MATH 261 MATH 261: Elementary Differential Equations MATH 261 FALL 2005 FINAL EXAM FALL 2005 FINAL EXAM EXAMINATION COVER PAGE Professor Moseley

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1 MATH 6 MATH 6: Elementary Differential Equations MATH 6 FALL 5 FINAL EXAM FALL 5 FINAL EXAM EXAMINATION COVER PAGE Professor Moseley PRINT NAME ( ) Last Name, First Name MI (What you wish to be called) ID # EXAM DATE Tuesday, May 3, 4, 8: am I swear and/or affirm that all of the work presented on this exam is my own and that I have neither given nor received any help during the exam. SIGNATURE DATE INSTRUCTIONS:.Besides this cover page, there are 33 pages of questions and problems on this exam. Page 34 contains Laplace transforms you need not memorize. MAKE SURE YOU HAVE ALL THE PAGES. If a page is missing, you will receive a grade of zero for that page. Read through the entire exam. If you cannot read anything, raise your hand and I will come to you..place your I.D. on your desk during the exam. Your I.D., this exam, and a straight edge are all that you may have on your desk during the exam. NO CALCULATORS! NO SCRATCH PAPER! Use the back of the exam sheets if necessary. You may remove the staple if you wish. Print your name on all sheets. 3.Pages -33 are multiple choice. Expect no part credit on these pages. There are no free response pages. However, to insure credit, you should explain your solutions fully and carefully. Your entire solution may be graded, not just your final answer. SHOW YOUR WORK! Every thought you have should be expressed in your best mathematics on this paper. Partial credit may be given as deemed appropriate. Proofread your solutions and check your computations as time allows. GOOD LUCK!! Scores Scores Scores page points score page points score page points score ST 75 ST Scores page points score ST ST 75 ST 5 ST3 5 ST4 3

2 Tot MATH 6 FINAL EXAM Fall 5 Professor Moseley Page PRINT NAME ( ) SS No. ( 5 pts.) Classify the following first order ordinary differential equations. The classification relates to the method of solution. Recall from class (attendance is mandatory) that the possible methods are: A) First order linear (y as a function of x).- Integrating factor = µ = exp( p(x) dx ) B) First order linear (x as a function of y).- Integrating factor = µ = exp( p(y) dy ) C) Separable. D) Exact Equation (Must be exact in one of the two forms discussed in class). E) Bernoulli, but not linear (y as a function of x). Use the substitution v = y -n. AB) Bernoulli, but not linear (x as a function of y). Use the substitution v = x -n. AC) Homogeneous, but not separable. Use the substitution v = y/x or v = x/y. AD) None of the above techniques works. Also recall the following: a. In this context, exact means exact as given (in either of the forms discussed in class). b. Bernoulli is not a correct method of solution if the original equation is linear. c. Homogeneous (use the substitution v=y/x) is not a correct method of solution if it converts a separable equation into another separable equation. Circle one and only one answer. Do not put more than one answer. If more than one method works, then any correct answer will receive full credit. Also remember that if I cannot read your answer, it is WRONG. DO NOT SOLVE.. (y + xy )dx + x dy = A B C D E AB AC AD. (e x + xy + x )dx + (x + y)dy = A B C D E AB AC AD 3. (y + x )dx + x dy = A B C D E AB AC AD 4. xye x+y dx + dy = A B C D E AB AC AD 5. (xy + cos(x) )dx + ( + x ) dy = A B C D E AB AC AD

3 Total points this page = 5. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Prof. Moseley Page PRINT NAME ( ) ID No. Last Name, First Name MI What you wish to be called 6. ( pts.) To solve the first order linear Ordinary Differential Equation (ODE) y' = y + x, we first put it in the standard form (for solving first order linear ODE s) (Circle the correct (first order linear) standard form for this ODE s. Be careful. No part credit for this problem. Hence if you miss this part, it may cause you to miss all parts): A. y = y + x (It s already in the appropriate form for solving a first order linear ODE) B. y + y = x, C. y + y + x, D. y y = x E. y y x = AB. None of the above 7. ( pts.) To solve the first order linear Ordinary Differential Equation (ODE) above, we must find an integrating factor. An integrating factor µ for this linear ODE is (Circle the correct integrating factor. Be careful. No part credit for this problem): A. µ = x, B. µ = x, C. µ = e x, D. µ = e x E. µ = e x, AB. µ = e x, AC. µ = e x, AD. µ = e -x, AE. None of the above 8. ( 3 pts.) In solving the linear Ordinary Differential Equation (ODE) above which of the following steps occurs. (Circle the step that is correct. Be careful. No part credit for this problem): x d(ye ) x x A. xe, B. xe, C. dx x d(ye ) xe dx x d(ye ) 3 x dx x d(ye ) dx -x d(ye ) x d(ye ) dx x e x -x x D., E. x e, AB. xe, dx x x d(ye ) AC., AD. None of the above steps ever appears in any solution of this problem. dx

