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# MATH 23 WEEK 1 NOTES Math 23

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This 10 page Class Notes was uploaded by Alice Hsu on Sunday September 18, 2016. The Class Notes belongs to Math 23 at Dartmouth College taught by Vardayani Ratti in Fall 2016. Since its upload, it has received 3 views. For similar materials see Differential Equations in Math at Dartmouth College.

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Date Created: 09/18/16

MATH 23 WEEK 1 INTRO TO DIFF EQ The simplest definition of a differential equation is that it is an unknown function described in terms of one or more of its derivatives. Ex 1: Newton’s second law of thermodynamics ???????????????? = ???????? Acceleration can also be described as a function of velocity v: ???????? = ????( ???????? ) where v is the dependent variable and t is the independent variable. This can also be described as a function of height: ???? ℎ = ????( 2) ???????? since velocity can be described as a function of height h. Force is also the sum of gravity, ???? = ????????, and air resistance, ????????. ???? = ????( ????????) = ???????? − ???????? ???????????? ???????? This is a specific example of the general form of a differential equation: ′ ???? = ???????? − ???? Ex 2: radioactive decay ???????? The rate of change of a substance ( ) is proportional to the amount of the substance A i.e. ???????? ???????? = −???????? ???????? If we integrate both sides, we obtain the equation that we more readily recognize as radioactive decay: ???????? ∫ = ∫???????? −???????? 1 − ????n ???????? = ???? + ???? ln ???????? = −???????? + ???? ???????? = ???????? −???????? ????(????) = ???????? −???????? That form above is related to the initial value problem, which will be explained after slope fields. Slope fields (aka direction fields) describe dy/dx, not the path of a function, but how the function itself is changing in relation to some variable. ???????? Problem 1: = 2 − ???? ???????? Charting the relationship, we get the table below, which can be graphed: y 0 1 2 3 4 ???????? 2 1 0 -1 -2 ???????? Observations: The solution approaches y=2. This value, where the slope equals 0 is known as the steady state, or equilibrium, solution. As t approaches infinity, the slope decreases when y >2, and increases when y<2 We can also solve for the steady state solution by setting dy/dt = 0. ???????? e.g.:????????= 0 = 2 − ???? ???? = 2 Utilizing the general form of a differential equation, we can obtain a formula: ???? = ???????? − ???? 0 = ???????? − ???? ???? = ???????? ???? ???? = ???? Problem 2:???????? = − ???? ???????? ???? ???? 0 1 1 1 -1 -1 ???? 1 0 1 -1 -1 1 ????????/???????? 0 Undefined -1 1 -1 1 Here we have a circle slope field. However, it is not really a circle, as the slopes are not unidirectional. It converges to 0 as x approaches infinity. Problem 3: we return to Newton’s law, but attach specific values to some of the variables. We know that gravity is 9.8 m/s and we le= . Thus, ???? 5 ???????? = 9.8 − ???? ???????? 5 The equilibrium solution is 49. Problem 4: involving mice and owls. The change in mice population is dependent on rate of growth (r) by the initial mice population (p) and the rate of predation by owls (k). ???????? = ???????? − ???? ???????? We allow the growth rate to be 0.5 and rate of predation to be 450. ???????? = 0.5???? − 450 ???????? The equilibrium solution is 900, which is how large the initial mice population must be to be sustainable. Solving differential equations involves going back to the initial value problems mentioned before: Using the example of problem 4, we solve for the initial value problem. ???????? = 0.5???? − 450 ???????? ???????? = ???? − 900 ???????? 