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Office hours: Mon, Jun 28, 2004, 14:00-16:00
Exam: Tue, Jun 29, 2004, 19:00-21:00; SE 3093
All term marks are posted. Please verify them.
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The home page for winter MAT212 course is http://www.math.toronto.edu/dkhmelev/2004/mat212a/courseoutline.html
INSTRUCTOR: Dmitry Khmelev
OFFICE: SB 4059B dkhmelev (at) math.toronto.edu
OFFICE HOURS: Tuesday 16:00-18:00 or by appointment.
REQUIRED TEXTBOOK: Elementary Differential Equations, 5th edition, H.Edwards and D.Penney.
EVALUATION: There would be one Midterm test worth 25%. There will be quizzes (tentatively each week on Tuesdays) worth 25%, and a comprehensive final exam worth 50%.
TIME&PLACE: Tue 19:00-22:00 & Thu 19:00-22:00, SE 2068. Notice that lecture starts exactly at 19:00, not at 19:10.
TUTORIALS: Your TA is Agustin Herrera-Morales. The tutorials times and places are:
| TUE 17-19 | ® | Rm SE 2068 |
| THU 17-19 | ® | Rm SE 2068 |
MARKS: Marks will be posted on the web page of this course.
MISSED QUIZ: There is no make-up quiz. If you miss a quiz due to illness, you will have to bring a dated doctor's certificate stating that you were too ill to take the quizzes. In that event we will re-rate your score based on the score you take. If, for example, you miss two quizzes (out of say five), and have a valid medical reason for both of missed quizzes, then your score will be computed based on the average on the three quizzes you took.
MID-TERM TEST: On Jun 8, 2004 (in-class for 50 minutes).
MISSED TEST: There is no make-up test. Students who cannot be present for a test must contact their instructor in person on or before their first day back at school. They must also bring a dated doctor's certificate stating that they were too ill to attend the school on the day of the test and giving the expected duration of their illness. Besides illness only very serious reason, properly documented, can be considered as valid excuses for missing a test. In these cases the course mark will be calculated by rescaling total quiz mark to 35% and rescaling the final exam mark to 65%. All other cases will receive a mark of zero on the missed test. There will be no supplemental exam.
DROP DATE: Jun 13, 2004
RE-MARKING THE TEST:
CHANGES: Changes to and clarification of details in this course outline will be announced to the class on at least two occasions to take effect.
COURSE OBJECTIVE: This course presents the basic ODE methodology used in many fields of applications. The emphasis of this course is on concepts and techniques and will be useful to students who seek to gain an understanding of the use of ODE in their own field.
SYLLABUS: 1.1, 1.2, 1.3 (Existence and Uniqueness), 1.4, 1.5, 1.6; 2.1, 2.2, 2.3, 2.5, 2.6, 2.7; 5.1, 5.2, 5.3, 5.4. If time permits we will do selected sections from Chapters 3, 4 and 7.
It is highly recommended to solve at least three problems per each subject listed.
Recommended problems for quiz #1, Tuesday, May 25, 2004. The quiz will begin at 18:30 in the end of tutorial section and will take 25 minutes.
Section 1.1. Verification by substitution (1-16); determining constants from initial condition (17-26); text problems (32-36); solving ODE by educated guess (37-42); investigation (43).
Section 1.2. Simple integration (1-18).
Section 1.3. Existence and Uniqueness (11-20).
Section 1.4. General solution by separating variables (1-18), initial value problems by separating variables (19-28), natural growth ODE and half-life (33-42, 45-48, 50-53).
Quiz 1 q/1.ps 1.pdf with solutions q/1sol.ps q/1sol.pdf
Recommended problems for quiz #2, Tuesday, Jun 1st, 2004. The quiz will begin at 18:30 in the end of tutorial section and will take 25 minutes.
Section 1.5. Integrating Linear 1st order ODE (1-25); swapping variables (26-28), mixture problems (33-37).
Section 1.6. Integrating using substitution methods (1-30); identifying and integrating exact equations (31-42).
Recommended problems for quiz #3, Thursday, Jun 3d, 2004. The quiz will begin at 18:30 in the end of tutorial section and will take 25 minutes.
Section 2.1. Superposition principle and finding particular solution from general for linear ODE of 2nd order (1-16), superposition fails for non-linear ODE (17-19), linear independence (20-26), non-homogeneous ODE, general sol'ns, Wronskian and Ex&Uniq Thm (27-31), general solutions for homogeneous ODE (33-42), detecting 2nd order homogeneous ODE by its general solution (43-48).
