Original post by fuzzybear
for part a, I was just wondering, the answer book has k1, k2, k3 as: 2, -2, -1, respectively
would it still be right to have them multiplied by a factor?
for example, by a factor of 2, would 4, -4, -2, be equally valid as an answer?
for part a, I was just wondering, the answer book has k1, k2, k3 as: 2, -2, -1, respectively
would it still be right to have them multiplied by a factor?
for example, by a factor of 2, would 4, -4, -2, be equally valid as an answer?
Yes.
As the poster above me pointed out, there are only unique solutions (k1 = ... = kn = 0) for linearly independent functionsQuick Reply
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