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This is the Exam of Introduction Differential Equations and its key important points are: Initial Value, Solution is Guaranteed, Constant Matrix, Possible Solution, Mixture, Stirred and Drains, Same Rate, Approximate, Constant Matrix, Linear Differential Equation
Typology: Exams
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Instruction: Please enter your NAME, ID NUMBER, FORM designation, and CRN NUMBER on your op-scan sheet. The CRN 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 question in row 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) (1, 2) (b) ( 0,1) (c) ( π / 2,3π / 2) (d) (1,3 π / 2)
and
. One possible solution of the system y ′^ = A y is:
(a) (b) (c)
2 2 2
t t t
e e e e
− −
2 t^ t
2 2 2 2
t t t
e e e e
− −
2 2 2
t t
e e −
(d)
2 2 2 2
t t t t
e te e te
(a) y (2) = 3 (b) y (2) = 5 (c) y (2) = e^2 (d) y (2) = 2 e^2
(a) t = e 180 / 200 (b) t = e 200 /180 (c) t = 200 ln 9 (d) t =ln(180 / 200)
e
. What is y (2)?
(a) y (2) = 4 (b) y (2) = 2 (c) y (2) = 2 e^2 (d) y (2) =ln 2
(a) − 5 (b) 6 − 7 (c) 6
− (d) 1 3
A ^ +^^ i^ ^ = + i + i .
The solution of , (0) 5 2
y y y is:
(a) e^3 (b) e 5cos 2 5sin 2 2 cos 2 3sin 2
t t^ t t t
3 5cos 2^ sin 2 2 cos 2 3sin 2
t t^ t t t
(c) e^3 (d) e 5cos 2 5sin 2 2 cos 2 sin 2
t t^ t t t
3 5cos 2^ sin 2 2 cos 2 sin 2
t t^ t t t
y '''' 2+ y ''' − 3 y '' = 0, y (0) = 1, y '(0) = −1, y ''(0) = 0, y '''(0) = 0 is
(a) 1 + 2 t − e −^3 t − et^ (b) 2 − e − t − e^3 t (c) 1 − t (d) (1/ 6) + (1/12) e −^3 t^ +(3 / 4) et
t y^2 '' + 2 ty ' − 2 y = 0, ( t ≠ 0 ),
is y 1 = t. Which of the following is the general solution?
(a) y = c t 1 + c t 2 −^3 (b) y = c t 1 + c t 2 −^1 (c) y = c t 1 + 2 c t 2 (d) y = c t 1 + c t 2 −^2