Mathematics · Differential Equations
Let x 1 (t) and x 2 (t) be the solution of the initial value problem: e -t dx 1 dt = -x 1 + x 2 e -t dx 2 dt = -x 1 - x 2 with x 1 (0) = 1, x 2 (0) = 0. If r(t) = √(x 1 2 (t) + x 2 2 (t)), then which of the following statements is true?
- 1r(t) → 0 as t → +∞✓
- 2r(ln 2) = e -1
- 3r(ln 2) = 2e -1
- 4r(t)e t → 0 as t → +∞
Correct answer
r(t) → 0 as t → +∞
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