# is there are specific way to solve coupled first-order differential equations with coefficients varying?

suppose I have "n" coupled differential equation represented by the matrix,

Y = A Y

, where Y is the column matrix containing first derivatives, namely, y1(t), y2(t), ... yn(t) . A is a square matrix whose each element contains some function dependent on "t" (not constants) and Y is the column matrix containing the solution set, namely, y1(t), y2(t), ... yn(t) .

If A, contained constants, then its easy to solve by Matrix Exponential method or Eigen-Value method. But, if it contains some varying functions, then is there any approach to solve this. Please, direct me to a good reference, if possible, with an example.

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 Sounds kind of like covariant differentiation. – ben Oct 9 '12 at 17:59 @ben am sorry, I didn't get u, but actually this is not a derivative, but a set of coupled differential equations, represented in a matrix form. But, till now I have seen the case for which 'A' is a square matrix containing constants, but what about the case if 'A' is a square matrix, which are functions of 't' – tsndiffopera Oct 9 '12 at 18:05

If your matrices commute at different times, that is $A(t)A(s) = A(s)A(t)$, then the solution is $Y(t) = \exp \Big( \int_0^t A(s) \mathrm{d}s \Big) Y(0)$. Here $\exp$ is just an ordinary matrix exponential.
Let $\mathscr{H}$ be a Hilbert space, $A: \mathbb{R} \to \mathscr{B}(\mathscr{H})$ a strongly continuous Hermitian operator valued function (i.e. each $A(t)$ is bounded and Hermitian). Then there is a unique solution of $$i \partial_t \psi(t) = A(t) \psi(t), \quad \psi(0) \in \mathscr{H}.$$ If $A$ is continuous w.r.t. the operator norm, then the solution is of the form $$\psi(t) = T \exp \Big( - i \int_0^t A(\tau) \mathrm{d}\tau \Big) \psi(0),$$ where $T \exp$ is the so called "time-ordered exponential".