Two-Point Boundary Value Problem - Differential Equations $y'' - \lambda y = 0$ where $y(0) + y'(0) = 0$ and $y(1) = 0$.
Find the possible eigenfunctions over the range of real values of $\lambda$.
Attempt:
We have three cases to check. The first one is if $\lambda = 0$. 
Case 1: $\lambda = 0$. Then, $y'' = 0$ which implies $y = c_{1}x + c_{2}$ and $y' = c_{1}$. Using the boundary conditions, $y(0) + y'(0) = 0$ implies $c_{2} + c_{1} = 0$, and $y(1) = 0$ implies $c_{1} + c_{2} = 0$. I am not sure what to conclude of this case.
Case 2: $\lambda > 0$. Then $y'' - \lambda y = 0$ has solutions $c_{1}e^{\sqrt{\lambda}x} + c_{2}e^{-\sqrt{\lambda}x}$. Using the boundary conditions, $y(0) + y'(0) = 0$ implies $c_{1} + c_{2} + c_{1}\sqrt{\lambda} - c_{2}\sqrt{\lambda} = 0$, and $y(1) = 0$ implies $c_{1}e^{\sqrt{\lambda}} + c_{2}e^{-\sqrt{\lambda}} = 0$. Again, I wasnt sure what to get out of this. 
Case 3: This case gives solutions in terms of sines and cosines.
The solution to this is $\phi_{0}(x) = 1 - x$ and when $n = 1,2,3, ...$ $\phi_{n}(x) = sin(\mu_{n} x) - \mu_{n} cos(\mu_{n} x)$ and $\lambda_{n} = -\mu_{n}^2$ where $\mu_{n}$ satisfies $\mu = tan(\mu)$.
Im thinking they substituted $\mu$ for lambda in the solution. I am not sure how to get to the solution.
 A: Case 1: $\lambda = 0$: 
You have $y = c_1 x + c_2$, which leads to:
$c_1+c_2 = 0$
$c_1+c_2 = 0$
This leads to $c_1=-c_2$,thus choose $c_2=1$, yielding $y =1-x$.
Case 2: $\lambda \gt 0$: 
You have $c_{1}e^{\sqrt{\lambda}x} + c_{2}e^{-\sqrt{\lambda}x}$, which leads to: 
$c_{1} + c_{2} + c_{1}\sqrt{\lambda} - c_{2}\sqrt{\lambda} = 0$, and  $c_{1}e^{\sqrt{\lambda}} + c_{2}e^{-\sqrt{\lambda}} = 0$
Lets write this in matrix form as:
$\begin{bmatrix}1+\lambda & 1-\lambda\\e^{\sqrt{\lambda}} & e^{-\sqrt{\lambda}}\end{bmatrix}.\begin{bmatrix}c_1\\c_2\end{bmatrix}=0$
If we calculate the determinant of this matrix, we get:
$$\displaystyle e^{-\sqrt{\lambda}}\lambda+e^{\sqrt{\lambda}}\lambda+e^{-\sqrt{\lambda}}-e^{\sqrt{\lambda}}$$
For $\lambda \gt 0$, is this value ever $0$? No, hence $c_1 = c_2=0$ and $y=0$.
Case 3: $\lambda \lt 0$:
You would get $y = c_1 \sin \sqrt{-\lambda}x + c_2 \cos \sqrt{-\lambda}x$ (note the $\lambda \lt 0$ in this case, so values are real and positive under the radical).
Now, try the boundary conditions and see if you get their result.
When doing these problems, you would need to merge the three solutions. $y=0$ is also a valid solution too, but the BCs may lead to another solution as you show in the answer above.