4 Possible points this page = 6. TOTAL POINTS EARNED MATH 6 FINAL EXAM Fall 5 Prof. Moseley Page 3 PRINT NAME ( ) ID No. Last Name, First Name MI What you wish to be called To solve the first order linear ODE, we isolate the unknown function on the left side of the equation. Recall that an ODE is really a vector equation with the infinite number of unknown variables being the values of the function for each value of the independent variable in the function s domain. The isolation of the dependent variable (or function) solves for all of the (infinite number of) unknowns simultaneously. In solving a particular first order linear ODE an x d(ye ) x integrating factor and the product rule were used to reach the following step: xe. dx 9. ( pts.) Circle the theorem from calculus that allows you to integrate the Left Hand Side of this equation: A. Intermediate Value Theorem, B. Mean Value Theorem C. Rolle's Theorem, D. Chain Rule, E. Fundamental Theorem of Calculus, AB. Product Rule, AC. Integration by Parts. AD. Partial Fractions, None off the above.. (5 pts.) Now complete the solution process to obtain y as a function of x. However, be sure that you have given all solutions; that is, that you have given parametrically the entire collection of solutions to this problem. Circle the correct solution or family of solutions below. A. y = x + + c e x, B. y = x + + c e x, C. y = x + c e x, D. y = x + + c e x, E. y = x + + c e x, AB. y = x + + c e x, AC. y = x + c e x, AD. y = x + + e x + c, AE. y = x + + e x + c, BC. y = x + e x + c, BD. y = x + + e x + c, BE. y = x + + e x + c, CD. y = x + e x + c, CE. y = x + e x + c, DE. None of the above families of solutions is correct..

5 Possible points this page = 7. TOTAL POINTS EARNED MATH 6 FINAL EXAM Fall 5 Prof. Moseley Page 4 i x You are to solve A x b where A, and. Be sure you write x x x i b i x y your answer according to the directions given in class for these kinds of problems (attendance is mandatory). A b U c =. i i i U c. (4 pts.) If is reduced to using Gauss elimination, then i i A., B., C., D., E., AB. None of the above. (4 pts.) The solution of is. A x b x x x i i i A. No Solution, B. x, C., D., ED., x y x x y i i i x y x y x y AB, AC., AD., AE. None of the above.

6 Total points this page = 8. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Prof. Moseley Page 5 ( 5 pts.) True or false. Solution of Abstract Linear Equations (having either R or C as the field of scalars). Assume T: V W is a linear operator from a vector space V to a vector space W. Now consider T(x) b. (*) Under these hypotheses, determine which of the following is true and which is false. If true, circle True. If false, circle False. True or False 3. If b, then (*) always has at least one solution. True or False 4. The vector equation (*) may have exactly two solutions. B {x,x,...,x n} x c x c x c x True or False 5. If the null space of T has a basis and b, then the general solution of (*) is given by c, c,..., c n are arbitrary constants. n n where True or False 6. Either (*) has no solutions, exactly one solution, or an infinite number of solutions. N(T) {} b True or False 7. If the null space of T is and is in the range space of T, then (*) has a unique solution.

7 Total points this page = 5. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Professor Moseley Page 6 (9 pts.) The dimension of the null space of the linear operator L[y] = y y that maps A(R,R) to A(R,R) is. Assuming a solution of the homogeneous equation L[y] = of the form y = e rx leads to the two linearly independent solutions y = and y = e x. Hence we can deduce that y h = c + c e x is the general solution of y y =. Use the method discussed in class to determine the proper (most efficient) form of the judicious guess for a particular solution y p of the following ode's. Circle the correct (most efficient) form of the judicious guess for a particular solution y p of the following ode's 8. y y = xe x A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE 9. y y = 3 sin x A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE. y y = 4e x A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE Possible Answers: A. Ae x B. Axe x C Ax e x D. Axe x + Be x E. Ax e x + Bxe x AB. Ae x AC. Axe x AD. Ax e x AE. Axe x + Be x BC. Ax e x + Bxe x BD. A sin x BE. cos x CD. A x sin x CE. A x cos x ABC. A sin x + B cos x ABD. A x sin x + B x cos x ABE. None of the above

8 Total points this page = 9. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 4 Professor Moseley Page 7 PRINT NAME ( ) SS No. You are to solve y" + y = tan(x) I = (, π/) (i.e. < x < π/). Let L[y] = y" + y.. ( pts.) The general solution of L[y] = is. A. c cos(x) + c sin(x), B. c cos(x) + c sin(x), C.c e x + c e x, D. c x + c, E. r = ±i, AB.r = ±, AC r = ±i, AD. None of the above.. (3 pts.) To find a particular solution y p to L[y] = tan(x) using the technique of variation of parameters, we let y p (x) = u (x) cos(x) + u (x) sin(x). Substituting into the ODE and making the appropriate assumption we obtain: A. u (x) cos(x) + u (x) sin(x) =, u (x) sin(x) + u (x) cos(x) =, B. u (x) cos (x) + u (x) sin(x) =, u (x) sin(x) + u (x) cos(x) = tan(x), C. u (x) cos (x) + u (x) sin(x) = tan(x), u (x) sin(x) + u (x) cos(x) =, D u (x) cos (x) + u (x) sin(x) =, u (x) sin(x) + u (x) cos(x) = sin(x), E. u (x) cos (x) + u (x) sin(x) =, u (x) sin(x) + u (x) cos(x) = cos(x), AB. None of the above. 3. (3 pts.) Solving we obtain: A. u (x) = sin (x)/cos(x), u (x) = sin(x), B. u (x) = sin(x), u (x) = sin (x)/cos(x), C. u (x) =, u (x) = sin(x), D. u (x) = sin (x)/cos(x), u (x) =, E. u (x) =, u (x) = sin(x), AB. None of the above. 4. (3 pts.) Hence we may choose : A. u (x) = ln(tan(x) +sec(x)), u (x) = cos(x), B. u (x) = cos(x), u (x) = ln(tan(x) +sec(x)), C. u (x) = x, u (x) = cos(x), D. u (x) = ln(tan(x) +sec(x)), u (x) = x, E. u (x) =, u (x) = cos(x), AB. None of the above. 5. ( pts.) Hence a particular solution to L[y] = tan(x) is: A. y p (x) = ln(tan(x) +sec(x)), B. [cos(x)] ln(tan(x) +sec(x)), C. [sin(x)]ln(tan(x) +sec(x)), D. [tan(x)] ln(tan(x) +sec(x)), E. sin(x) cos(x), AB. sin(x) cos(x), AC. [sin(x) cos(x)] ln(tan(x) +sec(x)), 6. ( pts.) Hence the general solution of L[y] = tan(x) is: A. ln(tan(x) +sec(x)) + c cos(x) + c sin(x), B. [cos(x)] ln(tan(x) +sec(x)) + c cos(x) + c sin(x), C.[sin(x)]ln(tan(x) +sec(x))+c cos(x) + c sin(x), D. sin(x) cos(x) + c e x + c e x, E. ln(tan(x) +sec(x)) + c e x + c e x AB. [cos(x)] ln(tan(x) +sec(x)) + c e x + c e x, AC. None of the above.

