MATH 2214 Midterm: MATH 2214 Virginia Tech 3.2 Exam
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= 4e3 6= 0, so they are a fundamental set: the general solution is y = 2c1et+c2e t+3. This yields c1 = e/4, c2 = e 4/2: a) yes. b) W = (cid:12) (cid:12) (cid:12) (cid:12) cos t. Sin t sin t cos t (cid:12) (cid:12) (cid:12) (cid:12) W = (cid:12) (cid:12) (cid:12) (cid:12) ln t. 0 (cid:12) (cid:12) (cid:12) (cid:12) ln 3 t. 6= 0, so they are a fundamental set: the general solution is y = c1 ln t+c2 ln 3. To satisfy the initial conditions, we need c2 ln 3 = 0, c1 = 3: a) yes. b) W = (cid:12) (cid:12) (cid:12) (cid:12: the general solution is e t/2. We have y = c1e t/2 + c2te t/2. y(1) = c1/ e + c2/ e = 1, 1 y (1) = c1/(2 e) + c2/(2 e) = 0, hence c1 = c2 = e/2.
