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
- does not change sign
- changes sign
- is positive
- is negative
Now,I have to decide which of the aforementioned options are correct ?
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?