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Course Number: MATH 124, Spring 2009

College/University: Washington

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AMATH 351 Homework 2 Due Oct 16(F) Section2.1 14,16,19,31,40 Section2.4 3,12,15,29 Section2.6 8,21,23,24 Section 2.1 In each of Problems 13 through 20 nd the solution of the given initial value problem. 14. y + 2y = te2t , y (1) = 0 16. y + (2/t)y = (cos t)/t2 , y ( ) = 0, t > 0 19. t3 y + 4t2 y = et , y (1) = 0, t < 0 31. Consider the initial value problem 3 y y = 3t + 2et , 2 y (0) = y0 . Find the...

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351 AMATH Homework 2 Due Oct 16(F) Section2.1 14,16,19,31,40 Section2.4 3,12,15,29 Section2.6 8,21,23,24 Section 2.1 In each of Problems 13 through 20 nd the solution of the given initial value problem. 14. y + 2y = te2t , y (1) = 0 16. y + (2/t)y = (cos t)/t2 , y ( ) = 0, t > 0 19. t3 y + 4t2 y = et , y (1) = 0, t < 0 31. Consider the initial value problem 3 y y = 3t + 2et , 2 y (0) = y0 . Find the value of y0 that separates solutions that grow positively as t from those that grow negatively. How does the solution that corresponds to this critical value of y0 behave as t ? In Problem 40 use the method of Problem 38 to solve the given dierential equation. 40. y + (1/t)y = 3 cos 2t, t>0 Section 2.4 In each of Problems 1 through 6 determine (without solving the problem) and interval in which the solution of the given initial value problem is certain to exist. 3. y + (tan t)y = sin t, y ( ) = 0 In each of Problems 7 through 12 state where in the ty -plane the hypotheses of Teorem 2.4.2 are satised. 12. dy = (cot t)y dt 1+y 1 In each of Problems 13 through 16 solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value y0 . 15. y + y 3 = 0, y (0) y0 = Bernoulli Equations. Sometimes it is possible to solve a nonlinear equation by making a change of the dependent variable that converts it into a linear equation. The most important such equation has the form y + p(t)y = q (t)y n . and is called a Bernoulli equation after Jakob Bernoulli. Problem 27 through 31 deal with equations of this type. 29. y = ry ky 2 , r > 0 and k > 0. This equation is important in population dynamics and is discussed in detail in Section 2.5. Section 2.6 Determin whether each of the equations in Problems 1 through 12 is exact. If it is exact, nd the solution. 8. (ex sin y + 3y )dx (3x ex sin y )dy = 0 Show that the equations in Problems 19 through 22 are not exact but become exact when multiplied by the given integrating factor. Then solve the equations. 21. ydx + (2x yey )dy = 0, (x, y ) = y 23. Show that if (Nx My )/M = Q, where Q is a function of y only, then the dierential equation M + Ny = 0 has an integrating factor of the form (y ) = exp Q(y )dy. 24. Show that if (Nx My )/(xM yN ) = R, where R depends on the quantity xy only, then the dierential equation M + Ny = 0 has an integrating factor of the form (xy ). Find a general formula for this integrating factor. 2