2 ???????? ???????? ∫ = ∫ ???? − 900 2 1 ln ???? − 900 = ???? + ???? 2???? ???? − 900 = ???? 2+???? ???? ????(????) = 900 + ???????? 2 Scenario 1: how long will it take for the mice to go extinct if p(0)=850? 850 = 900 + ???????? 0 −50 = ???? 2 ∴ ???? ???? = 900 − 50???? 2 0 = 900 − 50???? 2 −900 = −50???? 2 18 = ???? 2ln 18 = ???? Scenario 2: how long will it take for the mice to go extinct if ???? 0 = ???? ,0 < ???? < 900? 0 0 0 ????0= 900 + ???????? ????0− 900 = ???? ???? ???? ???? = 900 + ???? − 000 ???? ) 2 ????/2 0 = 900 + ???? −0900 ???? ) 2 −900 = (???? 0 900)???? −900 ???? ???? − 900 = ????2 0 2ln( 900 ) = ???? 900 − ???? 0 Scenario 3: If the mice go extinct in exactly 12 months, what was their initial population? 900 12 = 2ln( 900 − ???? ) 0 900 6 = ln(900 − ???? 0) 900 ???? = 900 − ????0 900 900 − ????0= ????6 900 900 − ????6 = ????0 Algorithm for solving simple diff eq problems: 1) Solve for initial value problem 2) Identify C for initial value 3) Solve for t SUMMARY: The solutions of a differential equation are functions Integration is an important task The solutions are non-unique. ′ From the general equation ???? = ???????? − ????, where ???? ≠ 0 and a and b are constants, when ???? ≠ ????/????, ???????? = ???????????? ???? − ???? ???? ???? ???????? ???? = ???? + ???????? when ???? = ±???? ≠ 0. If ???? = 0, then the solution is ???? = −???????? + ????. Why can we simply cross multiply? It’s not cross multiplication, it’s chain rule. DIFFERENTIAL EQUATIONS Classifications of Differential Equations: Ordinary DE: 2 ???? ???? 3 ????????2 + ???? = 0 Partial DE: e.g. heat and wave equations ???? ????(???? ????,???? )) ???????? ????,????) 2 = ???????? ???????? System of DE: e.g. Lotka (predator-prey) equations ???????? = ???????? − ???????????? ???????? ???????? ???????? = ???????? − ???????????? Ex: SDE: mice (x) and owls (y) ???????????? and ???????????? are both constant interaction terms. If y=0, then the owl equation is not factored in. The mice population grows exponentially. If y /= 0, then the owl equation is factored in. The mice population is affected, and the owl population is also affected. Order of a differential equation is the order of the highest order of derivatives present in the equation. Degree: the highest order’s power. Some differential equations are linear. Others are not. We always look at the dependent variable to determine order and linearity. Ex: linear DE, which is a linear combination of derivatives: ???? ????−1 ????0???? ???? + ???? ????1????( ) + ⋯+ ???? ???????????? = ???? ???? ( ) Ex: nonlinear DE, which can be linearized: ???? ???? ???? + ????sin(????) = 0 ???????? 2 We can linearize this equation around ???? ≪ 1 so that sin(????)~????. ???? ???? ∴ ???? + ???????? = 0 ????????2 ???? ???? 3 Ex:????????2 + ???? = 0 (nonlinear, second order) ???? ???????? 3 Ex: ???????? = ???? + ???? (linear, first order) ???? ???? ???????????? Ex: 2 − = cos ???? (nonlinear, second order) ???????? ???????? Consider the general form of the nth-order equation: ???? ???? (,????, ???????? ,…, ???? ???? ) = 0 ???????? ???????? ???? assume that the equation holds for all x in an open interval ????(???? < ???? < ????), where a and b approach infinity: Explicit solution: a function ???? ???? that substitutes into y in the above general equation that also satisfies the equation for all x in interval I. ???? ???? 2 Ex: prove that ???? is an explicit solutio2−of 2???? = 0. ???????? ???? 2 −1 ???? ???? = ???? − ???? ???? ???? = 2???? + ???? −2 ′′ −3 ???? (???? = 2 − 2???? Substitute back into the equation: 2 2 − 2????