This is a follow up question to this one.
Just to summarize. I want to find the kernel, eigenvalues and eigenvectors of the differential operator:
$$L(x):=x''+3x'-4x$$
In other words I want to find the eigenvalue and eigenvector solutions to $$Lx=\lambda x$$
I can write:
$$x''+3x'-4x=\lambda x \iff x''+3x'-(4+\lambda)x$$
I can factorize the characteristic Polynomial $p_{\lambda}$ to get the eigenvalues:
$$p_{\lambda}=x^2+3x-(4+\lambda)$$
The roots of this polynomial are:
$$\frac{-3 \pm \sqrt{25+4\lambda}}{2}$$
Therefore the general solution is:
$$x(t)=c_1 e^{-\frac{1}{2}(\sqrt{25+4\lambda}+3)t}+c_2 e^{\frac{1}{2}(\sqrt{25+4\lambda}-3)t}$$
The kernel is the solution to the equation $Lx=0$. Substituting $\lambda=0$ into my equation yields:
$$\ker(L)=c_1e^{-4t}+c_2e^{t}$$
My question: For the general solution, do I have to check what happens when $25+4\lambda \le 0$ i.e when the radicand is zero or negative. Will that change my solution?
For the case $\lambda<-\frac{25}{4}$ you can keep the notation, if you allow the use of complex exponential. If you want to stick to real functions - which might be the case, seeing that the problem wants you to work in $C^\infty(\mathbb R)$ - you can use real and imaginary part of the complex exponential, writing the eigenvectors as: $$x(t)=e^{-\frac32t}\left(c_1\sin\left(t\sqrt{-\lambda-\frac{25}{4}}\right)+c_2\cos\left(t\sqrt{-\lambda-\frac{25}{4}}\right)\right)$$
For the case $\lambda=-\frac{25}{4}$, the eigenvectors are $$x(t)=(c_1t+c_2)e^{-\frac32t}$$
The last case cannot be put in the same notation you used for the case $\lambda>-\frac{25}{4}$, as far as I know.