on repeated eigenvalues of systems of differential equations

Question

I am trying to understand the principle underlying the general solution for repeated eigenvalues in systems of differential equations. (assume the eigenvalue isn't 0). The general solution is Y(t) = e^(a*t)*V0 + t*e^(a*t)*V1 such that a is the repeated eigenvalue and V1=(A-a*I)V0 (A=2x2 matrix such that A*V1=a*V1, I=2x2 identity matrix) or V1=0 vector. I really would like to master my understanding of this equation, but the only reasoning behind this solution is that you can directly show it satisfies (dY/dt)=A*Y(t). My text book chose to use guess-and-check to derive this

solution, which from time to time is acceptable, but it is simply conceptually-challenged and for mastery I need further insights. Can you explain the origin of the independent variable, t, in the solution of Y(t)? Please, please be as rigorous as you can (rigor is the impetus for submission of this question).

Work Done

I checked Paul's online math notes (http://tutorial.math.lamar.edu/). The file is called DE_Complete.pdf. Paul justifies the use of t by referring back to another section, repeated roots in second order ordinary differential equations. Unfortunately, there is a grave typo, where Paul claims (correctly) that "a constant times a solution to a linear homogeneous differential equation is also a solution" (Page 102). He, however, goes on to multiply his homogenous differential equation by v(t), which is NOT a constant. Thus, in his conditional statement (IF you multiply a constant times a solution to a linear homogeneous differential, THEN the resulting equation is also a solution), Paul accidentally fails to satisfiy the antecedent (hypothesis, "IF you multiply a constant times a solution to a linear homogeneous differential"); hence, we say nothing about why t is present. Again, I ask that you refrain from guess-and-check. Guess and check is a barbaric method because it is an existential generalization that afford no universal insights. In other words, it works, but you'll never know why.