If $y_1$ and $y_2$ satisfy

$y_1 '' + \lambda_1 y_1 = 0 \ \ \ y_1 ' (0) = 0, \ \ y_1(L) = 0$

$y_2 '' + \lambda_2 y_2 = 0 \ \ \ y_2 ' (0) = 0, \ \ y_2(L) = 0$

Without determining the explicit form of both $y_1$ and $y_2$, show that

$\int_0^L y_1 y_2 \ dx = 0$ if $\lambda_1 \neq \lambda_2$ .

2. Multiply the first by $y_2$ and the second by $y_1$ and substract so

$y_2 y_1'' - y_1 y_2'' + (\lambda_1-\lambda_2) y_1 y_2 = 0$.

Now integrate [0,L]

$\displaystyle \int_0^L y_2 y_1'' - y_1 y_2''dx + (\lambda_1-\lambda_2) \int_0^L y_1 y_2dx = 0,$

$\displaystyle y_2 y_1' - y_1 y_2'|_0^L + (\lambda_1-\lambda_2) \int_0^L y_1 y_2dx = 0$,

3. To start off, when this question asks without determining the explicit form of both $y_1$ and $y_2$, what does that mean exactly? In other words, what is the explicit form?

Anyway we have....

$y_2(L)y_1'(L) - y_1(L)y_2'(L) - y_2(0)y_1'(0) + y_1(0)y_2'(0) + (\lambda_1 - \lambda_2) \int_0^L y_1y_2 \ dx = 0$

Since all these BC's equal zero we therefore have

$\int_0^L y_1 y_2 \ dx = 0$ if $\lambda_1 \neq \lambda_2$

This is right, right?
Thanks very much for your help.

4. Exactly!