PDE Problem on Eigenvalues and the Heat Equation Let $a$ be any fixed real number. Consider the eigenvalue system:
$$
\begin{cases}
X'' + \lambda X = 0, & 0 ≤ x ≤ 1\\[0.1cm]
aX(0) = X(1)\\[0.1cm]
aX'(0) = -X'(1)
\end{cases}.
$$
Prove that if $a = \pm 1$, then every real $\lambda$ is an eigenvalue. 
Prove that if $a \not= 1$, then no real $\lambda$ is an eigenvalue.
Approach: 
So I looked for solutions of the ODE in the form:
$$X(x) = e^{bx}$$
Where I need to solve for b. Differentiation of $e^{bx}$ just multiplies the function by b, so then $X(x)$ becomes 
$(b^2 + λ)*e^{bx} = 0$. And since $e^{bx}$ cannot be zero, 
$$b^2 + λ = 0$$
So, $b=\pm\sqrt{-λ}x$
This gives the general form of
$X(x) = c_1e^{\sqrt{-\lambda}x} + c_2e^{-\sqrt{-\lambda}x}$
Where $X'(x) = c_1\sqrt{-\lambda}e^{\sqrt{-\lambda}x} + c_2\sqrt{-\lambda}e^{\sqrt{-\lambda}x}$ is the 2nd derivative. Plugging in the boundary conditions gives:
$X(0) = c_1+c_2$
$X'(0) = c_1-c_2$
$X(1) = c_1e^{\sqrt{-\lambda}} + c_2e^{-\sqrt{-\lambda}}$
$X'(1) = c_1\sqrt{-\lambda}e^{\sqrt{-\lambda}} - c_1\sqrt{-\lambda}e^{\sqrt{-\lambda}}$
I then tried solving for $a$
$a = \frac{X(1)}{X(0)} = \frac{c_1e^{\sqrt{-\lambda}} + c_2e^{-\sqrt{-\lambda}}}{c_1+c_2}$
$a = \frac{X'(1)}{X'(0)} = \frac{c_1\sqrt{-\lambda}e^{-\sqrt{-\lambda}} - c_2\sqrt{-\lambda}e^{-\sqrt{-\lambda}}}{c_1-c_2} $
 A: Note that $b^2+\lambda=0$ has solutions $b=\pm\sqrt{-\lambda}$. So the general solution is
$$ X(x)=C_1e^{\sqrt{-\lambda}x}+C_2e^{-\sqrt{-\lambda}x}. $$
Using the initial conditions $aX(0) = X(1),aX'(0) = -X'(1)$, one has
$$ 
a(C_1+C_2)=C_1e^{\sqrt{-\lambda}}+C_2e^{-\sqrt{-\lambda}},\tag{1}$$
and
$$a\sqrt{-\lambda}(C_1-C_2)=-\sqrt{-\lambda}(C_1e^{\sqrt{-\lambda}}-C_2e^{-\sqrt{-\lambda}}).\tag{2}$$
The second equation above can be simplified as
$$ a(C_1-C_2)=-C_1e^{\sqrt{-\lambda}}+C_2e^{-\sqrt{-\lambda}}. \tag{3}$$
Now (1)+(3) and (1)-(3) give
$$ aC_1=C_2e^{-\sqrt{-\lambda}x}, aC_2=C_1e^{\sqrt{-\lambda}x}$$
from which it is easy to see
$$ (a^2-1)C_1C_2=0.$$
Case 1: $a^2-1=0$ or $a=\pm1$. Then $X(x)=C_1e^{\sqrt{-\lambda}x}+C_2e^{-\sqrt{-\lambda}x}$ is an eigenfunction (at least one of $C_1,C_2$ is not zero) with eigenvalue $\lambda$.
Case 2: $a^2-1\neq 0$ or $a\not=\pm1$. Then from (4), one has $C_1C_2=0$. So one of $C_1,C_2$ is zero. Suppose $C_1=0$. Then (1) and (2) become
$$ 
aC_2=C_2e^{-\sqrt{-\lambda}},-aC_2=C_2e^{-\sqrt{-\lambda}} $$
from which one has $C_2e^{-\sqrt{-\lambda}}=0$ or $C_2=0$. Thus $X(x)=0$. Similarly $C_2=0$ implies $C_1=0$. Thus $\lambda$ is not an eigenvalue.
