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San Jose State | MATH 133
63 sample documents related to MATH 133
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San Jose State MATH 133
Integrals of simple functions 1 Rational functions () () () () dx = xn dx = 1 dx = x du = 2 + u2 a x+C xn+1 +C n+1 ln |x| + C 1 u arctan + C a a if n = 1 2 Logarithms ln x dx logb x dx = = x ln x x + C x logb x x logb e + C 3 Exponential func -
San Jose State MATH 133
Math 133A, Fall 2005 Solutions and answers to the homework problems from Sections 3.2-6.3 3.2.6. (a) The eigenvalues are 1 = 4 and 2 = 9. (b) For 1 = 4 we nd that the eigenvectors are all [x1 , y1 ] such that 9x1 = 4y1 . For 2 = 9 we have x2 = y2 . -
San Jose State MATH 133
Math 133A, Fall 2005 Solutions and answers to the homework problems from Sections 2.1-3.1 2.1.2. For (i) the equilibrium points are x = y = 0 and x = 10, y = 0. For the latter equilibrium point, prey alone exist and predators are absent. For (ii), t -
San Jose State MATH 133
Math 133A Ordinary Differential Equations Fall 2005 Instructor: Telephone: Office Hours: Dimitar Grantcharov 924-5175 MW 3-5 pm, or by appointment. Office: MH 312 Email: grantcharov@math.sjsu.edu Web page: http:/www.math.sjsu.edu/~grantcharov P -
San Jose State MATH 133
San Jos State University e Math 133A, Fall 2005 Midterm 1 Solutions Section 03 September 28, 2005 Name: XYZ Score 1 25 2 25 3 25 4 25 Total 100 Explain your work 1. (25 points) Consider the dierential equation dy = (y 2 y 2) log(1 + y 2 ), dt -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Solutions to graded Homework 8 problems Ex. 3.1, #17. The linear system corresponding to the given second order ODE is dy =v dt dv = qy pv. dt (a) If q = 0 but p = 0, then equilibrium points (y, v) a -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Solutions to graded Homework 5 problems Ex. 1.9, #1. We rewrite the equation as dy = (y 4t) + (y 4t)2 + 4. dt The substitution u = y 4t gives dy/dt = du/dt + 4, so if we substitute, we obtain du + -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Solutions to graded Homework 9 problems Ex. 3.3, #1. The system is dY = AY, dt where A= 32 . 0 2 A is lower-triangular, so its eigenvalues are 1 = 3 and 2 = 2. Recall from one of the previous homewor -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Solutions to graded Homework 13 problems Ex. 6.1 #2. We have: L [t] = 0 T test dt test dt est s t=T T = lim = lim T 0 T t t=0 0 T 0 est dt s () = lim T 1 1 T esT + s s 0 est dt if s -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Solutions to graded Homework 12 problems Ex. 3.7, #2. (a) The trace is T = a and the determinant D = 2. Therefore, the corresponding curve C in the trace-determinant plane is the straight line paralle -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Solutions to graded Homework 11 problems Ex. 3.5, #13. If AY = Y for every vector Y , then in particular, A 1 1 = . 0 0 The left-hand side of the equation equals the rst column of A. Therefore, a = -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Midterm 2 Solutions Section 03 November 2, 2005 Name: Granwyth Hulatberi Score 1 25 2 25 3 25 4 25 Total 100 Explain your work 1. (25 points) Consider the dierential equation dy = y 2 2ty + t2 3 -
San Jose State MATH 133
San Jos State University e Math 133A, Fall 2005 Quiz 2 Solution Suppose we know that the graph below is the graph of a solution to dy/dt = f (y). y y(0) = 3 t (a) How much of the slope eld can you sketch from this information? [Hint: Note that t -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Solutions to graded Homework 2 problems Ex. 1.3, #14. Denote the values of y for which f (y) = 0 by y0 , y1 , and y2 , where y0 < 0 < y1 < y2 . These are the equilibria of the dierential equation dy/ -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Solutions to graded Homework 14 problems Ex. 6.2 #5. First use partial fractions to write F (s) = 1 A B = + . (s 1)(s 2) s1 s2 Multiplying both sides by (s 1)(s 2), we obtain 1 = A(s 2) + B(s 1 -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Sample Final Exam Section 3 Name: Score 1 2 3 4 5 6 Total Explain your work 1. Solve the following initial value problem: dy = 2ty 2 + 3t2 y 2 , dt Solution: y(1) = 1. 2. Consider the following di -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Solutions to graded Homework 7 problems Ex. 2.3, #5. The function Y (t) = (e2t et , e2t ) is not a solution to the system dx = 2x + y dt dy = y dt because the second component of Y (t), e2t , is not -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Solutions to graded Homework 3 problems Ex. 1.5, #8. Let f (y) = (y 2)(y 3)y. Then our ODE is dy/dt = f (y). The equilibria are 0, 2, and 3, since f equals zero there. Does the (unique) solution y( -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Basic Integrals and Methods of Integration This is a list of basic integrals and methods of integration that I expect you to know. You should be able to combine the integration formulas with the metho -
San Jose State MATH 133
