Topics delayed from Week 1.Discussion of answer checks implicit solution ln|y|=2x+c for y'=2y explicit solution y = C exp(2x) for y'=2y Troubles with explicit solutions of y'= 3 sqrt(xy) [1.4-6].Key ExamplesSeparable DE with no equilibrium solutions. Separable DE with finitely many equilibrium solutions. Separable DE with infinitely many equilibrium solutions. The list of answers to a separable DE. Influence of an initial condition to extract just one solution formula from the list. Examples for Midterm 1 problem 2: y'=x+y, y'=x+y^2, y'=x^2+y^2

Review and Drill Section 1.4Variables separable method. Discuss remaining exercises 1.4-6,12,18. Problem Notes 1.4 at the web site. Equilibrium solutions and how to find them.

Lecture on Section 1.5Theory of linear DE y'=-p(x)y+q(x). Integrating factor W=e^Q(x), Q(x) = int( p(x),x) (Wy)'/W, the fraction that replaces two-termed expression y'+py.Classification of y'=f(x,y)quadrature [Q], separable [S], linear [L]. Venn diagram of classes Q, S, L. Examples of various types. Test for quadrature (f_y=0) Test for linear (f_y indep of y) Test for not separable (f_y/f depends on x ==> not sep) Finding F and G in a separable equation y'=F(x)G(y)Linear integrating factor method 1.5Application to y'+2y=1 and y'+y=e^x. Examples: Testing linear DE y'=f(x,y) by f_y independent of y. Classifying linear equations and non-linear equations. Picard's theorem implies a linear DE has a unique solution. Main theorem on linear DE and explicit general solution.

- References for linear DE:

Review and DrillMethod of Quadrature Variables Separable method Equilibrium solutions from G(y)=0 and Non-equilibrium solutions from G(y) nonzero.

Detailed derivations for 1.4-6y' = 3 sqrt(-x) sqrt(-y) on quadrant 3, x<0, y<0 y' = 3 sqrt(x) sqrt(y) on quadrant 1, x>0, y>0 Equilibrium solution Found by sybstitution of y=c into the DE y'=3 sqrt(xy) Ans: y=0 is an equilibrium solution Non-equilibrium solution Found from y'=F(x)G(y) by division by G(y), followed by the method of quadrature. Applied to quadrant 1 y = ( x^(3/2)+c)^2 Applied to quadrant 3 y = - ((-x)^(3/2)+c)^2 List of 3 solutions cannot be reduced in number Graphic showing threaded solutions: quadrants 2,4 empty

How to deal with separable equationsTheorem. If f_y/f depends on x, then y'=f(x,y) is not separable Theorem. If f_x/f depends on y, then y'=f(x,y) is not separable Theorem. If y'=f(x,y) is separable, then f(x,y)=F(x)G(y) is the separation, where F and G are defined by the formulas F(x) = f(x,y0)/f(x0,y0) G(y) = f(x0,y). The invented point (x0,y0) may be chosen conveniently, subject to f(x0,y0) nonzero.

Partial fractionsHow to solve y'=(1-y)y Def: a partial fraction = constant / polynomial with one root Theorem. A polynomial quotient p(x)/q(x) is a sum of partial fractions, provided degree(p) < degree(q). The possible partial fractions have denominator dividing the denominator of q(x). How to solve for partial fraction constants A,B,C,... Clear the fractions Substitute invented values for x to get a system of equations for A,B,C,..., then solve the system. Methods: Sampling method [described above] Method of atoms [multiply out, match powers] Heaviside's coverup method

General Verhulst DESolving y'=(a-by)y by a substitution Let u=y/(a-by). Then substitution into the DE gives u'=au Solve u'=au to get u=u0 exp(ax). Back-substitute u(x) into u=y/(a-by), then solve for y. Solving y'=(a-by)y by partial fractions Divide the DE by (a-by)y Apply the method of quadrature. Find the constants in the partial fractions on the left. Integrate to get the answer a y0 y(x) = -------------------------- b y0 + (a - b y0) exp(-ax) where y0=y(0)=initial population size.

Linear Differential Equation y'+p(x)y=q(x)Section 1.5 Definition: Linear DE Test: y'=f(x,y) is linear if and only if the partial derivative f_y is independent of y. Algorithm Test the DE for linear Identify p(x), q(x) in the standard form y'+py=q. Determine an integrating factor W(x)=exp(int(p(x)dx)) Replace y'+py in the standard form y'+py=q by the quotient (Wy)' / W and then clear fractions to get the quadrature equation (Wy)' = qW Solve by the method of quadrature. Divide by W to find an explicit solution y(x). Three linear examples: y'+(1/x)y=1, y'+y=e^x, y'+2y=1. classification: separable, quadrature, linear. Two Methods for solving first order equations: Linear integrating factor method, Superposition + equilibrium solution for constant-coefficient linear,

Exam 1 date is 17 Feb 1-3pm in WEB 104 or 18 Feb 6:50am in JTB 140. Other exam times were pre-set by agreement at the start of the semester, on an individual basis. The plan was created to provide extra time to write the exam, which is designed for 50 min.

Sample Exam: Exam 1 key from F2009. See also S2009, Exam 1.

Questions on textbook sections 1.3, 1.4.

Review and drill Ch1.

Superposition TheorySuperposition for y'+p(x)y=0. Superposition for y'+p(x)y=q(x) A faster way to solve y'+2y=1

Problem 1.5-34The expected model is x'=1/4-x/16, x(0)=20, using units of millions of cubic feet. The answer is x(t)=4+16 exp(-t/16). Model Derivation Law: x'=input rate - output rate. Definition: concentration == amt/volume. Use of percentages 0.25% concentration means 0.25/100 concentration

Drill Section 1.5Three linear examples: y'+(1/x)y=1, y'+y=e^x, y'+2y=1. classification: separable, quadrature, linear. Methods for solving first order equations: Linear integrating factor method, Superposition + equilibrium solution for constant-coefficient linear DE Drill: worksheet distributed in class, for the example y' + 2y = 6. Solved in class y'+3y=6, y'+y=e^x, and several homogeneous equations like y'+3y=0, y'+2y=0. Solved for equilibrium solutions in more complicated examples like 2y' + Pi y = e^2.

- References for 2.1, 2.2:

To date, Murphy has covered problems 1,2,3 in the exam review sessions on Wed-Thu.