# How to find solution of this eigenfuction?

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.

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 $y(x)$.