I am just wondering for the ODE $$y^{\prime\prime} - \lambda y = 0,$$ with the given condition $y'(0)=0$.
Why is it that when $\lambda=0$, the eigenfunction is found to be $y_0=1$? i.e. the solution to that ODE when $\lambda=0$ is $y=1$.
Could someone explain why?
Also if the eigenfunction is $1$, does it mean any multiple of $1$ is also a solution? such as $y =1, y =-3, y =5, y = 9192, y =-1, y = 0$ (another question: does $y = 0$ actually give a valid solution)?
It comes because if you solve the ODE you get:
$$\begin{eqnarray} y'' - \lambda y & = & 0 \\ y'' & = & 0 & \text{since } \lambda = 0 \\ \int y'' \ dx & = & \int 0 \ dx \\ y' & = & A \\ & = & 0 & \text{since } y'(0) = 0 \\ \int y' \ dx & = & \int 0 \ dx \\ y & = & B \end{eqnarray}$$
So the eigenfunction of the second derivative operator, with the given boundary condition, is any constant function, as you suspected. (Without the boundary condition, the eigenfunctions would be of the form $y = Ax + B$.) It is common to choose a representative eigenfunction from which we can parametrise the full set, so choosing $y = 1$ just makes the numbers look nicer (and similarly, the family of functions $y = Ax + B$ can be written as a linear combination of $y = x$ and $y = 1$, although we could just as equally choose $y = 43x + 9$ and $y = 17 - x$ or some other pair of linearly independent equations).
As for your last question - does $y = 0$ give a valid solution? The answer is yes - if you differentiate it twice, you get $y'' = 0$ which clearly satisfies the original DE. We wouldn't choose it as our representative function just because you can't get any of the other solutions from it by linear combinations.