Do 3 steepest descent steps when: f(x) = 3XI2 + 2X22 - 12xl + 16x2, Xo = [I l]T

Math 340 Study Guide Exam 1 - March 03 Introduction to Ordinary Differential Equations Section 2.2: Solutions to Separable Equations Definitions: ′ o Exponential Equations: = − o Separable Equations: = − 1 = − Like-variables are separated onto either side of the equation Simplistic Example: = 2 = 2 1 = 2 1 ∫ 2 = ∫ −1 −2 = 1 2 = + 2 2 + General Method: o It takes three steps: 1. Separate the variables 2. Integrate both sides 3. Solve for the solution y(t), if possible Practice Problem: (#33 in the textbook)Amurder victim is discovered at midnight and the temperature of the body is recorded at 31 degrees C. One hour later, the temperature of the body is 29 degrees C.Assume that the surrounding air temperature remains constant at 21 degrees C. Use Newton’s law of cooling to calculate the victim’s time of death. o Answer- 9:54 PM Section 2.3: Models of Motion Main Idea: o Apply separable equations to linear motion and air resistance Simplistic Example: o Linear Motion = − = 2 = − o Air Resistance = − ) ( ) = − + = − − Practice Problem: (#17 in the textbook)A2-foot length of a 10-foot chain hangs off the end of a high table. Neglecting friction, find the time required for the chain to slide off the table. Hint: Model the problem with a second order differential equation and then solve using = ⁄ ) Section 2.4: Linear Equations Definitions: o Linear: ′ ( ) = + () o Homogeneous: = o Coefficients: ) () o Integrating Factor: ( ) = −∫ Simplistic Example: = ln = ∫ + | = ∫ += ∫ ( ) = ∫ Summary of Method ′ o Solving the equation = + takes four steps Page 02 1. Rewrite the equation as − = 2. Multiply by the integrating factor 3. Integrate the equation 4. Solve for x(t) Practice Problem: (#29 in the textbook) In Exercise 33 of Section 2.2, the time of death of a murder victim is determined using Newton’s law of cooling. In particular it was discovered that the proportionality constant in 5 Newton’s law was = ln( ⁄4 ) ≈ 0.223. Suppose we discover another murder victim at midnight with a body temperature of 31 degrees C. However, this time the air temperature at midnight is 0 degrees C, and is falling at a constant rate of 1 degree C per hour.At what time did the victim die (Remember that the normal body temperature is 37 degrees C) o Answer: 11:12 PM Section 2.6: Exact Differential Equations Definitions o Differential Form: = , + , ) o Differentials: o Integral Curves: , = o Exact: it is the differential of a continuously differentiable function o Integrating Factor: = , , + , , ) o Homogeneous of Degree n , = (,) Simplistic Example: 2 ( − 2 + − = 0 = = 2 − 1 1 [ ( )] 1 ℎ = − ) = − − 2 − = − General Method: Page 03 o The form has an integrating factor depending on one of the variables under the following conditions If ℎ = ( − ) is a function of x only, then = ∫ ℎ is an integrating factor If = ( − ) is a function of y only, then = − ∫ is an integrating factor Practice Problem: (#31 in the textbook) The equation given has the form , + , = 0. Show that P and Q are homogeneous of the same degree and state that degree: + + − = 0 o Answer: Degree One Section 2.7: Existence and Uniqueness of Solutions Definitions o Interval of Existence: the largest interval in which the solution can be defined o Deterministic: if there is only one solution, the physical system acts the same way each time it is started from the same set of initial conditions Simplistic Example: ′ ′ = , ) = (,) 0) = 0 = 0 = (, ) (, ) Theorems ′ o 1. The equation is in normal form = (,) o 2. The right-hand side (,) and its derivative are both continuous in the rectangle T o 3. The initial point (0 ,0 ) is in the rectangle R o 4. There is one and only one solution to the initial value problem o 5. The solution exists until the solution curve t→ (t,y(t)) leaves the rectangle R Page 04 0, < 0, Practice Problem: (#23 in the textbook) Show that4 = { is ≥ 0 a solution of the initial value problem − 4,ℎ 0 = 0. Find a second solution and explain why this lack of uniqueness does not contradict the theorem 0, < 0 o Answer: a second solution is = 5 , ≥ 0 Section 2.9:Autonomous Equations and Stability Definitions o Autonomous: ′ = () o Equilibrium Point: ℎ ℎ ) = 0 0 0 o Equilibrium Solution: = 0 o Phase Line: the dynamics involved in the motion of y(t) along the line o Asymptotically Stable: where the solution curves approach the point o Unstable: where the solution curve moves away from the point Simplistic Example: Find the equilibrium points and solutions for the equation || = − − 2 | − − ≥ 0, = − − = { 2 − + < 0. = − √ 1 = − √ Page 05 General