−3 − 2 (???? − ???? −1) = 0 ???? 2 − 2????−3 − 2 + 2???? −3= 0 0 = 0 Thus the solution lies on (−∞,0) ∪ (0,∞). Ex: but prove that ???? ???? = ???? is not an explicit solution. ′ 2 ???? ???? = 3???? ???? ′(???? = 6???? 2 6???? − (????3) = 0 ????2 6???? − 2???? = 0 ???? = 0 This is not a solution because it does not lie on an interval. An explicit solution is only a solution if it is true on an interval of values. Implicit solution: the non-isolated form of the solution. A relation ???? ????,???? = 0 on I that defines one or more explicit solutions. (1+????????????)???????? ???????? Ex: prove that is ???????? + 1 + ???????? = 0 an implicit solution of ???? + ???? + ???????????? = 0 Partially differentiate: ???????? ???????? ???????? ???????? ???????? 1 + ???????? + ???????? + ???????? = 0 (1 + ???????????????? ???????? 1 + ???????????????? + = 0 ???????? So it is an implicit solution. Initial value problem: find a solution to the DE on an interval I that satisfies x_0, the nth initial conditions of a nth-ODE. ???? ????0 = ????0 ???????? (????0 = ????1 ???????? ⋮ ???? ????−1???? (???? 0)= ???? ????−1 ???????? ????−1 Ex: ???? ???? = sin ???? − cos(????) is a solution to the IVP ???? ???? 2+ ???? = 0 ???????? ???????? Where ???? 0 = −1 and (0 = 1 ???????? ′ ???? ???? = cos ???? + sin ???? ( ) ???? ′(???? = −sin ???? + cos ???? ( ) Substitute back in: −sin ???? + cos ???? + sin ???? − cos ???? = 0 0 = 0 Verify using the initial conditions: ???? 0 = sin 0 − cos 0( ) = 0 − 1 = 1 ???? 0 = cos 0 + sin 0 ( ) = 1 + 0 = 1 Yep. For practical examples, refer back to Intro to DE. INTEGRATING FACTORS The algorithm for solving linear first-order differential equations: ′ 1) Manipulate equation into standard form ???? + ???? ???? ???? = ????(????) 2) Identify the integrating factor ???? ???? = exp(∫ ???? ???? ) 3) Multiply both sides of the standard equation by IF ???? 4) Identify that the LH side is the product rule for(???? ???? ???? . Integrate and solve for y. ???????? Note: if the problem is IVP, then you must solve for C. SEPARABLE EQUATIONS ′ Given ???? = ????(????,????), if ???? ????,???? = ???? ???? ????(????), then the equation is separable by separating and integrating, and we can attain a solution. Method: ???????? = ???? ???? ????(????) Ex 1: ???????? ???? = ???????? ???????? ???? ′ = ???? ???? ???????? = ???????????? ???? ???? ) ???? Let ???? ???? = 1 . 1 ???? ???? ) ∫ ( )???????? = ∫ ???????????? ???? ∫???? ???? ???????? = ∫????(????)???????? ln ???? = ???????? + ???? ̃ ???? = ???????? ???????? ???? ???? = ???? ???? + ???? EXISTENCE AND UNIQUENESS For first-order differential equations: Theorem 1: Given ???? + ???? ???? ???? = ????(????) with ???? ????( 0 = ???? 0 If P(t) and Q(t) are continuous in the neighborhood of t_0, I (an open interval), then the initial value problem has a unique solution through interval I. Ex: 2 ′ 2 (4 − ???? ???? + 2???????? = 3???? ,???? 1 = −3 2 ′ 2???? 3???? ???? + 4 − ????2 ???? = 4 − ????2 P and Q are continuous everywhere expect 2 and -2. We do have an interval where there exists a unique solution for this differential equation. For nonlinear differential equations: Theorem 2: Given ???? = ???? ????,???? ,???? ???? ) = ???? . If f (CONDITION 1) and????????(CONDITION 2) are continuous in some 0 0 ???????? rectangle containing (t_0, y_0), say R where ???? < ???? < ???? and ???? < ???? < ????, but not in the boundary, then the initial value problem has a unique solution in some interval ???? < ???? < ????. Note: existence of a solution can be deduced as long as f is continuous. However, uniqueness is not guaranteed unless the second condition is met.

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