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What was the solution of the differential equation? Was it y = ln(2/3-e^2x)
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lambda was 3 and 1, and the distance was 6.Wat answer did u get fr the last question?
ebda can only had one value which was 2lambda was 3 and 1, and the distance was 6.
I got lambda = 3, how did you get lambda = 1? Did you take mod as +/- or the root? Or is there a completely different method to do that?lambda was 3 and 1, and the distance was 6.
no u get a quadratic equation. u solve it and u get 3 and 1.I got lambda = 3, how did you get lambda = 1? Did you take mod as +/- or the root? Or is there a completely different method to do that?
Alright, do you remember the question? The equations of the planes and the line?no u get a quadratic equation. u solve it and u get 3 and 1.
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