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The october 4, 2007 math 251 examination i. The exam consists of 12 questions covering various topics in first order and second order differential equations, including finding integrating factors, determining equilibrium solutions, and solving initial value problems. Students are not allowed to use calculators or cell phones during the exam.
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Examination I
October 4, 2007
Name: Student Number: Section:
This exam has 12 questions for a total of 100 points. In order to obtain full credit for partial credit problems, all work must be shown. Credit will not be given for an answer not supported by work. The point value for each question is in parentheses to the right of the question number.
You may not use a calculator on this exam. Please turn off and put away your cell phone.
Total:
Do not write in this box.
t^2 y′^ − 4 ty = 0?
(a) t^4
(b) e^2 t
2
(c)
t^4 (d) e−^4 t
y′^ = −t^3.
Which statement below is false?
(a) The equation is linear.
(b) The equation is separable.
(c) The equation is exact.
(d) The equation is autonomous.
(t^2 + 4)y′′^ + 2ty′^ − t^3 y = 0
In what general form must their Wronskian, W (y 1 , y 2 )(t), appear?
(a) C(t^2 + 4)
(b) C
t^2 + 4
(c)
(t^2 + 4)
(d) C(t^2 + 4)^2
(a) Q′^ = 40 −
400 + t
(b) Q′^ = 40 −
400 + t
(c) Q′^ = 40 −
(d) Q′^ = 40 −
400 − t
y′^ = y(y − 5)(10 − y).
(a) (3 points) Find all equilibrium solutions.
(b) (5 points) Classify the stability of each equilibrium solution. Justify your answer.
(c) (2 points) If y(5000) = 6, what is (^) tlim→ ∞ y(t)?
(d) (2 points) If y(7) = 10, find y(21). (You do not need to solve the equation to find the answer.)
y′′^ − 7 y′^ + 6y = 0, y(0) = 5, y′(0) = 0.
(a) (8 points) Find the solution, y(t), of this initial value problem.
(b) (2 points) What is lim t→∞ y(t)?
y′′^ − 4 y′^ + 8y = g(t).
(a) (3 points) Find its complementary solution, yc(t).
For each of parts (b) through (d), write down the correct choice of the form of particular solution that you would use to solve the given equation using the Method of Undetermined Coefficients. DO NOT ATTEMPT TO SOLVE THE COEFFICIENTS.
(b) (3 points) y′′^ − 4 y′^ + 8y = 2e^2 t^ − 5 t^2 + sin 2t
(c) (3 points) y′′^ − 4 y′^ + 8y = −e^2 t^ sin 2t + 1
(d) (3 points) y′′^ − 4 y′^ + 8y = t^2 e−t^ cos 5t
Find the general solution of the equation.