show directly That the eigenfunction of the given Sturm-Liouville problem are orthogonal without explicitly solving for therm , and state the orthogonality conditions
${y}''+\lambda (1+x)y=0$
$y(0)=0$
${y}'(1)=0$
I'm solving it this way
$r^{2}+\lambda (1+x)=0$
$r^{2}=-\lambda (1+x)$
$r=\sqrt{-\lambda (1+x)}$
$\lambda (1+x)=k$
$r=\sqrt{-k}$
$r=\sqrt{k}i$
How do I use the suggestion you give me?. I've to confess that I'm a bit lost, can anyone give me a hint ?
Suppose you have a solution $y_1$ with $\lambda=\lambda_1$ and a solution $y_2$ with $\lambda=\lambda_2$ and $\lambda_1-\lambda_2 \ne 0$. Then the eigenfunctions are orthogonal with respect to the weight function $1+x$, but not necessarily with respect to the weight function $1$: $$ (\lambda_1-\lambda_2)\int_0^1 y_1(x)y_2(x)(1+x) dx = \\ =\int_0^1 y_1y_2''-y_1''y_2 dx \\ = \int_0^1(y_1y_2'-y_1'y_2)'dx \\ = y_1y_2'-y_1'y_2|_0^1 =0. $$