Some useful trigonometric identities Start with the identities sin(a + b) = sin a cos b + cos a sin b, sin(a b) = sin a cos b cos a sin b, cos(a + b) = cos a cos b sin a sin b, cos(a b) = cos a cos b + sin a sin b. Adding (1) and (2), then (3) an -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 1 Name: Determine for which values of m the function (x) = emx is a solution to the given equation. Explain your work. (a) (b) Solution: d2 y dy + 6 + 5y = 0 2 dx dx d3 y d2 y dy +3 2 +2 = 0. 3 d -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Midterm 1 Solutions Section 04 Score 1 25 2 25 3 25 4 25 Total 100 1. (25 points) Consider a dierential equation yy 2x = 0. cos y (1) (a) If x0 = 0, show that (1) cannot possibly have a solution d -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Midterm 2 Solutions Section 05 Score 1 25 2 25 3 25 4 25 Total 100 1. (25 points) Determine the form of a particular solution to the given ODE, but do not solve for the coecients: y + 6y + 9y = 2t2 -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 5 Solve the given initial value problem: y 4y + 4y = 0, Solution: The characteristic equation r2 4r + 4 = 0, has a double root r = 2. The general solution is therefore y(t) = (C0 + C1 t)e2t . S -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Homework 4: solutions to even numbered problems Ex. 2.5 8. = y 4 , x2 y 3 y 1 = C, and y = 0 12. exact, x2 y 3 + x log |y| = C 16. y = Cx Ex. 2.6 6. homogeneous 8. y = G(ax + by) 10. y = x Clog |x| -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 7 Find the solution to the initial value problem: y + 9y = 27, y(0) = 4, y (0) = 6. Solution: First we need to solve the associated homogeneous equation y + 9y = 0. Its characteristic equation i -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 2 Solve the equation (x + xy 2 ) dx + ex y dy = 0. Solution: The equation is separable. It can be solved by separating the variables, as follows (we use the same symbol, C, for the arbitrary cons -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 9 Express the given function using unit step functions and compute the Laplace transform. 0, 0 < t < 1, 2, 1 < t < 2, g(t) = 1, 2 < t < 3, 3, 3 < t. (Hint: L{u(t a)}(s) = eas /s.) Solutio -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 6 Solve the given initial value problem: y 2y + 2y = 0, y() = e , y () = 0. Solution: The associated characteristic equation is r2 2r + 2 = 0. Its roots are 1 i, so the general solution is y -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 4 Use the method under Equations of the form dy/dx = G(ax + by) to solve dy = (x + y + 2)2 . dx Solution: Let z = x + y + 2. Then y = z 2 , so z = 1 + y = 1 + z2. This equation is separable. By s -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 3 Name: Solve the initial value problem dy + 4y ex = 0, dx Solution: 4 y(0) = . 3 -
San Jose State MATH 133
San Jose State University Department of Mathematics Fall 2004 Math 133A: Ordinary Dierential Equations Instructor: Slobodan Simi c Oce: 318A MacQuarrie Hall Phone: (408) 924-7485 Email: simic@math.sjsu.edu Course web page: http:/www.math.sjsu.edu/ -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 9 Name: Express the given function using unit step functions and compute the Laplace transform. 0, 0 < t < 1, 2, 1 < t < 2, g(t) = 1, 2 < t < 3, 3, 3 < t. (Hint: L{u(t a)}(s) = eas /s.) S -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 8 Name: Determine the inverse Laplace transform of the given function: 3 . (2s + 5)3 Solution: -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 2 Name: Solve the equation (x + xy 2 ) dx + ex y dy = 0. Solution: 2 -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 5 Name: Solve the given initial value problem: y 4y + 4y = 0, Solution: y(1) = 1, y (1) = 1. Dont forget to register to vote! -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Solutions to graded Homework 10 problems Ex. 3.4, #5. (a) The characteristic polynomial associated with dY 1 2 = Y 1 1 dt is 2 + 2 + 3, so the eigenvalues are = 1 i 2. (b) The eigenvalues are compl -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Quiz 5 Solution Consider the linear system dY 1 2 = Y. 1 1 dt Determine if the origin is a spiral sink, a spiral source, or a center, and approximately sketch the phase portrait (paying attention to t -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Quiz 3 Change the dependent variable from y to u using the change of variables indicated. Describe the equation in the new variable (such as separable, linear, autonomous,.). dy = t(y + ty 2 ) + cos t -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Quiz 6 Solution Consider the following initial value problem: dy + 4y = 6, dt y(0) = 0. (a) Compute the Laplace transform of both sides of the equation. (b) Substitute the initial condition and solve -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Quiz 4 Solution Find the equilibrium points of the given system: dx =y dt dy = x x3 y. dt Solution: The equilibrium points of the given system of ODEs are solutions of the system of equations y=0 x -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Midterm 2 Solutions