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Washington - MATH - 124
AMATH 351 Homework 2 KeysJuly 8, 2009Section2.1 14,16,19,31,40 Section2.4 3,12,15,29 Section2.6 8,21,23,24Section 2.1In each of Problems 13 through 20 nd the solution of the given initial value problem. 14. y + 2y = te2t , y (1) = 0 Multiply (t) = e2t
Washington - MATH - 124
AMATH 351 Homework 3Due Oct 23(F)Section3.1 Section3.2 Section3.316, 21 3, 18, 22, 24 9, 211
Washington - MATH - 124
AMATH 351 Homework 3 KeysSection3.1 16,21 Section3.2 3,18,22,24 Section3.3 9,21Section 3.116.4y y = 0,y (2) = 1, y (2) = 1 42 1 = 0 = 1 2Characteristic equation So the general solution isy (t) = C1 e 2 + C2 e 2ttApply given conditionsy (2) = C1
Washington - MATH - 124
AMATH 351 Homework 4Due Oct 30(F)Section3.4 Section3.5 Section3.6 Section3.79,20,40 12,20,26 1,14 4,121
Washington - MATH - 124
AMATH 351 Homework 4Due July 22, 2009Section3.4 9,20,40 Section3.5 12,20,26 Section3.6 1,14 Section3.7 4,12 Section 3.4 9y + 2 y 8y = 0Characteristic equation is2 + 2 8 = = 0 2 or 4So the general solution isy (t) = C1 e2t + C2 e4t20y + y = 0, y (
Washington - MATH - 124
AMATH 351 Homework 5Due Nov 13 (F)Section7.1 Section7,2 Section7.3 Section7.56 15, 21, 22 22 2, 11, 311
Washington - MATH - 124
AMATH 351 Homework 5Section7.1 6 Section7,2 15, 21, 22 Section7.3 22 Section7.5 2, 11, 31 Section 7.1 6u + p(t)u + q (t)u = g (t), u(0) = u0 , u (0) = u0Let u1 = u, u2 = u , then the dierential eqn can be written asu2 + p(t)u2 + q (t)u1 = g (t)So the
Washington - MATH - 124
AMATH 351 Homework 6Due Nov 20 (F)Section7.6 Section7,7 Section7.8 Section7.93, 8, 28 5 3 1,71
Washington - MATH - 124
Washington - MATH - 124
AMATH 351 Homework 7Due Nov 27 (F)Section6.1 Section6.213, 19 4, 7, 11, 16, 231
Washington - MATH - 124
AMATH 351 Homework 7Due Aug 12, 2009Section6.1 13, 19 Section6.2 4, 7, 11, 16, 23 Section 6.1 13. This was the example showed in class. 19.L t2 sin atSince L cfw_sin at = page 319,a s2 +a2 ,s &gt; 0, according to the last row of the table onL t2 sin a
Washington - MATH - 124
AMATH 351 Homework 8Due Dec 9 (W)Section5.1 Section5.2 Section5.4 Section5.5 Section5.6 Section5.73, 9, 13 4, 12 6 10 6 71
Washington - MATH - 124
AMATH 351 Homework 8Due August 19, 2009Section5.1 3, 9, 13 Section5.2 4, 12 Section5.4 6 Section5.5 10 Section5.6 6 Section5.7 7Section 5.13.Determine the radius of convergence.x2n n! n=0For the series to be convergent,limx2(n+1) (n+1)! x2n n!n
Washington - MATH - 124
Introduction to Continuous Mathematical ModelingIntroduction my name: Manuel Torrilhon my email:matorril@amath.washington.edu Visiting Faculty from ETH Zurich, Switzerland (until Feb 2010)Introduction my name: Manuel Torrilhon my email:matorril@ama
Washington - MATH - 124
Introduction to dierential equations and applicationsBernard Deconinck Department of Applied Mathematics University of Washington Campus Box 352420 Seattle, WA, 98195, USA August 13, 2006Prolegomenon These are the lecture notes for Amath 351: Introducti
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Final Project GuidelinesAMATH 383, Autumn 2009Course ProjectsA major feature of this introductory mathematical modeling course is that students develop course projects and write term papers on those projects. The purpose of the project is that you obta
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Math 111 Exam 1 January 29, 2009Name: Quiz Section: Student ID Number:1 2 3 4 Total13 13 13 11 50 You are allowed to use a calculator, a ruler, and one hand-written 8.5 by 11 inch page of notes. Put your name on your sheet of notes and turn it in with
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EN017510 / 10 / 06Strain in a solid2v dx2 v dx1v uv d y2v d y1v xv y13Consider an arbitrary fiber within the elastic body, In the undeformed configuration, we can represent the fiber as a small vetor: dx = mdl0 wherevvv dl 0 is the length a
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EN017510 / 19 / 06Mechanical Behavior of Solids Linear Elastic solids = E (1D) ij = Cijkl kl or ij = S ijkl kl (3D)where C is sometimes called the stiffness tensor and S is sometimes called the compliance tensor. Both of them are 4th order elastic mod
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EN017510 / 26 / 06Plastic material behaviorYield condition: Plastic loading: = Y = Y , d &gt; 0Y Y0E1E10 PWe will now denote the initial yield stress asE Y 0 and the current yield stress as Y ; see above figure.Decompose strain into elastic &amp;
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EN017511 / 09 / 06Continue on the Airy stress function method in elasticity: 2 2 = 0 ( : Airy stress function) xx = 2 2 2 , xy = , yy = y 2 xy x 2Example 4:P0ythickness : 1Consider bending of a beam (height: 2c ; thickness: 1) caused by a conce
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EN017511 / 14 / 06Linear elasticity solution in polar coordinates Typical problems: Stress around a circular hole in an elastic solid.0aBoundary conditions: Traction free @ r = a :0 er = rr er + r e = 0 , i.e. rr = 0 , r = 0vvvbaBoundary cond
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EN017511 / 16 / 06Continue on the problem of circular hole under uniaxial tension (remote). Stress concentration occurs at r = a , =2.3TaTGoverning equation is: 22 =0The stress components in polar coordinates are: rr =1 1 2 + r r r 2 2 1 r
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EN017511 / 21/ 06Principle of minimum potential energy (continued) The potential energy of a system isV = wdV f i ui dV ti ui dSV V SPrinciple of minimum potential energy states that for all kinematically admissible u i , the actual displacement fiel
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