-\alpha)=0

so

r=-1\pm\sqrt{1-(\lambda-\alpha)}

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

X(0)=A+B

so A=-B[/tex]<br /> <br /> X&amp;#039;(1)=A[r_{+}e^{r+}-r_{-}e^{r_{-}}]=0<br /> <br /> A=0 would lead to the trivial solution, so <br /> <br /> r_{+}e^{\sqrt{1-(\lambda-\alpha)}}=r_{-}e^{-\sqrt{1-(\lambda-\alpha)}}<br /> <br /> \lambda=1+\alpha is clearly an eigenvalue<br /> <br /> and \lambda_{m} satisfy the equation<br /> <br /> \frac{-1+\sqrt{1-(\lambda_{m}-\alpha)}}{-1-\sqrt{1-(\lambda_{m}-\alpha)}}e^{2\sqrt{1-(\lambda_{m}-\alpha)}}=1