Also, please find general solution for the following ODE's:
A1. y¢¢+16y=0.1
A2. y¢¢-2y¢+5y=0.2
A3. 4y¢¢+8y¢+5y=0.3
In the following problems, find a homogeneous ODE of 2nd order with constant real coefficients, which has the given particular solution, or explain why such an ODE does not exists.
A4. pcos(3x)e-2x.4
A5. (sin(2x)+1)e-2x.5
A6. (sin(2x)+cos(2x))e-2x. 6
A7. xe-2x. 7
A8. e-2x+x.8
Recommended problems for Midterm test, Tuesday, Jun 8, 2004. The quiz will begin at 19:00 in the end of tutorial section and will take 50 minutes.
It will be on all sections covered so far plus
Section 2.2. Linear dependence by linear combination (1-6), linear independence via Wronskian (7-12), various problems (21-34).
Midterm test with solutions. q/Midterm.ps, q/Midterm.pdf, q/Midtermsol.ps, q/Midtermsol.pdf.
Recommended problems for quiz #4, Tuesday, Jun 15, 2004.
Section 2.3: General solutions for constant-coefficients ODE (1-20, 27-32), initial value problems (21-26,37,38), identifying a root by particular solution (33-36), idenifying ODE by general solution (39-42).
Section 2.5, Nonhomogeneous ODE's: finding particular solution by variation of parameters (1-20), identifying form of particular solution (21-30).
Recommended problems for quiz #5, Thursday, Jun 17: Section 2.6: Representing particular solution as a sum of two oscillations (1-6), finding steady periodic solution (7-14), identifying practical resonance frequency in mechanical systems (15-18).
Recommended problems for quiz #6, Tuesday, Jun 22:
Section 5.1: reduction to first-order system (1-10).
Section 5.2: elimination method for solving first-order systems (1-19).
Section 5.3: reviewing matrix arithmetics (1-8), representing first-order systems in matrix form (11-20), verification of particular solutions, and constucting the general solution, solving initial value problem (31-40), various problems on linear dependence (41-45).
Recommended problems for exam preparation (in addition to the above questions).
Section 5.4: finding general solutions using eigenvalue method for 2×2 matrices (1-16), 3×3 matrices (17-26).
Section 4.1: computing Laplace transform by integration (1-6), by using table of precomputed transforms (13, 15-22); basic inversion (23, 26-32); miscellaneous problems (36).
Section 4.2: application of Laplace transform for solving initial value problem (1-16).
| Last 5 dig of SN | Q1 | Q2 | Q3 | Mid-Term | Q4 | Q5 | Q6 | ||
| 00649 | L6001 | T6001 | 6 | 9 | 10 | 70 | 5 | 10 | 9 |
| 05586 | L6001 | T6001 | 8 | 7 | 8 | 65 | 7 | 7 | 8 |
| 07440 | L6001 | T6002 | 6 | 10 | 8 | 70 | 3 | 6 | 6 |
| 08820 | L6001 | T6002 | 4 | 5 | 8 | 35 | |||
| 08840 | L6001 | T6002 | 10 | ||||||
| 21386 | L6001 | T6001 | 10 | 10 | 10 | 100 | 10 | 10 | 10 |
| 22820 | L6001 | T6001 | 10 | 8 | 10 | 100 | 9 | 9 | 8 |
| 23453 | L6001 | T6002 | 7 | ||||||
| 24524 | L6001 | T6001 | 9 | 8 | 6 | 80 | 3 | 7 | 7 |
| 24767 | L6001 | T6001 | 10 | 10 | 10 | 90 | 9 | 7 | 9 |
| 24836 | L6001 | T6002 | 8 | 5 | 10 | 70 | 5 | 7 | 7 |
| 26210 | L6001 | T6001 | 8 | 8 | 10 | 100 | 9 | 10 | 10 |
| 29902 | L6001 | T6001 | 7 | 7 | 9 | 75 | 4 | 9 | 8 |
| 30930 | L6001 | ||||||||
| 33950 | L6001 | T6001 | 6 | ||||||
