This is an example of a linear system of ODEs. If we have x'=Mx, where M is a matrix of coefficients and x is a vector of unknowns, then the substitution of x=Xe^{lambda*t), gives (M-lambda*I)X=0, from which we deduce that the condition for nonzero solutions for the constant vector, X, corresponds to solving det(M-lambda*I)=0. Thus the lambda values are the eigenvalues, the computed values of X are the eigenvectors, and (M-lambda*I) is the characteristic matrix.
In the problem which you are solving you have second derivatives and this is why lambda^2 appears, rather than lambda in my example. This author has also substituted e^{lambda*t) into (3.8) and written the consequence of that substitution in a matrix/vector form which is analogous to my (M-lambda*I)X=0.
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