The eigen function boundary value problem is ($y'=\frac{dy}{dx}$ and similar for $y''$) $$ y'' - \lambda y = 0,\\ y'(0) = y(2)=0 $$ I think the solution is like this
Case 1: $\lambda <0 $ putting $ \lambda = -m^2 \implies y = C_1 cos(mx) + C_2 sin(mx)$
Case 2: $\lambda >0 $ putting $ \lambda = m^2 \implies y = C_3 e^{mx} + C_4 e^{-mx}$
Case 3: $\lambda =0 \implies y = C_5 x + C_6 $
Case 1 gives $$ C_2=0\\ C_1 cos(2m)=0 \implies m=\frac{(2n+1) \pi}{4},n=0,1,2,3.... $$ so, case 1 eigenfunction is $$ y_n=C_n cos \bigg ( \frac{(2n+1) \pi}{4} x\bigg ) $$
Case 2 gives $$ y = C_3 e^{mx} + C_4 e^{-mx}\\ y' = C_3 m e^{mx} + C_4 m e^{-mx} \\ y'(0)=0 \implies C_3 m - C_4 m = 0 \implies C_3 - C_4 = 0, m=0 \\ y(1)=0 \implies C_3 e^{2m} + C_4 =0 $$
Here I am stuck at last two conditions. What is eigenvalue and eigenfuction from the last two conditions?
Case 3 gives $$ C_5 = C_6 =0 $$ So, are there no eigenvalues and eignefunctions?
Any help or suggestion are appreciated.
Maybe this helps.
\begin{eqnarray*} -\partial _{x}^{2}y(x) &=&-\lambda y(x) \\ \int_{0}^{2}dx\{-\partial _{x}^{2}y(x)\}y(x) &=&\left[ \{-\partial _{x}^{{}}y(x)\}y(x)\right] _{0}^{2}+\int_{0}^{2}dx\{\partial _{x}y(x)\}^{2} \\ &=&\int_{0}^{2}dx\{\partial _{x}y(x)\}^{2}=-\lambda \int_{0}^{2}dxy(x)^{2} \end{eqnarray*}
Since the left hand side is non-negative and if $y(x)$ not identically $0$, $\lambda $ cannot be positive. If it is $0$ and $y(x)$ not identically zero then $\partial _{x}y(x)$ must vanish for all $x$ leaving constant $y(x)$ and hence, given $y(2)=0$, vanishing $y(x)$.