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Form A
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.
- If y(t) satisfies ty′^ + y = t with y(1) =1, then y(2) equals
(a)
(b) 2 (c) e^2 (d)
e
- If y′^ − 2 ty^3 = 0 with y(2) =1, then y satisfies
(a) y^2 =
9 − 2 t^2
(b) y^2 = t^3 − t
(c) y^3 =
t^2
(d) y =
12 − 2 t^2
- Suppose A has eigenvectors
and
satisfying A
, 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
- The general solution of y′′′′′^ + y′′′^ = 0 is
(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
- Consider the inhomogeneous differential equation y′′^ − 3 y′^ + 2y =
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 particular solution of y′′′^ − 4 y′′^ + 4y′^ = te^2 t^ + t may be found of the following form (where the Ki have yet to be determined):
(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
- The largest interval on which the initial value problem y′^ =
t y
with y(2) =1 has a
unique solution is the interval
(a) (−∞, ∞) (b) (0, ∞) (c) (
3 , ∞) (d) (1, 3)
- Let A =
. You are given that λ =4 is an eigenvalue for A. Then a corre-
sponding eigenvector is
(a)
(b)
(c)
(d)
- Suppose N grams of a radioactive material decays according to the usual decay law N ′^ = −kN , where k is the decay rate. If N 1 = N (1) is the number of grams of undecayed material after 1 time unit, and N 10 = N (10) is the number of grams of undecayed material after 10 time units, then a formula for the decay rate k is
(a) k =
ln 2 10
(b) k =
ln 2 N 1 N 10
(c) k = −9 + ln
N 10
N 1
(d) k = −
ln
( N 10
N 1
- Suppose you decide to approximate the solution of the initial value problem y′^ = 2 y + 4t, y(0) =1, using the Improved Euler Method with step size h = 12. Then your approximate value for y(1) will be
(a) y(1) ≈ 3 (b) y(1) ≈ 5 (c) y(1) ≈
(d) y(1) ≈
