Well, that's pretty much the definition you will find in any book. I don't know exactly what your question is. Perhaps some examples would help.
The problem d^2y/dx^2+ \lambda y= 0 with boundary conditions y(0)= 0, y(1)= 0 has the "trivial solution" y(x)= 0 for all x. Does it have any other, "non-trivial", solutions?
The way we write the general solution depends on \lambda. If, for example, \lambda= 0, the equation is d^2y/dx^2= 0 and integrating twice,y= C_1x+ C_2. Then y(0)= C_2= 0 and y(1)= C_1+ C_2= C_1= 0 so both constants are 0 and y(x)= 0 for all x, the trivial solution. 0 is NOT an eigevalue.
If \lambda is negative then we can simplify the problem by writing \lambda= -\alpha^2 where \alpha is any non-zero number. The equation becomes d^2y/dx^2- \alpha^2 y= 0. That has characteristic equation r^2- \alpha= 0 which has roots r= \alpha and r= -\alpha so the general solution is y(x)= C_1e^{\alpha x}+ C_2e^{-\alpha x} so y(0)= C_1+ C_2= 0 and y(1)= C_1e^{\alpha}+ C_2e^{-alpha}= 0. From the first equation, C_2= -C_1. Putting that into the second equation and factoring out C_1, we have C_1(e^{\alpha}- e^{-\alpha}= 0. Because \alpha is not 0, \alpha and -\alpha are different and, since e^x is a "one-to-one" function, e^\alpha and e^{-\alpha} are different. (e^{\alpha}- e^{-\alpha} is not 0 so C_1 must be 0 which means C_2= -C_1 is also zero. That is y(0)= 0(e^{\alpha x})+ 0(e^{-\alpha x})= 0 for all x. Again, y(x)= 0 is the only solution so NO negative number is an eigenvalue.
If \lambda is positive, we can write \lambda= \alpha^2 so the equation becomes d^2y/dx^2+ \alpha^2y= 0 which has characteristic equation r^2- \alpha^2 with solutions r= i\alpha and r= -i\alpha so the general solution is of the form y(x)= C_1 cos(\alpha x)+ C_2 sin(\alpha x). Now, since cos(0)= 1 and sin(0)= 0, y(0)= C_1= 0. y(1)= C_2 sin(\alpha)= 0. If sin(\alpha) is not 0, then C_2 is 0 and again we get y(x)= 0 for all x. But sine is 0 for any multiple of \pi, so if \alpha is, say, n\pi, C_2 does NOT have to be 0: y(x)= C sin(n\pi x) satisfies both the differential equation and the boundary conditions. n\pi is an eigenvalue for for any integer n. The corresponding eigenfunctions are y(x)= C sin(n\pi x) for C any non-zero number.
Notice by the way, that all of this depends not just on the equation but also on the boundary conditions. Also, as I said, y(x)= 0 for all x always satifies this equation so being an eigenvalue requires that the solution NOT be unique.