9 Total points this page = 5. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Professor Moseley Page 8 PRINT NAME ( ) IDo. You solved y" + y = tan(x) I = (π/, π/) (i.e π/ < x < π/) on the previous page. Now let L: A((,π/),R)A(R,R) be defined by L[y] = y" + y and answer the following questions.. ( pts.) The general solution of L[y] = is. A. c cos(x) + c sin(x), B. c cos(x) + c sin(x), C. c e x + c e x, D. c x + c, E. r = ±i, AB. r = ±, AC. r = ±i, AD. None of the above.. (3 pts.) To find a particular solution y p to L[y] = tan(x) using the technique of variation of parameters, you let y p (x) = u (x) cos(x) + u (x) sin(x). Substituting into the ODE and making the appropriate assumption you obtained: A. u (x) cos(x) + u (x) sin(x) =, u (x) sin(x) + u (x) cos(x) =, B. u (x) cos (x) + u (x) sin(x) =, u (x) sin(x) + u (x) cos(x) = tan(x), C. u (x) cos (x) + u (x) sin(x) = tan(x), u (x) sin(x) + u (x) cos(x) =, D u (x) cos (x) + u (x) sin(x) =, u (x) sin(x) + u (x) cos(x) = sin(x), E. u (x) cos (x) + u (x) sin(x) =, u (x) sin(x) + u (x) cos(x) = cos(x), AB. None of the above. 3. (3 pts.) Solving we obtain: A. u (x) = sin (x)/cos(x), u (x) = sin(x), B. u (x) = sin(x), u (x) = sin (x)/cos(x), C. u (x) =, u (x) = sin(x), D. u (x) = sin (x)/cos(x), u (x) =, E. u (x) =, u (x) = sin(x), AB. u (x) = sin(x), u (x) =, AC. None of the above. 4. (3 pts.) Hence we may choose : A. u (x) = ln(tan(x) +sec(x)) + sin x, u (x) = cos(x), B. u (x) = cos(x), u (x) = ln(tan(x) +sec(x)), C. u (x) = x, u (x) = cos(x), D. u (x) = ln(tan(x) +sec(x)), u (x) = x, E. u (x) =, u (x) = cos(x), AB. u (x) = cos(x), u (x) = x, AC. None of the above. 5. ( pts.) Hence a particular solution to L[y] = tan(x) is: A. y p (x) = ln(tan(x) +sec(x)), B. [cos(x)] ln(tan(x) +sec(x)), C. [sin(x)]ln(tan(x) +sec(x)), D. [tan(x)] ln(tan(x) +sec(x)), E. sin(x) cos(x), AB. sin(x) cos(x), AC. [sin(x) cos(x)] ln(tan(x) +sec(x)), AD. None of the above. 6. ( pts.) Hence the general solution of L[y] = tan(x) may be written as: A. ln(tan(x) +sec(x)) + c cos(x) + c sin(x), B.[cos(x)] ln(tan(x) +sec(x)) + c cos(x) + c sin(x), C.[sin(x)]ln(tan(x) +sec(x))+c cos(x) + c sin(x), D. sin(x) cos(x) + c e x + c e x, E. ln(tan(x) +sec(x)) + c e x + c e x AB. [cos(x)] ln(tan(x) +sec(x)) + c e x + c e x, AC. None of the above.

10 Total points this page = 5. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Prof. Moseley Page 9 Solve y IV 4 y + 4y= on this page and answer the questions on the next page. Be careful as once you make a mistake, the rest is wrong.

11 MATH 6 FINAL EXAM Fall 5 Prof. Moseley Page You solved y IV 4 y + 4y= on the previous page. Now let L:A(R,R) A(R,R) be defined by L[y] = y IV 4 y + 4y and answer the following questions by circling the correct answers. 7. ( pt). The order of the ODE is: A. B. C. 3 D. 4 E. 5 AB. 6 AC. 7 AD) None of the above. 8. ( pt). The dimension of the null space for the linear operator L is: A. B. C. 3 D. 4 E. 5 AB. 6 AC. 7 AD) None of the above. 9. ( pts). The auxiliary equation for the ODE is: A. r 4r + 4 =, B. r 4 4r + 4 =, C. r 4 4r 3 + 4r =, D. r 6 + 4r 3 + 4r =, E. r 6 4r 3 + 4r =, AB. None of the above. 3. ( pts). Listing repeated roots, the roots of the auxiliary equation are: A. r =,, B. r =,,,, C. r =,, D. r =,4, E. r =,,4, AB. r =,,, AC. None of the above. 3. ( pts). A basis for the null space of the linear operator L is : A. {, x, e x, xe x }, B. {, x, e x, xe x }, C. {, x, x, e x }, D. {, e x }, E. {, x, x, x 3 }, AB. {e x, xe x, e x, xe x }, AC. {, x, x, e x }, AD. {, e x }, AE. None of the above. 3. ( pt). The general solution of y IV 4 y + 4y= is: A. y(x) = c + c x + c 3 e x + c 4 xe x, B. y(x) = c + c x +c 3 e x + c 4 xe x, C. y)x) = c + c x + c 3 x + c 4 e x, D. y(x) = c + c e x E. y(x) = c + c x + c 3 x + c 4 x 3, AB. y(x) = c e x + c xe x + c 3 e x + c 4 xe x AC. y(x) = c + c x + c 3 x +c 4 e x AD. y(x) = c + c e x. AE. None of the above.

