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Math 340 - Week 1

by: Susan Ossareh

Math 340 - Week 1 Math 340

Susan Ossareh
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Intro-Ordinary Differen Equatn

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These notes cover the basic material needed for Diff Eq which is essentially all of chapter 1 and section 2.1
Intro-Ordinary Differen Equatn
Class Notes
Math, Differential Equations




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This 8 page Class Notes was uploaded by Susan Ossareh on Sunday January 24, 2016. The Class Notes belongs to Math 340 at Colorado State University taught by in Spring 2016. Since its upload, it has received 30 views. For similar materials see Intro-Ordinary Differen Equatn in Math at Colorado State University.


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Date Created: 01/24/16
Math 340 Lab IntrodUCtion to Ordinary Differential EqUations JanL1ary 19th 2016 What We Covered 1 Viewed the course syllabus a Highlights i Main Website httpwwwmathcolostateeduvzhoucoursemath34o ii Use Canvas for grades and course announcements iii Homework is not required but strongly suggested The problems can be found on the main course website iv There will be a quiz every Wednesday starting next week v There will be 2 midterms and one final 1 Exam 1 March 3 2 Exam 2 April 3 3 Final May 10 vi Grading 1 Quizzes 30 2 Exams are 20 each 3 Final 30 2 Class Break Down a Highlights i Course Material Leading Up to the First Midterm 1 Review of content from Math 261 2 Integrals 3 First Order Diff Eq ii Course Material Leading Up to the Second Midterm 1 High Order Diff Eq 2 Systems 3 Course Content Chapter 1 Introduction to Differential Equations a Essentially differential equations is all about manipulating variables functions We will use them to solve applied problems make predictions and analyze solutions b Sections 11 13 uses derivatives integrals and concepts that make up calculus c Section 11 Differential Equation Models i There is a theme through Diff Eq of which relates math to the scientific application In math it s the derivative In science through experimentation it s the rate of change When they are equated we get a differential equation ii Mechanics 1 Newton s Second Law states a force acting on a mass is equal to the of momentum with respect to time where momentum is mass m times velocity v Therefore the force F is equal to the derivative of the momentum mv If the mass is constant then d dv dtmv mdt ma Fma 2 Newton s Universal Law of Gravitation states any body with mass M attracts any other body with mass m directly toward the mass M with a magnitude to the product of the two masses and inversely proportional to the square of the distance separating them There is the constant G which is universal and where the distant is r thus the magnitude of the force is GMm 3 Example 1 see page 3 in required textbook Newton s laws of mechanics creates general instructions for nearly every model of motion This helps us understand the forces acting on a body without having to know them explicitly Suppose then you are given a model of a ball that s thrown in the air near the surface of the Earth near with respect to the diameter of the Earth If x measures the distance the ball is above Earth then we can make a few assumptions a xt is the FUNCTION displacement of motion dx b v where vveloc1ty dt dv dzx c a where a acceleration dt d12 GM d g r z where g Earth s acceleration due to grav1ty e Then Newton s second law is dzx F mg ma m f And when the masses cancel we arrive at our differential equation dzx g dt2 i This example is a differential equation because it has the unknown function x t and at least one of its derivatives iii Population Models 1 As there are two ways to consider rate of change there is a correlation to population models a Through math we consider the with respect to time as dP dt b But through biology we say the rate of change is to the population rP c Putting the mathematical rate of change and scientific proportion together and we get a differential equation i Where rreproductive rate so the rate of change is rP 2 We can then re write the solution to solve for the function Pt where P0 is a constant that represents the