All Educational Materials for MAT244H1 at University of Toronto St. George (UTSG)

UTSGMAT244H1Fall

MAT244H1 Study Guide - Fall 2018, Comprehensive Midterm Notes -

54 Page
12 Oct 2018
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UTSGMAT244H1Fall

MAT244H1 Study Guide - Fall 2018, Comprehensive Midterm Notes -

14 Page
12 Oct 2018
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UTSGMAT244H1Fall

MAT244H1- Midterm Exam Guide - Comprehensive Notes for the exam ( 34 pages long!)

34 Page
11 Oct 2017
A differential equation is an equation involving some hypothetical function and its derivatives. As such, the differential equation is a description of
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UTSGMAT244H1Fall

MAT244H1 Study Guide - Comprehensive Final Exam Guide -

54 Page
20 Nov 2018
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UTSGMAT244H1Summer

Summary notes

32 Page
12 Sep 2010
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UTSGMAT244H1Winter

apm346

15 Page
27 Apr 2011
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UTSGMAT244H1Fall

MAT244H1 Study Guide - Midterm Guide: Jordan Bell, Wronskian, The Roots

5 Page
25 Oct 2018
July 9, 2013: the characteristic equation is r2+3r 10 = 0, and solving gives r = 5, 2. So solutions of the homogeneous equation are y1 = e 5t and y2 =
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UTSGMAT244H1Fall

MAT244H1 Study Guide - Midterm Guide: Jordan Bell, Integrating Factor, Integral Equation

3 Page
25 Oct 2018
June 2, 2013: to have (y )(cid:48) = y(cid:48) 2 t y, we need (cid:48)y = 2 t y and so. Integrating, ln = 2 ln t = ln(t 2). Since the left hand side of
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UTSGMAT244H1Fall

MAT244H1 Study Guide - Midterm Guide: System Of Linear Equations, Wronskian, Scilab

4 Page
25 Oct 2018
Solve the initial value problem y(cid:48)(cid:48) y(cid:48) 6y = 6 y(0) = 1, y(cid:48)(0) = 1. Solution: the homogeneous linear equation is y(cid:48)(c
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UTSGMAT244H1Fall

MAT244H1 Study Guide - Midterm Guide: Saddle Point, Phase Portrait, Lagrange Multiplier

7 Page
25 Oct 2018
Number of test pages: (including this cover sheet) Page 2 of 7: (6 points) calculus of variations. Find the extremal of the functional y(0) = 1, y(1) =
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UTSGMAT244H1Winter

MAT244H1 Lecture Notes - Lecture 1: Integral Curve, Product Rule

2 Page
1 May 2019
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UTSGMAT244H1Fall

MAT244H1 Study Guide - Midterm Guide: Jordan Bell, Wronskian, The Roots

5 Page
25 Oct 2018
July 9, 2013: the characteristic equation is r2+3r 10 = 0, and solving gives r = 5, 2. So solutions of the homogeneous equation are y1 = e 5t and y2 =
View Document
UTSGMAT244H1Fall

MAT244H1 Study Guide - Midterm Guide: Jordan Bell, Integrating Factor, Integral Equation

3 Page
25 Oct 2018
June 2, 2013: to have (y )(cid:48) = y(cid:48) 2 t y, we need (cid:48)y = 2 t y and so. Integrating, ln = 2 ln t = ln(t 2). Since the left hand side of
View Document
UTSGMAT244H1Fall

MAT244H1 Study Guide - Midterm Guide: System Of Linear Equations, Wronskian, Scilab

4 Page
25 Oct 2018
Solve the initial value problem y(cid:48)(cid:48) y(cid:48) 6y = 6 y(0) = 1, y(cid:48)(0) = 1. Solution: the homogeneous linear equation is y(cid:48)(c
View Document
UTSGMAT244H1Fall

MAT244H1 Study Guide - Midterm Guide: Saddle Point, Phase Portrait, Lagrange Multiplier

7 Page
25 Oct 2018
Number of test pages: (including this cover sheet) Page 2 of 7: (6 points) calculus of variations. Find the extremal of the functional y(0) = 1, y(1) =
View Document
UTSGMAT244H1Fall

MAT244H1 Study Guide - Midterm Guide: Jordan Bell, Wronskian, Stationary Point

5 Page
25 Oct 2018
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UTSGMAT244H1Fall

MAT244H1 Study Guide - Midterm Guide: Wronskian, Jordan Bell

3 Page
25 Oct 2018
Mat 244 solved examples relevant to test 2. July 2, 2013: let y1(x) = sin(ecos sin log x), and let y2(x) = sin(ecos sin log x). We don"t have to comput
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UTSGMAT244H1Fall

MAT244H1 Study Guide - Midterm Guide: Integrating Factor

4 Page
25 Oct 2018
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UTSGMAT244H1Fall

MAT244H1 Study Guide - Midterm Guide: Jordan Bell, Phase Portrait

3 Page
25 Oct 2018
August 9, 2013: f = y(cid:48)2 x3 . 2y(cid:48) x3 . (cid:18) 2y(cid:48) (cid:19) x3 so. Using y(0) = 1 and y(1) = 1 gives us c1 = 1 and c2 = 2. Hence y
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UTSGMAT244H1Fall

MAT244H1 Final: Practice Final Summer 2013 Solution

15 Page
25 Oct 2018
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