June 2011 PURDUE UNIVERSITY Study Guide for the Credit Exam in (MA 262) Linear Algebra and Differential Equations

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1 June 20 PURDUE UNIVERSITY Study Guide for the Credit Exam in (MA 262) Linear Algebra and Differential Equations The topics covered in this exam can be found in An introduction to differential equations and linear algebra by Stephen Goode, Chapters. See also for recent course information and materials. The exam consists of 25 multiple choice questions with two hour time limit. No books, notes, calculators or any other electronic devices are allowed. IMPORTANT. Read this material thoroughly if you contemplate trying for advanced placement (and extra credit which counts toward graduation). 2. Study all the material listed in the outline.. Work the practice problems below. The correct answers are at the end. Topics covered in the exam Linear Algebra:. Complex numbers, polar representation, roots of complex numbers. 2. Vector spaces: subspace, basis, spanning set, linear combination, linear independence.. Matrix operations: addition, multiplication, inverse, determinants. 4. Row reduction of matrices: row-echelon normal form, rank. 5. Systems of linear equations: solve using matrix methods, augmented matrices, Cramer s rule, solution space. 6. Eigenvalues and eigenvectors. Differential Equations:. First-order differential equations with and without initial conditions. Types include - separable variables, exact, linear. 2. Applications of first-order E. s to mixture problems, growth and decay problems, falling bodies, electrical circuits, orthogonal trajectories.. Second and higher-order linear E. s with constant coefficients: Undetermined coefficients, initial value problems. Variation of Parameters. 4. Applications of second-order E. s to Newton s 2nd law, spring mass systems, electrical circuits, etc. Systems of Differential Equations:. Matrix Formulation. 2. Solutions to linear systems with constant coefficient using eigenvalues and eigenvectors.. Variation of parameters for systems.

2 MA 262 PRACTICE PROBLEMS page /8 Circle the letter corresponding to your choice of correct answer.. Find the determinant of A if A = If A is a 4matrixandP = solutions of A a solution? x y z w = p p 2 p p 4 and Q = q q 2 q q 4 are 2 which of the following is also. Which of the following complex numbers z satisfies z 2 = ri where r is some real number? E. P + Q P Q 2P Q Q 2P E. all of the above i + i +i +i E. none of these 4. Which of the following are real vector spaces (in each case multiplication by a real number and addition have the usual meanings)? (i) The set of all real valued functions f such that f(x +)=2f(x) for all x. (ii) The set of all real polynomials p such that p() = p(0) +. (iii) The set of all 4 4 matrices whose first and last rows are equal. (i) and (iii) none (i) and (ii) only (iii) E. (ii) and (iii)

3 5. The smallest subspace of R containing the vectors (, 2, ) and (5,, ) is the set of all (x, y, z) satisfying which of the following? x, y, z 0 x 2 + y 2 + z 2 =0 x + y z =0 x +5y +7z =0 E. x 4y +7z =0 6. Which of the following sets of vectors is linearly dependent? (i) (, 0, ), (,, 0), (0,, ) (ii) (, 0, ), (,, 0), (0,, ) (iii) (, 0, ), (0,, 0), (0, 0, 0) none (ii) only (i) and (ii) (i) and (iii) 7. The characteristic values of A = if k are real k + E. (ii) and (iii) k 2 > k 2 k 2 = k 2 E. k 2 < 8. The system of equations nonzero solution x y = 2 z a b c has a for all nonzero a, b, c if a = b + c if 2c = a + b if b a c E. if c 2a b

4 /2 / If the 2 2 matrices B = and A = x y 0 satisfy B 2 = A, then the bottom row of B is equal to (, 0) ( 2, ) 2 E. ( ) 2, 2 ( ) 2, 2 ( 2, ) 0. If A = then which of the following is not an entry 4 2 of A? all are entries E. none are entries. The matrix A = has as a characteristic value. What is the dimension of the vector space spanned by the characteristic vectors corresponding to? 2 0 E. cannot be determined

5 2. If A and B are 4 4 matrices and deta =4,detB =6, then det(a B T )= E. Undefined. If y + x y = ex ex and y() = 0, then y = xe x (x )e x x 2 (ex ) x (ex ) E. (2x 2)e x 4. The solution of (cos 2 x)y = y 2 satisfying y(π/4) = is y =2 tan x y =anx y = 2 cot x y = +sinx +cosx E. y = cos x 2cosx sin x

6 5. The general solution of is dy dx = 2xy +x2 y 2 x 2 y + x2 y + x = c y x2 y x = c 2xy + y 2 +x 2 = c log(x 2 + y 2 )=c E. none of the above 6. A tank initially contains 50 gal of water. Alcohol enters at the rate of 2 gal/min and the mixture leaves the tank at the same rate. The time t 0 when there is 25 gal of alcohol in the tank satisfies 25 = e t 0/2 = t 0 /50 t 0 =25ln2 25 = et 0/25 E. t 0 =25 7. The general solution of y (4) +4y = 0 is e x (c + c 2 x)+e x (c cos x + c 4 sin x) e x (c cos x+c 2 sin x)+e x (c cos x+c 4 sin x) (c + c 2 x)e x +(c + c 4 x)e x e x (c + c 2 x + c x 2 + c 4 x ) E. none of the above 8. For a particular solution of y +y +2y = xe x one should try a solution of the form Ae 2x + Be x Ae x + Bxe x (A + Bx)e x (Ax + Bx 2 )e x E. Ax 2 e x

7 9. A body attached to the lower end of a vertical spring has acceleration d2 x dt 2 = 6x ft/sec2,wherex = x(t) isthe distance of the body from the equlibrium position at time t. If the body passes through the equilibrium position at t = 0 with v = 2 ft/sec, then x = cos 4t cos 4t +sin4t sin 4t 20. The substitution t = e x transforms y +(e x )y + e 2x y = e x into e 4t d 2 y dt 2 + dy dt + y = t E. cos 4t sin 4t t 2 d2 y +(t )dy dt2 dt + t2 y = t t 2 d2 y dt 2 + tdy dt + t2 y = t d 2 y +(t )dy dt2 dt + t2 y = t E. none of the above 2. The smallest order of a linear homogeneous differential equation with constant coefficients which has y = xe 2x cos x as a solution is E. more than 6 22.A2 2 matrix A has as a characteristic value with multiplicity 2. Then two linearly independent solutions of x = Ax are of the form k e x, (k 2 + k )e x k e x, (k 2 + tk )e x k e x, k 2 te 2x k e x, k 2 e x

8 2. If a fundamental matrix for x = Ax is e 0 U(t) = 0 e t, then a particular solution u p (t) e of x = Ax +, such that u p (0) = 0 is e t 24. A 2 2 matrix A has as a characteristic value with 0 corresponding characteristic vector ]; and a characteristic value 2 with corresponding characteristic vector. Then a solution to x 0 = Ax is E. [ te t te ] te t te te t t 2 e te t e e t [ e e t ] e t e [ e e ] e t e t E. none of these 25. The real 2 2 matrix A has characteristic values [ + ] i and i i with corresponding characteristic vectors and i x (t). If x(t) = satisfies x x 2 (t) = Ax and x 2 (t) = e t (cos t +sint) thenx (t) = e t (sin t cost) e t (sin t 2cost) e t (cos t sin t) e t (cos t sin t) E. none of these

9 ANSWER KEY. D 4. E 2. C 5. B. C 6. C 4. A 7. B 5. E 8. D 6. D 9. C 7. B 20. A 8. C 2. B 9. B 22. B 0. A 2. C. A 24. A 2. C 25. D.