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## Math 340 - Week 4

by: Susan Ossareh

17

1

6

# Math 340 - Week 4 Math 340

Susan Ossareh
CSU

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Notes covering topics from 2.6 to 2.9
COURSE
Intro-Ordinary Differen Equatn
PROF.
TYPE
Class Notes
PAGES
6
WORDS
CONCEPTS
Math, Differential Equations
KARMA
25 ?

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This 6 page Class Notes was uploaded by Susan Ossareh on Sunday February 14, 2016. The Class Notes belongs to Math 340 at Colorado State University taught by in Spring 2016. Since its upload, it has received 17 views. For similar materials see Intro-Ordinary Differen Equatn in Math at Colorado State University.

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Date Created: 02/14/16
th Math 340 Lecture – Introduction to Ordinary Differential Equations – February 8 , 2016 What We Covered: 1. Quiz #03 a. Highlights i. Will cover section 2.6: exact differential equations 2. Worksheet 04 a. Literally all we did was work through the problems i. For Problem #4, use the integrating factor 1. µ(x,y) is an integrating factory of Pdx+Qdy=0 if µ(x,y)Pdx+µ(x,y)Qdy=0 is exact 2. How can we find µ(x,y), you ask? I’ll show you… If µ is a function of x or y alone, then: ℎ = 1(???????? − ????????) is a function of x ???? ???????? ???????? ???? ???? = ???? ∫ℎ ???? ???????? ???? = 1(???????? − ????????) is a function of y, then ???? ???? = ????∫???? ???? ???????? ???? ???????? ???????? Suggested Homework:  Section 2.6: 4, 9, 10, 12, 19, 23, 26, 28 Math 340 Lab – Introduction to Ordinary Differential Equations – February 9, 2016 What We Covered: 1. Announcements a. Quiz #3 i. Covers section 2.6 2. Course Content – Chapter 2: First Order Equations a. Section 2.6: Exact Differential Equations Continued i. The steps needed to solve a differential equations 1. First, check if it’s exact 2. Integrating Factor 3. F(x,y)=C b. Example: ???????? − 2 ???????? + ???? − ???????? ???????? = 0 1 ℎ = ???? (????????− ???? ???? → ???????????????? ???????????????????????????? ???????? ???? 1 ???? = ???? (????????− ???? ???? → ???????????????? ???????????????????????????? ???????? ???? 1 ℎ ???? = − ???? 1 1 1 ???????????????????????????????????????????? ????????????????????????: ???? ???? = ???? ∫ℎ ???? ????????= ???? −∫???????????? = ???? −ln(????)= ????ln???? )= ???? 1 [(???????? − 2 ???????? + ???? − ???????? ???????? = 0] ???? 2 (???? − )???????? + ???? − ???? ???????? = 0 ???? ???? ???????????? + ???????????? = 0 **Because we know µ=1/x, (Pdx+Qdy) has to be equal to zero** ???????????? ???????????? = 1 = 1 ????ℎ????????ℎ ???????????????????? ???????? ???? ???????????????????? ???????? ???????? Our exact equation means that there is F(x,y) ???????? ???????? = ???????? = ???????? ???????? ???????? ???? = ∫???????? ???????? = ∫ ???? − ???? ???????? ) ????2 ???? = ???????? − + ????(????) 2 ????2 ???????? ????(???????? − + ???? ???? ) = 2 ???????? ???????? ???????? = ???? + ????′(????) ???????? 2 ???? + ???? ???? = ???? − ???? 2 ???? ???? = ???? ???? ???? = 2ln(????) 2 ???? ???? ????,???? = ???????? − + 2ln(????) ???? 1. Our exact equation means that there is F(x,y) ii. Homogenous Equations 1. Homogenous- a function G(x,y) is homogeneous G(x,y) is homogeneous of degree n if ????(????????,????????) = ???? ????(????,????) 2. An ODE, P(x,y)dx+Q(xmy)dy=0 is homogeneous if P and Q are homogeneous of the same degree 3. They can be put into a form in which they can be solved by using the substitution y=xv, where v is a new variable 4. Example: 1 ???? ????,???? = ???? + ???? 2 ????(???? ,???? ) = 1 ???? ???? (???? ) + (???? ) 2 ???? ???? 1 1 −2 ????(????????,????????) = 2 ∗ 2 2 = ???? ????(????,????) ???? ???? + ???? ???? ????,???? = ???? + ???????? 2 2 ????(????????,????????) = ???????? + ???? ???????? )????