12 Points this page = 8. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Professor Moseley Page (9 pts.) Let L:A(R,R) A(R,R) be defined by L[y] = y + y. The dimension of the null space of the linear operator L is 3. Assuming a solution of the homogeneous equation L[y] = of the form y = e rx leads to the three linearly independent solutions y = and y = cos(x) and y 3 = sin(x) so that a basis of the null space of L is {, cos(x), sin(x)}. Hence we can deduce from the linear theory that y c = c + c cos(x) + c 3 sin(x) is the general solution of y + y = on R. Use the method discussed in class (attendance is mandatory) to determine the proper (most efficient) form of the judicious guess for a particular solution y p of the following ODE's. Circle the answer that corresponds to your final guess. 33. y + y = sin(x) A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE BCD BCE BDE. CDE. ABCD 34. y + y = 4 x A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE BCD BCE BDE. CDE. ABCD 35. y + y = 4xe -x A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE BCD BCE BDE. CDE. ABCD

13 Possible Answers A. A B. Ax C. Ax D. Ax + B E. Ax + Bx AB. Ax + Bx + C AC. Ae x AD. Axe x AE. Ax e x BC. Axe x + Be x BD. Ax e x + Bxe x BE. Ae x CD. Axe x CE. Ax e x DE. Axe x + Be x ABC. Ax e x + Bxe x ABD. A sin x ABE. cos x BCD A x sin x BCE A x cos x BDE. A sin x + B cos x CDE. A x sin x + B x cos x ABCD None of the above Total points this page = 9. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Prof. Moseley Page (9 pts.) Compute the Laplace transform of the following functions. Then circle the letter corresponding to the Laplace transform of the function. 36. f(t) = + 3t A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE. 37. f(t) = e t + 3 e -3t A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE. 38 f(t) = sin(t) + 3 cos(3t) A B C D E AB AC AD AE BC BD BE CD CE DE ABC ABD ABE. Possible Answers 3 s s 3 s s 3/ s s s s s s A., B., C., D., E., AB. 3 s

14 AC., AD., AE., BC. 3, s s 3 s s 3 s 3 s (s 3) (s 3) 4 3s s 3 3 BD., BE., CD., CE., s 4 s s s 4 s 4 s 9 s s 4 3 s s 3 DE., ABC., ABD. ABE. None of the above. s 4 s 9 s 4 s 9 s 4 s 9 Total points this page = 9. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Prof. Moseley Page 3 ( pts.) Compute the inverse Laplace transform of the following functions: F(s) = + A B C D E AB AC AD AE BC BD BE CD s s + CE DE s F(s) = A B C D E AB AC AD AE BC BD BE CD s + 9 CE DE s F(s) = A B C D E AB AC AD AE BC BD BE CD s - s + CE DE

15 Possible Answers A. + e t, B. 3 + e t, C. + e 3t, D. +3 e t, E. + e t, AB. cos 3t + (4/3) sin 3t, AC. cos 3t + 4 sin 3t, AD. cos t + (4/3) sin t, AE. 3 cos 3t + (4/3) sin 3t, BC. cos 3t + 3 sin 3t, BD. e t cos 3t + 5 e t sin 3t, BE. e t cos t + 5 e t sin t, CD. e t cos 3t + e t sin 3t, CE. 5 e t cos 3t + 5 e t sin 3t, DE. None of the above Total points this page =. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Prof. Moseley Page 4 i Consider the matrix A = C x. 4. ( pt.) The degree of the polynomial where the solution of p(λ) = yields the eigenvalues of A is. A., B., C. 3, D. 4, E. 5, AB. 6, AC.7, AD. None of the above. 43. ( pt.) The polynomial p(λ) where the solution of p(λ) = yields the eigenvalues of A is.. A. p(λ) = (λ)(λ i), B. p(λ) = (λ)(λ +i), C. p(λ) = (λ+)(λ i), D. p(λ) = (λ+)(λ +i), E. p(λ) = (λ)(λ i), AB. p(λ) = (λ)(λ +I), AC. p(λ) = (λ+)(λ i), AD. p(λ) = (λ)(λ +i), AE. p(λ) = (λ+)(λ + i), BC. p(λ) = (λ+)(λ +i), BD. p(λ) = (λ+)(λ i), BE. p(λ) = (λ)(λ +i), CD. None of the above. 44. ( pt.) Counting repeated roots, the matrix A given above has how many eigenvalues? A., B, C., D. 3, E 4, AB. 5, AC. 6, AD. 7, AE ( pts.) The eigenvalues of A may be chosen as. Circle the correct answer. A. λ =, λ = i, B. λ =, λ = i, C. λ =, λ = i, D. λ =, λ = i, E. λ =, λ = i, AB. λ =, λ = i, AC. λ =, λ = i, AD. λ =, λ = i, AE. λ =, λ = i, BC. λ =, λ = i, BD. λ =, λ = i, BE. λ =, λ = i, CD. None of the above