initial population This then accounts for if the population will grow expontentially This is called a logistic equation Pt Poe 3 The aw with that model is that the reproductive rate r is not a constant so it s not accurately applied We can better model population through the equation dP 1 P P dt r k a Where k carrying capacity and r is a constant 1 Section 12 The Derivative i What is a derivative 1 The rate of change of a function dv dzx a Ex 61 v 2 dt dt b aka the modellng functlon 2 The slope of the tangent line to a graph a EX y fx0 f x0x x0 b aka the geometric de nition 3 The best linear approximation of a function a EX The Taylor s Theorem Lx f x0 f x0x x0 b aka the algebraic definition 4 The limit of difference quotients a Ex f x0 lim w x gtxo 95 950 b aka the limit quotient de nition 5 Table of formulas a Ex f gt f b aka formulaic definition i Essential for understanding Diff Eq ii What is an integral 1 The area under the graph of a function a Ex I fxdx 2 The antiderivative a EX fgxdx fx C b aka indefinite integral do NOT forget constant C 3 A table of formulas a Ex f gt ffxdx iii Solution by Integration 1 The format for writing a first order differential equation is y f t y a Suppose you re given y cost solve the differential equation i To do this you take the integral antiderivative of both the right hand and left hand sides Because there are no given bounds this is an indefinite integral which means the solution must include the constant C yt I cost dt sint C Suggested Homework Section 11 5 Section 12 6 1o 14 Section 13 2 6 12 14 2o 24 27 no sketches Review integrals o U substitution 0 Integration by parts 0 etc Math 340 Lecture Introduction to Ordinary Differential Equations January 20th 2016 What We Covered 1 Short Review from where we left off in Lab a Solution by Integration i The format for a first order differential equation can be written as y f t y where y is the unknown function and t is the independent variable ii Family of functions y cost y jcost dt y sint C 1 This example is a general solution because C is not defined 2 Course Content Chapter 1 Introduction to Differential Equations a Section 13 Integration Continued i There are general solutions and particular solutions The general solution gives the solution to y t without de ning constants A particular solution satis es the initial condition 1 Example 1 Find the solution to y t tet that satisfies y0 2 a What we can take away from this immediately is that our initial condition is at the point 02 b Now we can look at the equation given we need to take the antiderivative of both the RHS and LHS To take the integral of the right hand side we need to use the integration by parts method This will give us the general solution y t te yt tetdt u tdu dtv etdv etdt uv jvdu tet jetdt yt tet 6 C c To get the particular solution we need to solve for C so we can input the point 02 yO OeO e C2 e C2 1C2 C3 d To visualize the solution we use solution curves which is a graph that translates the functions vertically such that yt F t 2 Example 2 Find the solution to the initial value problem y i with y1 3 a To nd the general solution we integrate both the RHS and LHS I 1 y x j 1 d y x 9 yx lnx C b Then we nd the particular solution by inputting the initial condition y1 ln1 C 3 C 3 yx lnlxl 3 c We then need to find the domain or interval of existence for yx We have two conditions we need to take into account 1 natural log is undefined at zero and 2 the function yx needs to be continuous Therefore we should pick 0 00 ii The motion of a ball I l l l l l l l l l l l l l l l l l l l l l l l l A yttet et3 00 l l l l l l l l l l l l l l l l l l 1 The model derived from the motion of a ball example in section 11 is l l l l l l Q at2 vo2oft s 2 We want to solve for x this can be done by using the methods we ve covered so far Remember differential equations is all about g where g is the universal gravity constant 32 fts Xo6ft and manipulating variables dzx dv dt2 dt g v gdt gtC v 32t C v020 gtC20 dx 32t 20 dt x 32t 20dt x 16t2 20tC x0 6 gt C 6 xt 16t2 ZOt 6 3 Course Content Chapter 2 First Order Equations a Section 21 Differential Equations and Solutions 1 ii iii iv Ordinary Differential Equations ODE involves an unknown function of a single variable generally the