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 S2010. See also F2009, Exam 1.

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 Testf_x/f depends on y ==> y'=f(x,y) not separable 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. Troubles with explicit solutions of y'= 3 sqrt(xy) [1.4-6]. 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 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

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 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 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 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.

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,

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.

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

Examples and ApplicationsGrowth-Decay model y'=ky and its algebraic model y=y(0)exp(kx). Pharmokinetics of drug transport [ibuprofen] Pollution models. Three lake pollution model [Erie, Huron, Ontario]. Brine tanks. One-tank model. Two-tank and three-tank models. Recycled brine tanks and limits of chapter 1 methods. Linear cascades and how to solve them. Method 1: Linear integrating factor method. Method 2: Superposition and equilibrium solutions for constant-coefficient y'+py=q. Uses the shortcut for homogeneous DE y'+py=0. Separation of variables The equation y'=7y(y-13), y(0)=17 F(x) = 7, G(y) = y(y-13) Separated form y'/G(y) = F(x) Answer check using the Verhulst solution P(t) = aP_0/(bP_0 + (a-b P_0)exp(-at)) Separation of variables details. Review of partial fractions. Partial fraction details for 1/((u(u-13)) = A/u + B/(u-13)

Partial fractions.DEFINITION: partial fraction=constant/polynomial with exactly one root THEOREM: P(x)/q(x) = a sum of partial fractions Finding the coefficients. Method of sampling clear fractions, substitute samples, solve for A,B, ... Method of atoms clear fractions, multiply out and match powers, solve for A,B,... Heaviside's cover-up method partially clear fraction, substitute root, find one constant Separation of variable solutions with partial fractions. Exercise solutions to the problems due in 2.1.

- References for 2.1, 2.2, 2.3. Includes the rabbit problem, partial fraction examples, phase diagram illustrations.