16 Total points this page = 6. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Prof. Moseley Page 5 PRINT NAME ( ) ID No. 3 λ = is an eigenvalue of the matrix A = (5 pts.) Using the conventions discussed in class (attendance is mandatory), find a basis B for the eigenspace associated with this eigenvalue. Circle the correct answer. A. B = {[,] T, [4,4] T }, B. B = {[,] T }, C. B = {[,] T }, D. B = {[,] T, [4,8] T } E. B = {[,] T } AB. B = {[,3] T } AC. B = {[,4] T }, AD. B = {[4,] T } AE. B = {[3,] T }, BC. B = {[,] T, [4,4] T }, BD. B = {[,] T }, BE. B = {[,] T }, CD. B = {[,] T, [4,8] T }, CE. B = {[,] T } DE. B = {[,3] T } ABC. B = {[,4] T }, ABD. B = {[4,] T } ABE. B = {[3,] T }, BC. None of the above.

17 Total points this page = 5. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Prof. Moseley Page ( 3 pts.) Consider the scalar equation u + 4 u u = where u = u(t) (i.e. the dependent variable u is a function of the independent variable t so that u = du/dt and u = du/dt). Convert this to a system of two first order equations by letting u = x and u = y (i.e. obtain two first order scalar equations in x and y; you may think of x as the position and y as the velocity of a point particle). Now write this system of two scalar equations in the vector form x Ax x where x and A is a x matrix. Be sure to give A explicitly. Choose the correct answer y from the following: x x x x x x A., B., C. y 4 y y 4 y y 4 y x x x x x x D., E., AB. y 4 y y 4 y y 4 y x x x x AC., AD., AE. None of the above. y 4 y y 4 y

18 Total points this page = 3. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Professor Moseley Page 7 Solve x 3 Ax where A and on this page. Provide appropriate steps x y and a table as explained in class (attendance is mandatory). Remember, once you make a mistake, the rest is wrong. Then answer the questions on the next page. x

19 MATH 6 FINAL EXAM Fall 5 Professor Moseley Page 8 PRINT NAME ( ) ID No. 3 Let A and L:A(R,R )A(R,R ) be defined by. Let r be an eigenvalue L[x] x Ax of A with associated eigenvector. Similarly, let r be a second eigenvalue with associated eigenvector. 48. ( 3 pts.) We may choose r and r as follows: A. r =, r =, B. r =, r =, C. r =, r =, D. r =, r =, E. r =, r =, AB. r =, r =, AC. r =, r =, AD. r =, r = 3, AE. r =, r = 3, BC. r =, r = 3, BD. None of the above are possible. 49. ( 5 pts.) Following the conventions discussed in class (attendance is mandatory) the eigenvectors associated with the eigenvalues selected above should be chosen as: A.,, B.,, C.,, D.,, E.,, AB.,, AC.,, AD.,, AE.,, BC.,, BD.,, BE., CD. None of the above are the eigenvectors for the eigenvalues chosen in question 48 above. (This includes the case where BD was chosen in question 48.) t t 5. (3 pts.) A basis for the null space of the linear operator L is: A., B e, e,, t t t t t t B. B e, e,, C. B e, e,, D. B e, e,, t t t t t t E. B e, e,, AB. B e, e,, AC. B e, e,,

20 t t t t AD. B e, e,, AE. B e, e,, BC BD. t t B e, e,, CD. None of the above t t B e, e, t t 5. ( pts.) The general solution of x Ax is: A. x(t) c e c e, t t t t x(t) c e c e t t B. x(t) c e c e, C. x(t) c e c e, D. t t t t t t E. x(t) c e c e, AB. x(t) c e c e, AC. x(t) c e c e, t t t t t t AD. x(t) c e c e, AE. x(t) c e c e, BC. x(t) c e c e, t t t t BD. x(t) c e c e, BE. x(t) c e c e, CD. None of the above. Total points this page = 3. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Professor Moseley Page 9 ( 5 pts.) True or False. Odd, even, and periodic functions. Let f and g be real valued functions of a real variable; that is, f:r R and g:r R. Circle True if the statement is true. Circle False if the statement is false. True False 5. The function f is even if f(-x) = f(x) x R. True False 53 The function f is odd if f(-x) = f(x) x R. True False 54. If f and g are both odd functions, then the product of f and g is an odd function. True False 55. The function f is periodic of period T if f(x + T) = f(x) x R. True False 56. If f and g are even functions, then we know that f (x)g(x)dx =. ( 5 pts.) Circle Odd if the function is odd. Circle Even if the function is even. Circle Neither if the function is neither odd nor even. 57. f(x) = x A. Odd B. Even C. Neither