Section 04 Score 1 25 2 25 3 25 4 25 Total 100 1. (25 points) Determine the form of a particular solution to the given ODE, but do not solve for the coecients: y 8y + 16y = 2t2 -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Midterm 1 Solutions Section 05 Score 1 25 2 25 3 25 4 25 Total 100 1. (25 points) Consider a dierential equation yy 2xey = 0. (1) (a) If x0 = 0, show that (1) cannot possibly have a solution dened -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 3 Solve the initial value problem dy + 4y ex = 0, dx Solution: The equation is linear with P (x) = 4, Therefore, the integrating factor is (x) = exp so the general solution is y= 1 (x) (x)Q(x) d -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 8 Determine the inverse Laplace transform of the given function: 3 . (2s + 5)3 Solution: Using L{tn }(s) = n!/sn+1 with n = 2 and L{eat f (t)}(s) = L{f }(sa), we obtain L1 3 (2s + 5)3 (t) = 3 1 2 -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Sample Final Exam Name: Score 1 2 3 4 5 6 Total Explain your answers 1. Solve the following initial value problem: y = (1 + y 2 ) tan x, Solution: y(0) = 3. 2. Consider the equation (y 2 + 2xy)dx -
San Jose State MATH 133
Math 133A Essentials I. Basic notions. Solutions (explicit and implicit), initial value problems (IVPs) Existence and uniqueness theorem Direction elds II. First order equations. Methods for solving: Separable equations y = u(x)v(y) Linear equat -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 4 Name: Use the method under Equations of the form dy/dx = G(ax + by) to solve dy = (x + y + 2)2 . dx Solution: -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 6 Name: Solve the given initial value problem: y 2y + 2y = 0, Solution: y() = e , y () = 0. -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 7 Name: Find the solution to the initial value problem: y + 9y = 27, Solution: y(0) = 4, y (0) = 6. -
San Jose State MATH 133
Summary for Midterm 2 I. First-order systems: General theory. 1. Notion of a solution (curve), equilibrium, phase portrait. 2. Converting 2nd order equations into systems. 3. Vector and direction elds vs. solution curves. 4. Solving decoupled and pa -
San Jose State MATH 133
San Jose State University Department of Mathematics Fall 2005 Math 133A: Ordinary Dierential Equations Instructor: Slobodan Simi c Oce: 318A MacQuarrie Hall Phone: (408) 924-7485 Email: simic@math.sjsu.edu Web: http:/www.math.sjsu.edu/simic/Fall05 -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Solutions to graded Homework 1 problems Ex. 1.1, #2. (a) Recall that a solution P (t) is called an equilibrium if it is constant. This means P (t) = 0, for all t. So the equilibria of the given equat -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Solutions to graded Homework 4 problems The solution to dy + p(t)y = q(t) dt y(t) = where (t) = e R p(t) dt . is 1 (t) (t)q(t) dt, Ex. 1.8, #2. We have p(t) = 3/t and q(t) = t5 , so (t) = t3 and y -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Quiz 1 Name: Solve the given initial value problem: dy t2 = , dt y + t3 y Solution: y(0) = 2. -
San Jose State MATH 133
San Jos State University e Math 133A, Fall 2005 Quiz 2 Name: Suppose we know that the graph below is the graph of a solution to dy/dt = f (y). y y(0) = 3 t (a) How much of the slope eld can you sketch from this information? [Hint: Note that the -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Quiz 3 Name: Change the dependent variable from y to u using the change of variables indicated. Describe the equation in the new variable (such as separable, linear, autonomous,.). dy = t(y + ty 2 ) + -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Quiz 4 Name: Find the equilibrium points of the given system: dx =y dt dy = x x3 y. dt Solution: -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Quiz 5 Name: Consider the linear system dY 1 2 = Y. 1 1 dt Determine if the origin is a spiral sink, a spiral source, or a center, and approximately sketch the phase portrait (paying attention to the -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Quiz 6 Name: Consider the following initial value problem: dy + 4y = 6, dt y(0) = 0. (a) Compute the Laplace transform of both sides of the equation. (b) Substitute the initial condition and solve fo -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2004 Quiz 1 Determine for which values of m the function (x) = emx is a solution to the given equation. Explain your work. (a) (b) d2 y dy + 6 + 5y = 0 2 dx dx d2 y dy d3 y +3 2 +2 = 0. 3 dx dx dx Solutio -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Solutions to graded Homework 6 problems Ex. 2.1, #5. Substituting x = 0 into the equation for dx/dt yields dx/dt = 0 for all t. Therefore, x(t) is constant, and since x(0) = 0, it follows that x(t) = -
San Jose State MATH 133
San Jose State University Math 133A, Fall 2005 Quiz 1 Solution Solve the given initial value problem: dy t2 = , dt y + t3 y y(0) = 2. Solution: Separating the variables and integrating, we obtain ydy = t2 dt. 1 + t3 The right-hand side can be so
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