Yes, that is correct. (Strictly speaking you should show that [itex]\lambda[/itex] must be greater than 1 to be an eigenvalue but that is fairly obvious.)
As for "solvable", yes, but not by any algebraic means.
Here's a plot of the tangent function and just plain ol' x. You can see that the solutions get closer and closer to odd-integer multiples of pi over two. The only way to find the solutions is numerically. Note, however, that you do have one exact known solution at lambda = 1.
Hmm. Typically, you'd find those via Fourier analysis, right? Trouble is, I'm not sure that your eigenfunctions here are orthogonal. If you look at the original DE, multiplying through by exp(2x) gets you a Stürm-Liouville form, to be sure. Your q(x)=0, and your weight function is exp(2x). So, that tells me that your eigenfunctions are orthogonal with respect to the weighted inner product
So you could do Fourier analysis that way.
Do you follow?
No, no. Your 's are the eigenfunctions that you found in the OP:
Stürm-Liouville theory gives you that those eigenfunctions are orthogonal w.r.t. to the weight function exp(2x). Hence, if you have an initial condition like
and you have the representation
then you're trying to solve the equation
for the 's. So do your Fourier analysis: multiply both sides by both your arbitrary eigenfunction and the weight function and integrate:
Because you have orthogonality of the eigenfunctions w.r.t. this weighted inner product, the RHS collapses down nicely to the following:
Hence, it follows that
That is what I call Fourier analysis.
Does that make sense?
[EDIT]: Double-check the signs of your DE with the form of the Stürm-Liouville type DE. I may have gotten a sign wrong, which might affect the weight function. The basic idea, however, is sound.