21 58. f(x) = -3 A. Odd B. Even C. Neither 59. f(x) = sin(x) A. Odd B. Even C. Neither 6. f(x) = *-x* A. Odd B. Even C. Neither 6. f(x) = -e x A. Odd B. Even C. Neither Total points this page =. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Professor Moseley Page Let PC fs( R, R; ) = {ff(r,r): f is periodic of period, f and f' are piecewise continuous on [),], and f(x) =( f( x+) + f( x))/ at points of discontinuity}, PC fs([, ], R; ) be the functions in PC fs ( R, R; ) with their domains restricted to [,] and PC ffs([, ], R; ) be the subspace of PC ([, ], R; ) for which the Fourier series is finite. Recall that fs k k B {/ } {cos( ) : k N} {sin( ) : k N} PC fs ([, ], R; ) PC ([, ], R; ) ffs and a Schauder basis of PC ([, ], R; ). fs is a Hamel basis of 6. ( pt.) A Hamel basis for PC ([,], R;) is A. {/ } {cos( ) : k N} {sin( ) : k N}, k k k k {/ } {cos( ) : k N} {sin( ) : k N} B. {/ } {cos( ) : k N} {sin( ) : k N}, C. D., E. AB. None of te above. ffs k k k k {/ } {cos( ) : k N} {sin( ) : k N} k k {/ } {cos( ) : k N} {sin( ) : k N} 63. ( pts.) Determine the Fourier series for the function f(x) which has period 4 and is defined on the interval [-,) by f(x) = 3 + cos(πx) + sin(3π x). (k ) 4 (k ) A. f(x) = 3 + sin( x), B. f(x) = + sin( x), (k ) (k ) n n

22 4 (k ) C. f(x) = 3 + cos( x), D. f(x) = + sin(k x), n (k ) n k 4 (k ) E. f(x) = + cos(k x), AB. f(x) = + sin( x), n k n (k ) (k ) (k ) AC. f(x) = + cos( x), AD. f(x) = + sin( x), (k ) (k ) n n AE cos(πx) + sin(3π x), BD. 3 + cos(πx) + sin(3π x). CD. + cos(πx) + sin(3π x), DE. 3 + cos(πx) + sin(3π x), BC. 3 + cos(πx) + 3 sin(3π x), BE. 3 + cos(πx) + sin(π x), CE. + cos(πx) + sin(3π x). ABC. None of the above. Total points this page = 3. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Professor Moseley Page Determine the Fourier series for the function f(x) which has period 4 and is defined on the interval [-,) by x f(x) x Sketch f for several periods. As discussed in class, indicate on your sketch the function to which the Fourier series converges. Next write down the general formulas for a Fourier series and its coefficients that you memorized.

23 MATH 6 FINAL EXAM Fall 5 Professor Moseley Page 64. ( pts.) Suppose that a function f is periodic with period. Then the formula for the general Fourier series for f given in our text is N n n A. f(x) = a + a n cos( x) bn sin( x), n n n B. f(x) = a + a n cos( x) bn sin( x), a n n n C. f(x) = + a n cos( x) bn sin( x), a D. f(x) = + N n n a cos(nx) b sin(nx) n a E. f(x) = n n + a n cos( x) bn sin( x), a n n AB. f(x) = + a n cos(nx) bn cos(nx) AC. None of the above. n 65. (pts.) where n A. a n = n f (x) cos( x)dx, n =,,,..., b n = f (x) sin( x)dx n =,,... n B. a n = n f (x) cos( x)dx,n =,,,... b n = f (x)sin( x)dx n =,,...

24 n C. a n = n f (x) cos( x)dx, n =,,,... b n = f (x)sin( x)dx n =,,... n D. a n = n f (x) cos( x)dx,n =,,,... b n = f (x)sin( x)dx n =,,... E. a n = f (x) cos(nx)dx, n =,,,... b n = f (x)sin(n x)dx n =,,... AB a n = f (x) cos(x)dx,n =,,,... b n = f (x) sin(x)dx n =,,... AC. None of the above. 66. ( pt.) On the previous page you determined the Fourier series for the function f(x) which x has period 4 and is defined on the interval [-,) by f(x). To apply the x above formulas, we see that we should choose: A. =, B. =, C. = 3, D. = 4, E. =, AB. =, AC. = 3, AD. = 4, AE. None of the above. Total points this page = 5. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Professor Moseley Page 3 On the page before the previous page, you determined the Fourier series (i.e., the Fourier series coefficients) as given in your text for the function f(x) which has period 4 and is defined on x the interval [-,) by f(x). x 67.(3 pts.) a =. A., B., C., D. /, E. /3, AB., AC., AD. / AE. /3 BC. None of the above. 68. (5 pts.) a n =. A., B./n, C. /n, D. a k =/(k), a k+ =/(k+) E. a k =π/(k), a k+ =π/(k+), AB. a k =/(kπ), a k+ =/[(k+)π], AC. a k =/(kπ), a k+ =/[(k+)π], AD. a k =/(k), a k+ = AE. a k =π/(k), a k+ =, BC. a k =/(kπ), a k+ =, BD. a k =/(kπ), a k+ =, BE. a k =, a k+ =/(k+), CD. a k =, a k+ =π/(k+), CE. a k =, a k+ =/[(k+)π], DE. a k =, a k+ =/[(k+)π], ABC. None of the above. 69. (5 pts.) b n =. A., B./n, C. /n, D. b k =/(k), b k+ =/(k+)

