Theory of separable equations section 1.4.Separation test:Define F(x)=f(x,y0)/f(x0,y0), G(y)=f(x0,y), then FG=f if and only if y'=f(x,y) is separable.Non-Separable TestTEST I. f_x/f depends on y ==> y'=f(x,y) not separable TEST II. f_y/f depends on x ==> y'=f(x,y) not separableReview: Basic theory of y'=F(x)G(y):y(x) = H^(-1)( C1 + int(F)), H(u)=int(1/G,u0..u). Solutions y=constant are called equilibrium solutions. Find them using G(c)=0. Non-equilibrium solutions arise from y'/G(y)=F(x) and a quadrature step.

Implicit and explicit solutions.Discussion of answer checks for implicit solutions and also explicit solutions. Exercise 1.4-6: Trouble with explicit solutions of y'= 3 sqrt(xy) Separable DE with no 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 Example 1: Show that y'=x+y is not separable using TEST I or II (partial derivative tests). Example 2: Find the factorization f=F(x)G(y) for y'=f(x,y), given (1) f(x,y)=2xy+4y+3x+6 [ans: F=x+2, G=2y+3]. (2) f(x,y)=(1-x^2+y^2-x^2y^2)/x^2 [ans: F=(1-x^2)/x^2, G=1+y^2].

Answer Checks and Key Examples.Discussion of answer checks implicit solution ln|y|=2x+c for y'=2y explicit solution y = C exp(2x) for y'=2y Answer check for 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

Lecture on Section 1.5Theory of linear DE y'=-p(x)y+q(x).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)

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.

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 substitution 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 test separable and non-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.Linear integrating factor method 1.5Def. Integrating factor W=e^Q(x), Q(x) = int( p(x),x) (Wy)'/W, the fraction that replaces two-termed expression y'+py. Application 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:

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,

Maple lab 1: quadratics, partial derivatives. Present problem 1 from the midterm 1 sample [S2010 midterm 1 key]. Exam 1 date is in the syllabus and also the online due dates page. Questions on textbook sections 1.3, 1.4. Review and drill Ch1. Sample Exam: Exam 1 key from F2010. See also S2010, Exam 1.

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

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.

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

Introduction to Ch 2 topicsAutonoumus DE y'=f(y) Solution of the Verhulst DE y'=(a-by)y Numerical solutions of DE. No exercises, but: Maple lab 3 Maple lab 4