Method: o 1. Graph the right hand side of f(x) and add the phase line information to the x-axis. Find, mark, and classify the equilibrium points where f(x)=0. In each of the intervals limited by the equilibrium points, find the sign of f and draw an arrow to the right if f is positive and to the left if f is negative o 2. Create a tx-plane, transfer the phase line information to the x- axis, draw the equilibrium solutions, and then use the phase line information to sketch nonequilibrium solutions in each interval limited by the equilibrium points Practice Problems: (#23 in the textbook) Perform each of the following tasks for = 6 − , 0 = 2 o Solve the initial value problem analytically o Use the analytical solution from part 1 and the theory of limits to find the behavior of the function as t→ +∞ o Without the aid of technology, use the theory of qualitative analysis presented in this section to predict the long-term behavior of the solution. Does your answer agree with that found in part 2 Which is the easier method Answer: = 6 − 4 , lim = 6 ) →∞ Section 7.3: Solving Systems of Equations Definitions o Pivot: a row vector in a matrix is the first nonzero element of that row o Row Echelon Form: in each row that contains a pivot, the pivot lies to the right of the pivot in the preceding row. Any rows that contain only zeros must be at the bottom of the matrix o Free Variable: a variable associated to a column without a pivot o Free Column: a column without a pivot Simplistic Example: Find the solution Page 06 2+ 3 3 2 +42 = 5 + 2 + 3 + 5 + 7 = 8 1 2 3 4 5 2 1 4 +26 + 3 + 14 = 2 5 0 1 3 2 2 1 1 2 3 5 7 8 2 4 6 9 15 2 This is in Row Echelon Form o Get a nonzero in the a11lace 1↔ 2 1 2 3 5 7 8 0 1 3 2 2 1 2 4 6 9 15 2 o Want a = 0 13 −21+ 3 1 2 3 5 7 8 0 1 3 2 2 1 0 0 0 -1 1 -14 We have pivots in every column except the 3 and 5 , we can call these Free Columns or free variables which means they can be anything o X3and x 5re free variable. x 1 x2, x4depend on x , x 3 5 3 5= 3: − + = −14 4 5 − 4 = −14 4= + 14 2: + 3 + 2 + 2 = 1 2 3 4 5 2 3 + 2 + 14 + 2 = 1 2= −27 − 3 − 4 1: 1+ 2 2 3 +35 + 4 = 85 1+ 2 −27 − 3 − 4 + 3 + 5 + 14 + 7 = 8 1 6 + 3 − 8 + 5 + 7 = 8 = 8 + 3 − 4 o So the solution is… Page 07 1= −8 + 3 − 4 = −27 − 3 − 4 2 3= 4= 14 + = 5 -8 3 -4 -27 -3 -4 0 t +s + 1 0 14 0 1 0 0 1 General Solution: The linear system Ax=b can be solved using the four steps: o 1. Form the augmented matrix M=[A,b] o 2. Use row operations to eliminate coefficients and reduce M to row echelon form o 3. Write down the simplified system o 4. Solve the simplified system by assigning arbitrary values to the free variables and back solving for the pivot variables Practice Problem: (#27 in textbook) Find the solution of the systemAy=b given: −3 −3 1 4 = 8 7 −2 = −8 8 6 −1 −5 o Answer: = ( 1 −2 1 ) Section 7.4: Homogeneous and Inhomogeneous Systems Definitions: o Nullspace: the set of all solutions to the homogeneous system of linear equations Ax=0. The nullspace of Ais denoted by null(A) Simplistic Example: Find the nullspace of the matrix 2 1 = 4 2 o To do this, we reduceAto row echelon form −2 +1 2 Page 08 2 1 0 0 1 − ⁄ − 1⁄ = = 2= 2 2 1 = ( 1 ) −2 Properties of Nullspace: o 1. Suppose x and y are vectors in null (A). Then x+y is also null (A) o 2. Suppose x is in null (A) and a is a number. Then ax is also in null (A) Practice Problem: (#27 in the textbook) Describe the nullspace of the given matrix parametrically: 1 2 = −2 −4 1 1 o Answer: = 2 = − = −1 3 1 Section 7.5: Bases of a Subspace Definitions o Nontrivial: at least one of the coefficients is different from zero o Linearly Independent: the only linear combination of them that is equal to the zero vector is the trivial one where all of the coefficients are equal to zero o Basis: a set of vectors in a subspace that span V and are linearly independent Simplistic Example: Prove the following vectors are linearly independent 1 −2 −2 1= −2 , 2= −1 , 3= −3 2 2 4 1 −2 −2 = [1,2,3] = −2 −1 −3 2 2 4 2R1 + R2, -2R1 + R3, 6/5R2 + R3… Page 09 1 −2 −2 = 0 −5 −7 2 0 0 − ⁄5 General Solution: Suppose that , ,…, are vectors in R . To 1 2 determine if they are linearly dependent or independent o 1. Form the matrix X=[ ,1,…2 ] wth columns , ,1,2. o 2. Find the nullspace null (X) If null(X)=0, then 1,2,…, re linearly independent If = ( , ,…, ) is a nonzero vector in null(X), then 1 2 1 1+ 2 2 ⋯+ = and , ,…,1 2re linarly dependent Practice Problem: (#21 in the textbook) Show whether the vectors are linearly independent or find a nontrivial linear combination that is equal to zero: 1= (−1,7,7) , =2(−3,7,−4) , = (−4,314,23) o Answer: Independent Page 10