25 E. b k =π/(k), b k+ =π/(k+), AB. b k =/(kπ), b k+ =/[(k+)π], AC. b k =/(kπ), a k+ =/[(k+)π], AD. b k =/(k), a k+ = AE. b k =π/(k), b k+ =, BC. b k =/(kπ), a k+ =, BD. b k =/(kπ), b k+ =, BE. b k =, b k+ =/(k+), CD. b k =, b k+ =π/(k+), CE. b k =, b k+ =/[(k+)π], DE. b k =, b k+ =/[(k+)π], ABC. None of the above. (k ) 7. ( pts.) The Fourier series may be written as: A. f(x) = + sin( x), n (k ) 4 (k ) 4 (k ) B. f(x) = + sin( x), C. f(x) = + cos( x) n (k ) n (k ) (k ) (k ) D. f(x) = + sin( x), E. f(x) = + cos( x) n (k ) n (k ) 4 (k ) AB. f(x) = + sin( x), AC. f(x) = + n (k ) (k ) cos( x) n (k ) (k ) (k) AD. f(x) = + sin( x), AE. f(x) = + cos( x) n (k ) n (k) 4 (k ) 4 (k) BC. f(x) = + cos( x), BD. f(x) = + cos( x) n (k ) n (k) BE. None of the above. Total points this page = 5. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Professor Moseley Page 4 PRINT NAME ( ) ID No. 7. ( 5 pts.) Using the method of separation of variables with separation constant λ, one of the following sets of two Ordinary Differential Equations (ODE's) can be obtained from the following Partial Differential Equation (PDE): PDE t u x x + x u t t = Circle a correct set of ODE's from the following answers. Recall that the process does not yield a unique set of ODE's. Follow the advice given in class as to the choice of separation constant (attendance is mandatory). A. x X + λ X =, t T λ T =, B. X + λ x X =, T λtt =, C. X + λ x X =, t T λ T =, D. λ X + x X =, λt t T =, E. x X + λ X =, t T + λ T =, AB. X + λ x X =, T + λ t T =, AC. X λ x X =, t T λ T =, AD. λ X + x X =, λt + t T =, AE. Separation of variables does not work on this PDE. BC. None of the above

26 Total points this page = 5. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Professor Moseley Page 5 Recall the definitions of PC fs( R, R; ), PC fs ([, ], R; ), PC ffs([, ], R; ), and B PC fs ([, ], R; ) given on page. Now let PC fs,o ( R, R; ) be the subspace of PC fs ( R, R; ) containing only odd functions, PC fs,o ([, ], R; ) be the functions in PC fs,o ( R, R; ) with their domains restricted to [,], and PC ([, ], R; ) be the subspace of PC ([, ], R; ) for which the Fourier series is finite. Recall that ffs,o k B {sin( ) : k N} PC fs,o ([, ], R; ) fs,o is a Hamel basis of PC ([, ], R; ) ffs,o and a Schauder basis of PC fs,o ([, ], R; ). To formulate the heat conduction in a rod problem as a linear mapping problem we let D = (,) (,), D = [,] [,), A HC ( D,R;) = {u(x,t)f( D,R): ua(d,r)c( D,R)} and A HC, ( D,R;) = {u(x,t)a HC ( D,R)): u(,t) = and u(,t) = for t > }. Now let L:A HC, ( D,R) A(D,R) be defined by L[u] = u t α u xx, Thus we incorporate the boundary conditions into the domain of the operator. Now let N L be the null space of L and A HC,,o,ffs (,R;α D,) = {u(x,t)n L : u(x,) PC } and A HC,,o,fs (,R;α ffs,o ([, ], R; ) D,) = {u(x,t)n L : u(x,) PC ([, ], R; ) }. Hence we see that A HC,,o,ffs ( D,R;α,) A HC,o,fs ( D,R) N L fs,o

27 k k A HC, ( D,R) A HC ( D,R;). Now recall that B {e sin( ) : k N} is a Hamel basis for PC fs,o ([, ], R; ) A HC,,o,ffs ( D,R;α,) and a Schauder basis for A HC,o,fs ( D,R). You are to solve: the following problem on this sheet and then answer the questions on the next two sheets. PDE u t = u xx < x <, t > BVP for a PDE BC u(,t) =, u(,t) =, t > IC u(x,) = 6 sin(6πx) < x <. MATH 6 FINAL EXAM Fall 5 Professor Moseley Page 6 7. ( 3 pts.) The "general" or formal solution of PDE u t = α u xx < x <, t > BC u(,t) =, u(,t) =, t > is given by. n t A. u(x,t) = cne sin( x), B. u(x,t) = n n n t n C. u(x,t) = cne sin( x), D. u(x,t) = n n t E. u(x,t) = cne sin(nx), AB. u(x,t) = n N n n n t n cne sin( x) n t n cne sin( x) n t n n n c e sin( x) n t n n n AC. u(x,t) = cne cos( x), AD. u(x,t) = a n sin( x) bn cos( x) n n

28 AE. None of the above. 73. (3 pts.) The "general" or formal solution of PDE u t = u xx < x <, t > BC u(,t) =, u(,t) =, t > is given by. n t 4 A. u(x,t) = cne sin( x), B. u(x,t) = n n 4n t n C. u(x,t) = cne sin( x), D. u(x,t) = n n t 4 E. u(x,t) = c e sin(nx), AB. u(x,t) = n n N n n n t n 4 AC. u(x,t) = cne cos( x), AD. u(x,t) = AE. None of the above. n t 4 n cne sin( x) n t 4 n cne sin( x) n n t 4 n cne sin( x) n n a sin( x) b cos( x) n n n n Total points this page = 6. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Professor Moseley Page ( 5 pts.) The solution of PDE u t = u xx < x <, t > BVP for a PDE BC u(,t) =, u(,t) =, t > IC u(x,) = 6 sin(6πx) < x < is. n t n 4 A. 6e sin( x) B. n t 4 6 e sin( x) 9 t t C. 6 e sin( x) D. 6 e sin(6x) n 36 t E. 6 e sin(6x) AB. t 6 e sin(6x)

29 AC. t 6 e sin(3x) AD. 44 t 6 e sin(6x) AE. None of the above. Total points this page = 5. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Professor Moseley Page 8 On this page you are to solve: the following problem: PDE u t = u xx < x <, t > BVP for a PDE BC u(,t) =, u(,t) =, t > IC u(x,) = < x < Begin by writing down the general solution for the PDE and the BC's. Then answer the questions on the next three pages.

