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  • $\begingroup$ As for DSolve[], try DSolveValue[{y'[t] == -{1, 2, 3}*y[t], y[0] == {1, 2, 3}}, y[t], {t, 0, 1}]. $\endgroup$ Commented 2 days ago
  • $\begingroup$ @MichaelE2 Have you read the addendum at the bottom? {1, 2, 3}*y[t] is not same as {1, 2, 3}.y[t]. $\endgroup$ Commented 2 days ago
  • $\begingroup$ Isn't the addendum wrong? They are not numerically identical. {1, 2, 3}*y[t] is a 3x3 matrix and {{1, 0, 0}, {0, 2, 0}, {0, 0, 3}} . y[t] is a 3-vector. Right? $\endgroup$ Commented 2 days ago
  • 1
    $\begingroup$ y[t] is vector {y1, y2, y3} in both cases. How is {1, 2, 3}*y[t] 3x3 matrix when {1, 2, 3}*{y1, y2, y3} produces vector {y1, 2 y2, 3 y3}??? $\endgroup$ Commented 2 days ago
  • 4
    $\begingroup$ "How is {1, 2, 3}*y[t] 3x3 matrix..." The key point is the evaluation order. If it's y[t]={y1, y2, y3}; {1, 2, 3}*y[t] then it won't be a matrix of course, but what's discussed here is amount to Clear[y]; {1,2,3} y[t]/.y[t]->{y1,y2,y3}. (NDSolve doesn't have HoldAll, HoldFirst, etc. attribute, so something like {1, 2, 3}*y[t] won't hold, it immediately evaluates to {y[t],2y[t],3y[t]}, but NDSolve internally reverses the procedure, this is quite unusual, and that's what Michael E2 wants to talk about in this post. ) $\endgroup$ Commented 2 days ago