Eigenvalues of an integral operator on $L^2[-1, 1]$ 
Find the eigenvalues of the integral operator $K: L^2[-1, 1] \to L^2[-1, 1]$ defined by $(Kx)(t) = \int_{-1}^1 (1 - 3t \tau)x(\tau) d\tau$.

I began with the fact that eigenvalues must be values of $\lambda$ that satisfy $\int_{-1}^1 (1 - 3t \tau)x(\tau) d\tau = \lambda x(t)$ for some $x(t) \in L^2[-1, 1]$, and reasoned that if $F(c) = \int_{-1}^c (1 - 3t \tau)x(\tau) d\tau$ is twice differentiable, then $F'(c) = x(c) - 3tx(c) - 3 \int_{-1}^c x(\tau) d\tau$ and $F''(c) = (1 - 3t)x'(c) - 6x(c)$, which makes the original equation $\lambda x''(t) = (1 - 3t)x'(1) - 6 x(1)$ if $x(t)$ is also twice differentiable.
First of all, I'm having trouble with solving the equation at the end of all that. It's been quite some time since I've actually dealt with differential equations, so my memory is rusty on how to do it. I'm inclined to just try and integrate both sides twice, but I'm not entirely sure what to make the bounds of integration. Should the left side be integrated something like $\lambda \int_{-1}^t \int_{-1}^\tau x''(z) dz d\tau$ (assuming the "just integrate both sides" strategy works at all)? 
Second, I'm not sure if my reasoning is right under the assumption that $F$ and $x$ are both twice differentiable; I'm trying to apply the fundamental theorem of calculus, but the first bound being $-1$ rather than $0$ is making me antsy. This might be a baseless fear since I can't quite find anything wrong, but I figure it's worth asking anyway.
Third, if I need to find all of the eigenvalues, is it sufficient to check for the functions "nice" enough for the above argument to work? Is it enough for me to do this work and then note that twice-differentiable functions are dense in $L^2[-1, 1]$, or do I need to do something else to be sure I have all of the eigenvalues?
 A: You seem to have missed the crucial property that the image of your operator is a linear function in $t$.  This means that the eigenfunction equation $K(x) = n x$ (I will write $n$ for eigenvalues) takes the following simple form: $$nx(t) = m_0 -3m_1 t$$ where $$m_0 = \int_{-1}^1 x(\tau)\,d\tau$$ and $$m_1 = \int_{-1}^1 \tau x(\tau)\, d\tau.$$
Thus we need to find all values $n$ for which a solution to this linear equation exists.  Note that since $x\in L^2$ the $m_0$ and $m_1$ exist, but in fact we are seeing that if $x(t)$ is an eigenfunction, it is a humble linear function, unless $n=0$.  Now, is $n=0$ an eigenvalue?  Yes.  You just need to find a nonzero function $x(t)$ for which $K(x)=0$.  I did the simplest thing: I tried $x(t) = At^2+Bt+C$ and tried to find $A,B,C$ that will make both $m_0$ and $m_1$ equal to zero.  I found $x(t) = -3t^2+1$.  There are many more.  But now we know $n=0$ is an eigenvalue.  Now we can assume $n\neq 0$.
We get  $$x(t) = \frac{m_0}{n} -3\frac{m_1}{n} t.$$  If $m_1=0$ (and then $m_0\neq 0$, otherwise we get the zero function) then $x(t)$ is a constant function $x(t)=c\neq 0$ and so $m_0=2c$ giving $c=2c/n$ so $n=2$ is an eigenvalue with constant eigenfunctions.
Now we assume $m_1\neq 0$, so the function is not constant.  We can exploit the fact that $t^2$ is even, so has non-zero integral: multiply both sides of the equation by $t$ and integrate: 
$$tx(t) = \frac{m_0}{n}t -3\frac{m_1}{n} t^2,$$
$$\int_{-1}^1 tx(t)dt =  \frac{m_0}{n}\int_{-1}^1tdt -3\frac{m_1}{n} \int_{-1}^1t^2 dt,$$
giving the equation
$$ m_1 = -\frac{2m_1}{n} $$ which forces $n=-2$; this eigenvalue actually occurs, for instance for $x(t)=t$.  
Since this analysis exhausts all linear functions $x(t)$ (i.e. either they are actually constants or genuinely linear), the possible eigenvalues we get for those are $2$ and $-2$.
Recapitulating: $n=0$ is an eigenvalue for a quadratic function $-3t^2+1$.  For $n\neq 0$ the function $x(t)$ must be linear; if it is actually constant, the only possible eigenvalue is $2$ which actually occurs for every constant function.  If genuinely linear, the only possible eigenvalue is $-2$, which again occurs for $x(t)=t$.