30 MATH 6 FINAL EXAM Fall 5 Professor Moseley Page ( 3 pts.) The solution of PDE u t = α u xx < x <, t > BC u(,t) =, u(,t) =, t > IC u(x,) = u (x) is given by. n t A. u(x,t) = cne sin( x), B. u(x,t) = n n n t n C. u(x,t) = cne sin( x), D. u(x,t) = n n t E. u(x,t) = cne sin(nx), AB. u(x,t) = n N n n n t n cne sin( x) n t n cne sin( x) n t n n n c e sin( x) n t n n n AC. u(x,t) = cne cos( x), AD. u(x,t) = a n cos( x) bn sin( x) n n

31 AE. None of the above. 76. (3 pts.) where n A. c u (x)sin x dx B. n n C. c u (x)sin x dx, D. n E. c u (x)sin nx dx, AB. n n n c u (x)sin x dx n n c u (x)sin x dx n n c u (x)sin x dx n AC. c u (x)sin x dx, AD. a u (x)cos x dx and n n n n b u (x)sin x dx n AE. None of the above. Total points this page = 6. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Professor Moseley Page ( 3 pts.) The solution of PDE u t = u xx < x <, t > BC u(,t) =, u(,t) =, t > IC u(x,) = is given by. n t 4 A. u(x,t) = cne sin( x), B. u(x,t) = n n 4n t n C. u(x,t) = cne sin( x), D. u(x,t) = n n t 4 E. u(x,t) = c e sin(nx), AB. u(x,t) = n n N n n n t n 4 AC. u(x,t) = cne cos( x), AD. u(x,t) = n t 4 n cne sin( x) n t 4 n cne sin( x) n n t 4 n cne sin( x) n n a sin( x) b cos( x) n n n n

32 AE. None of the above. 78. (3 pts.) where A. cn sin x dx B. n C. cn sin x dx, D. n E. cn sin nx dx, AB. n cn sin x dx n cn sin x dx n cn 4sin x dx AC. cn sin x dx, AD. a n cos x dx and n n n bn sin x dx AE. None of the above. Total points this page = 6. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Spring 5 Professor Moseley Page ( 5 pts.) The solution of PDE u t = u xx < x <, t > BVP for a PDE BC u(,t) =, u(,t) =, t > IC u(x,) = < x < is. (k) 4 t (k ) A. u(x,t) = 4 e sin( x), (k ) k (k) 8 t (k ) B. u(x,t) = 4 e sin( x), (k ) k (k) 4 t (k ) C. u(x,t) = 4 e cos( x), (k ) k

33 (k) 4 t (k ) D. u(x,t) = 4 e sin( x), (k ) k (k) 8 t (k ) E. u(x,t) = 4 e cos( x), (k ) k (k) 4 t (k ) AB. u(x,t) = 4 e sin( x), (k ) 4 k (k) 4 t (k ) AC. u(x,t) = 4 e cos( x), (k ) 4 k (k) 8 t (k ) AD. u(x,t) = 4 e sin( x), (k ) k (k) 8 t (k) AE. u(x,t) = 4 e cos( x), (k) n k 4 t k BC. u(x,t) = 4 e sin( x), k k k t k BD. u(x,t) = e sin( x), k k BE. None of the above. Total points this page = 5. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Professor Moseley Page 3 PRINT NAME ( ) SS No. 8. ( 4 pts.) If both ends of a rod are insulated (recall that we assume that the lateral sides are also insulated so that the temperature does not vary over a cross section), then a good mathematical model of the physical heat conduction problem is given by: PDE u t = α u xx < x <, t > BVP for a PDE BC u x (,t) =, u x (,t) =, t > IC u(x,) = f(x) < x < where f(x) is the initial temperature distribution in the rod. ////////////////////////////////////////////// - O O - / ////////////////////////////////////////////// \ insulated insulated * *

34 x = x = Applying the separation of variables process to the PDE results in the two ODE's:. X" + σ X =. T' + α σ T = where σ is the separation constant. Determine the spacial eigenvalue problem that results from applying the BC given above. Circle the correct answer. A. X" + σ X = B. X" + σ X = C. X" + σ X = X() =, X() = X() =, X() = X() =, X() = D. X" + σ X = E. X" + σ X = AB. X" + σ X = X() =, X() = X() =, X() = X() =, X) = AC. X" + σ X = AD. X" + σ X = AE. X" + σ X = X() =, X() = X() =, X() = X() =, X() = BC. None of the above. Total points this page = 4. TOTAL POINTS EARNED THIS PAGE MATH 6 FINAL EXAM Fall 5 Professor Moseley Page 33 PRINT NAME ( ) SS No. TABLE OF LAPLACE TRANSFORMS THAT NEED NOT BE MEMORIZED f(t) = - {F(s)} F(s) = {f(t)} Domain F(s) ))))))))))) )))))))))) ))))))))) n! t n n = positive integer ))))) s > s n+ a sinh (at) ))))))) s > *a * s - a s cosh (at) ))))))) s > *a * s - a b e at sin (bt) )))))))) s > a

35 (s - a) + b s - a e at cos(bt) )))))))) s > a (s - a) + b n! t n e at n = positive integer )))))) s > a (s - a) n+ u(t) )) s > s e -cs u(t - c) ))) s > s e ct f(t) F(s - c) s f(ct) c > ) F () ) c c δ(t) δ(t - c) e -cs