to solve the following system, a and b are constants where $$ x(1)= 3$$ $$ y(0) = 0$$
$${dx \over dt} = {-2ax+by}$$ $${dy \over dt} = {ax-2by}$$
how do you approach it? using eigenvalue and eigenvector directly won't lead my anywhere in regard to a and b, I tried to solve each each differential equation separately first with the initial condition and it is a mess.
Hint: Don't be afraid of matrices containing parameters. Let $\lambda_1$ and $\lambda_2$ be the eigenvalues of $\pmatrix{-2a & b\cr a & -2b\cr}$ (these can be computed easily, but it may be simpler to express things in terms of them rather than carry around some complicated expressions). The corresponding eigenvectors can be written as $\pmatrix{b\cr \lambda_1 + 2a}$ and $\pmatrix{b\cr \lambda_2 + 2a}$. Next you need to express the initial vector as a linear combination of these. (I'm assuming that should be $x(0)=3$ rather than $x(1)=3$).
There is one complication: the case where the two eigenvalues coincide, but if $a$ and $b$ are real, that only happens when $a=b=0$.