MATH 202 Final, Spring 2002

FINAL EXAMINATION, MATH 202 June 7, 2006; 8:00-10:00 P.M. VERSION I Name: Signature: Number: Section: 13, 14, 15, 16 1. Instructions: • No calculators are allowed. • There are two types of questions: PART I consists of SIX sub- jective questions, and PART II consists of EIGHT multiple- choice questions of which each has exactly one correct answer. • Give detailed solutions for the problems of PART I in the provided space. • 10 points for each question of PART I. • Circle one answer for each question of PART II. • 5 points for each correct answer of PART II. • 0 point for no answer or more than one answer of PART II. GRADE OF PART I /60 GRADE OF PART II /40 TOTAL GRADE /100 1 2 Part I(1). solve, without using Laplace transforms, the initial-value problem dx dt = 2x + 4y dy dt = −x + 6y with initial conditions x(0) = −1, y(0) = 6. 3 Extra Page 4 Part I(2). Use Laplace transforms to find the solution of the initial- value problem y′′ + y = f(t), f(t) = � 1, 0 ≤ t < π/2 0, π/2 ≤ t ≤ ∞ , y(0) = 1, y′(0) = 0. 5 Part I(3). Use the substitution u = √x y to find the general solution of the differential equation d2u dx2 + � 1 − ν2 − 1/4 x2 � u = 0, x > 0. 6 Part I(4). Show that the differential equation 4x2 y′′ − 2x(x − 2) y′ − (3x + 1) y = 0, has a regular point and indicial roots r1/2 and r2 = −1/2 at x = 0, and use Frobenius method to find a series solution about x = 0 associated with r1 = 1/2. 7 Extra Page 8 Part I(5). Find the general solution of the differential equation x dy dx = (1 + x) y + x y2. by an appropriate substitution. 9 Part I(6). Find the general solution of the differential equation x2 y′′ − x y′ + y = x3, x > 0. 10 Extra Page 11 Part II: Multiple Choice Problems. Part II(1). The solution of the initial-value problem y′(t) = 1 − sin t − � t 0 y(τ)dτ, y(0) = 0, satisfies (a) y(t) = cos t + sin t ∗ cos t. (b) y(t) = cos t − sin t ∗ cos t. (c) y(t) = sin t + sin t ∗ cos t. (d) y(t) = sin t − sin t ∗ cos t. (e) None of the above. Part II(2). L−1 �e−s s2 + tan s � = (a) (t + 1)U(t + 1) + (sin t)/t. (b) (t − 1)U(t − 1) − (sin t)/t. (c) (t − 1)U(t − 1) − (cos t)/t. (d) (t + 1)U(t + 1) − (sin t)/t. (e) None of the above. Part II(3). The solution of the initial-value problem dy dx = −1 2(x + y − 1)3 − 1, y(1) = 1, satisfies (a) y(4) = −5/2. (b) y(4) = 5/2. (c) y(3) = 5/2. (d) y(3) = −5/2. (e) None of the above. 12 Part II(4). The solution of the initial-value problem �sin 2x y + x � dx + � y − sin2 x y2 � dy, y(0) = 1, is given implicitly by the equation x2y + 2 sin2 x = h(y), where (a) h(1) = 0. (b) h(1) = 2. (c) h(1) = 4. (d) h(1) = 6. (e) None of the above. Part II(5). The differential equation (x2 + 1)y′′ − 4xy′ + 6y = 0 has an ordinary point at x = 0 and two power series solutions �∞ n=0 cnxn, whose coefficients cn, n = 0, 1, 2, · · · , satisfy the recurrence relation (a) cn+2 = (n − 2)(n − 3) (n + 1)(n + 2)cn, n = 0, 1, 2, · · · . (b) cn+2 = −(n − 2)(n − 3) (n + 1)(n + 2)cn, n = 0, 1, 2, · · · . (c) cn+2 = −(n − 3)(n − 4) (n + 1)(n + 2)cn, n = 0, 1, 2, · · · . (d) cn+2 = (n − 1)(n − 2) (n + 1)(n + 2)cn, n = 0, 1, 2, · · · . (e) None of the above. Part II(6). The solution of the initial-value problem � x cos2 �y x � − y � dx + x dy = 0, y(1) = π 4 satisfies (a) y(e−2) = 0. (b) y(e−1) = 0. (c) y(e) = 0. (d) y(e2) = 0. (e) None of the above. 13 Part II(7). Answer TRUE (T) or (FALSE) only: (a) —– The Laplace transform of a function f is L(f(t)) = � ∞ 0 e−stf(t) dt. (b) —- The convolution of functions f and g on [0, ∞) is defined by (f ∗ g)(t) = � t 0 f(τ)g(t − τ) dτ. (c) —- The general solution of the parametric Bessel equation x2 y′′ + x y′ + (λ2x2 − ν2)y = 0 is y = c1Jν(λx) + c2Yν(λx). (d) —- If the set of two functions f1 and f2 is linearly independent on an interval I, then the Wronskian W(f1, f2) ̸= 0 for all x ∈ I. (e) —- The initial-value problem dy/dx = √y, y(x0) = y0 has no solution for y0 < 0. Part II(8). If y1 = (sin x)/x is one solution of the differential equa- tion, x y′′ + 2 y′ + x y = 0, then the solution of the associated boundary-value problem with bound- ary conditions y(π/2) = 0 and y(π) = 1/2 satisfies (a) y(π/4) = − √ 2. (b) y(π/4) = √ 2. (c) y(π/4) = − √ 2/2. (d) y(π/4) = √ 2/2. (e) None of the above.