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This is the Exam of Introduction Differential Equations and its key important points are: Machine Grading, Eigenvectors, Possible Solution, General Solution, Inhomogeneous, Differential Equation, Determined, Linear System, Transformed, Largest Interval
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Math 2214 Common Part of Final Exam May 6, 2002
Instructions: Please enter your NAME, ID NUMBER, Form designation, and INDEX NUMBER on your op-scan sheet. The index number should be written in the upper right-hand box labeled ”Course.” Do not include the course number. In the box labeled ”Form” write the appropriate test form letter A. Darken the appropriate circles below your ID number and Form designation.
Use a #2 pencil; machine grading may ignore faintly marked circles.
Mark your answers to the test questions in rows 1–12 of the op-scan sheet. You have 1 hour to complete this part of the final exam. Your score on this part of the final exam will be the number of correct answers. Please turn in your op-scan sheet and the question sheet at the end of this part of the final exam.
(a)
(b) 2 (c) e^2 (d)
e
(a) y^2 =
9 − 2 t^2
(b) y^2 = t^3 − t
(c) y^3 =
t^2
(d) y =
12 − 2 t^2
and
satisfying A
Then a possible solution to Y ′^ = AY is
(a) Y =
e^2 t^ + 2e−^2 t 2 e^2 t^ + 4e−^2 t
(b) Y =
5 e^2 t 3 e^2 t
(c) Y =
e^2 t^ + 4te^2 t e^2 t^ + te^2 t
(d) Y =
e^2 t^ + 4e−^2 t 2 e^2 t^ + e−^2 t
(a) y = c 1 + c 2 et^ + c 3 e−t^ + c 4 te−t^ + c 5 t^2 e−t
(b) y = c 1 + c 2 t + c 3 cos t + c 4 sin t
(c) y = c 1 + c 2 t + c 3 t^2 + c 4 cos t + c 5 sin t
(d) y = c 1 + c 2 t + c 3 t^2 + c 4 e−t^ + c 5 te−t
1 + et^
. If you try to
find a particular solution of the form yp = uet^ + ve^2 t, then u and v may satisfy:
(a) u′^ =^ −^
e−t 1 + et^
, v′^ =
e−^2 t 1 + et
(b) u′^ =
−e^2 t 1 + et^
, v′^ =
et 1 + et
(c) u′^ =
1 + et e^3 t^
, v′^ =
1 + et e^3 t
(d) u′^ =
(1 + et)e^3 t^
, v′^ =
(1 + et)e^3 t
(a) yp = K 1 t^2 + K 2 t + K 3 t^3 e^2 t^ + K 4 t^2 e^2 t
(b) yp = K 1 t^2 + K 2 t^2 e^2 t^ + K 3 te^2 t
(c) yp = K 1 + K 2 t + K 3 e^2 t^ + K 4 te^2 t
(d) yp = K 1 t + K 2 te^2 t^ + K 3 e^2 t
t y
with y(2) =1 has a
unique solution is the interval
(a) (−∞, ∞) (b) (0, ∞) (c) (
3 , ∞) (d) (1, 3)
. You are given that λ =4 is an eigenvalue for A. Then a corre-
sponding eigenvector is
(a)
(b)
(c)
(d)
(a) k =
ln 2 10
(b) k =
ln 2 N 1 N 10
(c) k = −9 + ln
(d) k = −
ln
(a) y(1) ≈ 3 (b) y(1) ≈ 5 (c) y(1) ≈
(d) y(1) ≈