| 34573 | L6001 | 8 | 8 | 9 | 70 | 3 | 10 | 8 | |
| 34579 | L6001 | T6001 | 8 | 9 | 10 | 72 | 3 | 9 | 7 |
| 34677 | L6001 | T6001 | 9 | 5 | 10 | 85 | 4 | 10 | 9 |
| 35102 | L6001 | T6002 | |||||||
| 37414 | L6001 | T6001 | 9 | 3 | 10 | 75 | 2 | 8 | 7 |
| 37913 | L6001 | T6001 | 10 | 8 | 10 | 90 | 9 | 10 | 10 |
| 38633 | L6001 | T6001 | 7 | 7 | 8 | 20 | 3 | 4 | 3 |
| 40794 | L6001 | T6001 | 10 | 6 | 10 | 80 | 6 | 9 | 7 |
| 44419 | L6001 | T6001 | 10 | 10 | 10 | 80 | 9 | 10 | 10 |
| 47224 | L6001 | T6002 | 9 | 6 | 10 | 75 | 9 | 9 | 9 |
| 52202 | L6001 | T6002 | 10 | 10 | 10 | 85 | 10 | 10 | 10 |
| 58139 | L6001 | T6002 | 10 | 7 | 10 | 50 | 9 | 0 | 9 |
| 59158 | L6001 | T6001 | |||||||
| 60770 | L6001 | T6002 | 8 | 6 | 9 | 80 | 10 | 10 | 6 |
| 67925 | L6001 | T6001 | 10 | 5 | 10 | 95 | 9 | 10 | 10 |
| 68832 | L6001 | T6002 | 5 | 60 | 3 | ||||
| 68887 | L6001 | T6002 | 10 | 8 | 10 | 85 | 9 | 10 | 8 |
| 69832 | L6001 | T6001 | 2 | 8 | |||||
| 71668 | L6001 | T6001 | 10 | 8 | 10 | 80 | 6 | 10 | 7 |
| 73935 | L6001 | T6001 | 10 | 8 | 5 | 100 | 8 | 10 | 7 |
| 77604 | L6001 | T6001 | 10 | 5 | 7 | 80 | 9 | 8 | 5 |
| 79133 | L6001 | T6001 | 7 | ||||||
| 80222 | L6001 | T6001 | 10 | 9 | 10 | 100 | 5 | 9 | 4 |
| 85576 | L6001 | T6001 | 10 | 5 | 4 | 40 | |||
| 86836 | L6001 | T6002 | 8 | 5 | 9 | 92 | 4 | 3 | |
| 89073 | L6001 | T6001 | 10 | 10 | 10 | 100 | 10 | 10 | 10 |
| 89650 | L6001 | T6002 | 6 | ||||||
| 91718 | L6001 | T6001 | 10 | 6 | 10 | 100 | 10 | 6 | 10 |
| 93011 | L6001 | T6002 | 7 | ||||||
| 94074 | L6001 | T6002 | 6 | 6 | 10 | 80 | 8 | 9 | 10 |
| 94261 | L6001 | T6002 | 7 | 7 | N | 65 | 3 | 5 | 9 |
| 97955 | L6001 | T6002 | 9 | 10 | 5 | 80 | 5 | 4 | N |
| 98402 | L6001 | T6001 | 10 | 10 | 10 | 80 | 4 | 9 | 9 |
| 99122 | L6001 | T6001 | 10 | 9 | 10 | 92 | 4 | 10 | 9 |
1C1cos4x+C2sin4x.
2(C1cos2x+C2sin2x)ex
3(C1cos(x/2)+C2sin(x/2))e-x
4From the form of the term we can read out the complex root for characteristic eqn -2+3i, which must appear together with the root -2-3i. Characteristic equation: (r-(-2+3i))(r-(-2-3i))=0 or r2+4r+13=0. Hence ODE is y¢¢+4y¢+13=0. (The missed sign here was noticed by Youssef Aboul-Naja. Thanks!).
5First, group expression with respect to exponent: (sin(2x)+1)e-2x=sin(2x)e-2x+e-2x. Term C1sin(2x)e-2x can only appear in a pair with term C2cos(2x)e-2x, since complex roots -2+2i and -2-2i come in a pair. To have the term C3e-2x we must also have root -2. Therefore the characteristic equation must have 3 roots -2±2i and -2, which is impossible for an equation of 2nd degree. Therefore there are no linear homogeneous ODE, having this particular solution.
6This particular solution fits general solution (C1sin(2x)+C2cos(2x))e-2x with C1=C2=1. Such general solution is introduced by complex roots -2±2.
7This particular solution fits general solution (C1+C2x)e-2x with C1=0, C2=1. Such general solution is generated by repeated root -2.
8Term e-2x appears from the root -2.
Term x appears in the expression (C1+C2x)e0x, corresponding
to the double root 0. Characteristic equation with 3 roots must be
cubic, while for homogeneous ODE of 2nd order the characteristic
equation is quadratic. Therefore such an ODE does not exists.