= ???????? + ???? ???????? = ???? ????(????,????) 2 2 (???? + ???? )???????? + ???????????????? = 0 a. So then is it homogeneous? 2 2 2 2 2 2 ????(????????,????????) = (????????) + (????????) = ???? (???? + ???? ) = ???? ????(????,????) 2 2 ????(????????,????????) = ???? ???????? )????= ???? ???????? = ???? ????(????,????) b. This equation isn’t exact, separable, or linear. So we need a strategy… Here’s an idea, let’s substitute v=y/x to get our final answer ???????? = ???????????? + ???????????? 2 2 (???? + ???? )???????? + ???????????????? = 0 (???? + ???? ????2 2)???????? + ???????????? ???????????? + ???????????? = 0 2 2 2 2 2 3 ???? ???????? + ???? ???? ???????? + ???? ???? ???????? + ???? ???????????? = 0 (???? + 2???? ???? 2)???????? + ???? ???????????? = 0 2[ 2 ] ???? (1 + 2???? )???????? + ???????????????? = 0 (1 + 2???? 2)???????? + ???????????????? = 0 c. This equation is now separable so it’s easier to find our solution (1 + 2???? 2)???????? + ???????????????? = 0 ???????????????? = − 1 + 2???? 2)???????? ???????????? ???????? ∫ 2= ∫− 1 + 2????2 ???? ???? = 1 + 2???? ???????? = 4???????????? 1 ln 1 + 2????2) = −ln|????| 4 2 1 ln((1 + 2???? ) + ln ???? = ???? 2 1 ln(???? 1 + 2???? )4) = ???? 1 ±???? = ????(1 + 2???? ) 2 4 ???????????????????? ???? = ???????? ???? = ???? ???? **having a point and continuous function gives you 1 solution** d. We know there are times when it doesn’t work because some functions don’t create rectangles where f is defined Suggested Homework:  Study for quiz 03 tomorrow!  Section 2.6: 4, 9, 10, 12, 19, 23, 26, 28 Math 340 Lecture – Introduction to Ordinary Differential Equations – February 10, 2016 What We Covered: 1. Course Content – Chapter 2: First Order Equations a. Section 2.7: Existence and Uniqueness of Solutions i. Let’s assume we have the given equation ???? = ′ ????+ 3???? that’s not defined in any R ???? containing x(0)=1 1. This has no solution ii. Let’s use it as an example and see if there is a unique solution iii. Example: ???? ???? = ???? ????,???? =) + 3????, ???? 1 = 2 ???? 1. So wait… can there be a solution? ( ) 1 ???????????????? ???? = { ????,???? : − ∞ ≤ ???? ≤ ∞, ???? ≥ 2 ???? ???? ????,???? = + 3???? ???????????????????????????? ???????????? ???????????????????????????????????????? ???????? ???? ???? (????0,????0) ± (1,2) ∈ ???? 2. There exists a solution!! iv. Interval of Existence of a Solution 1. In section 2.1, we said that the IoE is the largest interval that the solution can be defined in. Let’s look at this in an example  2. Example: a. Consider the initial value problem, find the solution and the interval of existence ′ 2 ???? = 1 + ???? , ???? 0 = 0 ???? ????,???? = 1 + ???? ,????ℎ???????? ???????? ???????????????????????????????????????? ′ (0 ,????0)= 0,0 ,????ℎ???????? ???????? ???????? ???????????????????????????????????????? 1. Here R is the entire tx-plane, so the IVP has a solution ′ 2 ???????? ???? = 1 + ???? → 1 + ????2 = ???????? ???????? ∫ =∫???????? 1 + ???? 2 −1 ???????????? (???? = ???? + ???? 2. So this gives us our general equation… ???? ???? = tan(???? + ????) ???? 0 = tan ???? = 0 → ???? = 0 3. And here is our particular solution! Cool! ???? ???? = tan(????) 4. So what’s our interval of existence? a. Note that f(t,x) is defined for all ???? ∈ ℝ ???? ???? b. But… ???? ???? = tan ???? , so… − 2< ???? < 2 3. Okay, Last example for today: ???? = ???? ????,???? = −2???? + ???? ???? ,????ℎ???????????? ???? 0 = 3( ) ???? ???? = {0; ???????? ???? < 1 {5; ???????? ???? ≥ 1 a. Notice here that f(t,x) is discontinuous… i. Let’s solve for t<1, which makes our equation a homogeneous linear equation ???? = −2???? + 0 ???? 0 = 3 ???? ???????????????????????????????? ???? ???? = ???????? −2???? 0 ???? 0 = ???????? = 3 → ???? = 3 ???? ???? = 3???? −2???? ii. And now let’s solve for ???? ≥ 1 ???? = −2???? + 5 −2 ???? 