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I was looking at previous year exam papers and was stuck on the following problem:

For the boundary value problem, $\,\,y''+\lambda y=0; y(0)=0,y(1)=0, \,\,\exists$ an eigenvalue $\lambda$ for which there corresponds an eigenfunction in $(0,1)$ that

  1. does not change sign
  2. changes sign
  3. is positive
  4. is negative

Now,I have to decide which of the aforementioned options are correct ?

My Attempt:

The general solution is :
$y(x)=a\cos(\sqrt \lambda x)+b\sin(\sqrt \lambda x).$

Then we apply the boundary values. From $y(0)=0$,we get $a=0.$

From $y(1)=0,a=0\,\,$ we get $\,\,0=b\sin(\sqrt \lambda )$.

We assume that $b \neq 0$ so that $\,\,0=b\sin(\sqrt \lambda )$.Then we have,$\sqrt \lambda=n \pi \implies \lambda =(n \pi)^2$.Let's write $\lambda_n=(n \pi)^2.$ Since we have $a=0,$ only sine term remains,so eigenfunctions are $y_n=\sin(\sqrt \lambda_n x)\,\,$ with eigenvalues $\lambda_n=(n \pi)^2,\,\,\,\, n=1,2,3, \dots$.

Now,when I look at the options ,I can not progress about which way to go. Can someone point me in the right direction keeping in mind for deciding which options are correct?

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2 Answers 2

You've got it! What can you say about the function $y_1(x)=\sin(\pi x)$ on the interval $(0,1)$? What about scalar multiples for various scalars? And what about $y_2(x)$? P.S. I find the problem confusing. I interpret it as $4$ separate T/F questions.

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You missed the case $\lambda<0$, which nevertheless does not have an eigenvalue!

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