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2

Here is an approach with Association (and Dataset for visualisation) : vars1 = {"u1", "v1", "u2", "v2"}; vars2 = {"u1", "v1", "u3", "v3"}; vars3 = {"u1", "v1", "u4", "v4"}; k1 = (2 10^11 6 10^-4)/3 {{0, 0, 0, 0}, {0, 1, 0, -1}, {0, 0, 0, 0}...


4

but without hardcoding? First of all, line[t_] = {t, 2t, 3t} is not an equation. Equation should have == in it. This is function definition. Second, better to use delayed := instead of immediate = when making functions, unless there is a good reason to use immediate definition. I do not know why you do not want to do normal explicit definitions for x[t_] ...


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