independent variable together with one or more of its derivatives 1 Example 1 y 4y e The order of a differential equation is the order of the highest derivative that occurs in the equation Therefore a first order differential equation has only up to the first derivative of the unknown function Partial derivatives cannot exist in an ODE instead they are considered Partial Differential Equations Notations 1 General Form Dtyy y yquot 0 2 Normal Form yquot ftyy yquot 1 a When given an equation that needs to be put into normal form you simply deduce what order the differential equation is then isolate that derivative to the LHS and shift the rest of the variables to the RHS 31 Solutions 1 A solution is a differentiable function yt such that Mt y y 0 for all t in the interval where yt is defined We can confirm a solution by inputting the function and it s derivative into the differential equation a Example 1 Show that yt C equot2 is a solution to the first order equation y 2ty i To solve this we take the given function yt and find it s derivative Then we will set it equal to y Note yt and y are two completely different functions RHS Zty 2tCet LHSy Ce t2 2tCet ii Because they both concede to the solution 2tCe t2the solution is satisfied Suggested Homework 0 Section 13 2 6 12 14 2o 24 27 0 Section 21 2 6 8 1o 12 14 Math 340 Lecture Introduction to Ordinary Differential Equations January 22nd 2016 What We Covered 1 Math 340 Worksheet 1 a Highlights i Problems were solved in class individually and then on the board ii Problem Set 1 Find a general solution to the equation y tsint and a particular solution that satisfies the condition 01 There is this solution de ned 2 Find a general solution to the equation x J17 Where is this solution de ned 3 Verify that yt tgfc solution satisfying y0 0 2 Course Content Chapter 2 First Order Equations a Section 21 Differential Equations and Solutions Continued i Concept Question Is the function yt 0 a particular solution 1 Answer Yes ii Example Find a solution to y ty2 dy E W is a solution to y tyz Is there a particular 2 We will divide both Sides by y2 and then multiply by dt and take the integral d jy 32 jtdt t2 1 y 26 Now we solve for y 1 2 Multiply both Sides by 1 and then the RHS by E y m 1 This kind of equation is called a separable equation because you can separate the variables on either side of the equation and then can be managed more easily iii Implicitly Defined Solutions 1 Example Find a solution for y with y0 1 dy ex dx1y 1 ydy Jexdx 2 y y equot C a y is an implicit solution b We now want to check that it is a solution using implicit differentiation 2y y2 26quot C Zy Zyy 26quot y 1 y ex ex y 1 y c We can also nd the explicit solution to solve for y using the quadratic formula 2 i J4 8ex C 3106 2 yx 1i 12exC 1 i The i creates a problem because it offers two solutions We need to pick either or to satisfy the initial condition For this specific situation we ll pick plus to get yo 1 y0 1 1 2e0 C 1 f3 zc 2 3 2C 4 2C 1 1 C ii If the condition was yo 4 then y0 1 12e C 4 1 f3 2 4 3 zc 3 V 3 2C 3 3 2C 9 C 3 iv Initial Value Problems 1 Initial value problems essentially just mean they want the particular solution because an ODE has infinitely many solutions 2 A first order differential equation together with an initial condition y f t yt yto yo is called an initial value problem v Interval of Existence 1 Interval of Existence a solution to a diff eq is defined to be the largest interval over which the solution can be defined and remain a solution a Solutions to differential equations are required to be differentiable 2 Example y y2 y0 1 has solution o oooOOOOOoooOOOOmono OOooooooOOAwe OoOOOOOOOOOooOoOOOOOOOooo O 0 Log 707 o 0 O OO i0 7 O OO 0 O OO i0 0 O M70 O iooiiooiiooiiooiiooo 0 0 0 Section 21 0 o 0 Suggested Homework O i io i io a 2 6 8 1o 12 14 O i io O o i io i io The IoE is 00 0 7i0 O O io i io i io 7io O 0 i io o i io i io i io O Ooiiooiiooiiooiiooiiooiiooiiooiiooiiooiiooiiooiiooiiooiiooiiooiiooiiooiiooiiooiiooiiooiiooiiooiiooiiooiioo O o i O 00 0 77 O 00 000 go A O O O O O O O O O O O O O O O O O O O O O O 0 O O 0 O O O 0 O O O 0 O 0 0 O O O O O O O 0 O 0 0 O O 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 OOOOO


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