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 [itex]A=-B[/tex]<br /> <br /> $$X'(1)=A[r_{+}e^{r+}-r_{-}e^{r_{-}}]=0$$<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 /> $$\frac{-1+\sqrt{1-(\lambda_{m}-\alpha)}}{-1-\sqrt{1-(\lambda_{m}-\alpha)}}e^{2\sqrt{1-(\lambda_{m}-\alpha)}}=1$$[/itex]