1 = 3???? 5 5???? 2 ???? ???? = + (3 − )????−2???? 2 2 ???? ???? = {3???? −????, ???? < 1 5 5 { + 3(− ???? )???? 2 −2????, ???? ≥ 1 2 2 Suggest Homework:  Section 2.7: 6, 7, 9, 10, 18(ii only), 19(ii only), 31, 32 Math 340 Lecture – Introduction to Ordinary Differential Equations – February 12, 2016 What We Covered: 1. Survey a. Highlights i. Anonymous survey input on which topics we understand in class and how we’d like to see class improve 2. Course Content – Chapter 2: First Order Equations] a. Section 2.7: Existence and Uniqueness of Solutions i. Uniqueness of Solutions 1. If there is only one solution, then the physical system acts the same way each time it is started from the same set of initial conditions. This is considered deterministic. 2. Example: ′ 1 ???? = ???? ????,???? = ???? , ????????????ℎ ???? 0 = 0 a. For any rectangle that contains (0,0), in fact, for the whole plane, we have… 1 ???? ????,???? = ???? 3 b. This is defined and continuous which means (0,0) is contained. So there exists a solution! Let’s find it, shall we? ′ 1 ???? = ???? , ???? 0 = 0 ???????? 1 = ???? 3 ???????? −1 ∫???? 3???????? = ∫???????? 3 2 ???? = ???? + ???? 2 3 2 (0) = 0 + ???? → 0 = ???? 2 c. So this means the particular solution is… 3 2 ???? = ???? 2 2 3 2???? ???? = ( ????) = √ ( )3 3 3 d. Now the interval of existence is [0,∞] e. A solution in ℝ 2???? 3 ???? ???? = {( ) ,???????? ???? ≥ 0 3 {0,???????????????? f. Let’s check if x(t)≡ 0 is a solution given the initial condition x(0)=0 i. Left Hand Side: ???? ???? = 0 1 1 ii. Right Hand Side: (???? ???? ) = (0) = 0 1 g. Also note: ???? ????,???? = ???? 3 2 ???????? 1 −3 1 ???????? = 3 ???? = 2, ???? = 0 3???? 3 h. Hence, x is continuous but the partial derivative isn’t ′ 3. Example: Let’s consider ???????? = ???? + 3????? Does the Unique Theorem guarantee a unique solution with x(1)=2? a. So we need to find a rectangle R where… i. (1,2) is contained ii. f(t,x) is continuous ???? 1. ???? ????,???? = + 3???? is defined and continuous for ???? every x and ???? ≠ 0 ???????? ???????? 1 iii. and ????????is continuous ???????? = ????is continuous for ???? ≠ 0 1 ???????????????? ???? = { ????,???? : ) ≤ ???? ≤ 100,−10 ≤ ???? ≤ 10 5 b. In R, these 3 conditions are true, so there is only one solution in R ii. Applying the Existence and Uniqueness Theorems 1. For every theorem, it’s necessary for us to find a rectangle R in which the equation can exist within, as seen in the last example 2. So consider this equation, ???? = ???? − 1 cos(????????), and suppose we have a solution x(t) that satisfies x(0)=1. So we can claim that x(t)=1 for all t. bu how can we prove it? a. We’ll use the uniqueness theorem! Genius i. ???? ????,???? = ???? − 1 cos ???????? is continuous everywhere ???????? ii. = cos ???????? − ???? ???? − 1 sin(????????) ???????? iii. It’s continuous everywhere and (0,1) is contained b. By the Unique Theorem there is only one solution to this IVP. Let’s check if (x(t)=1 is a solution: i. LHS: ???? ???? = 0 ii. RHS: ???? ???? − 1 ∗ cos ???? ???? ???? = 1 − 1 cos ???? = 0 ( ) c. So then x(t) is a solution to ???? = ???? ????,???? and ???? 0 = 1 → (0,1) d. This means that all 3 conditions are satisfied so x(t)=1 is a solution! Woot! e. Something else to take into account: If y(t) is also a solution, the y(0)=1, but x(0)=1 as well. That means y(t)=x(t)=1… so! Any solution is x(t)=1 b. Section 2.9: Autonomous Equations and Stability i. A first-order autonomous equation is an equation of the special form ???? = ????(????) ′ ii. Let me show you a quick illustration of what that means… 1. ???? = cos ???? is autonomous ′ 2 2. ???? = 1 − ???? is autonomous 3. ???? = ???? + 3???? is not ???? Suggested Homework:  Section 2.7: 6, 7, 9, 10, 18(ii only), 19(ii only), 31, 32  Section 2.9: 5, 7, 10, 11, 17, 22

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