California State University Northridge MATH 280: Applied Differential Equations Midterm Exam 3

Size: px

Start display at page:

Download "California State University Northridge MATH 280: Applied Differential Equations Midterm Exam 3"

Benedict Carroll
5 years ago
Views:

1 California State University Northridge MATH 280: Applied Differential Equations Midterm Exam 3 April 29, Duration: 75 Minutes. Instructor: Jing Li Student Name: Signature: Do not write your student ID number on this front page. Please write your student ID number in the space provided on the second page. Take your time to read the entire paper before you begin to write, read each question carefully. Remember that certain questions are worth more points than others. Make a note of the questions that you feel confident you can do, then do those first: you do not have to proceed through the paper in the order given. You have 75 minutes to complete this exam. This is a closed book exam, no notes of any kind are allowed. The use of cell phones, pagers or any text storage or communication device is not permitted. Only the Faculty approved TI-30 calculator is allowed. The correct answer requires justification written legibly logically: you must convince me that you know why your solution is correct. Answer these questions in the space provided. Use the backs of pages if necessary. Where it is possible to check your work, do so. Good Luck! 1

2 Student number: Problem Total Points Your Marks Question 1. [14 points] Find the interval radius of convergence for the given power series ( 1) k (x 5)k 10k Solution: By the ratio test, lim (x 5)k+1 /10 k = lim k (x 5) k 10 k x 5 = x 5. k The series is absolutely convergent for 1 x 5 < 1, x 5 < 10, or on ( 5,15). The radius of 10 convergence is R = 10. At x = 5, the series ( 1)k ( 10) k /10 k = 1 diverges by the nth term test. At x = 15, the series ( 1)k 10 k /10 k = ( 1)k diverges by the nth term test. Thus, the series converges on ( 5,15), the radius of convergence is 10. 2

3 Question 2. [8 points] Determine the singular points of the given differential equation. Classify each singular points as regular or irregular. x 3 (x 2 25)(x 2) 2 y + 3x(x 2)y + 7(x + 5)y = 0. Solution: Irregular singular point: x = 0; regular singular points: x = 2,±5 3

4 Question 3. [8 points] For the differential equation (x 2 25)y + 2x y + y = 0. Without actually solving the above differential equation, find the minimum radius of convergence of power series solutions: (a)[4 points] about the ordinary point x = 0. (b)[4 points] about the ordinary point x = 1. Solution: The singular points of (x 2 25)y + 2x y + y = 0 are 5 5. The distance from 0 to either of these points is 5. The distance from 1 to the closest of these points is 4. 4

5 Question 4. [18 points] A force of 2 pounds stretches a spring 1 foot. A mass weighing 3.2 pounds is attached to the spring, the system is then immersed in a medium that offers a damping force that is numerically equal to 0.4 times the instantaneous velocity. (a) [12 points] Find the equation of motion if the mass is initially released from rest from a point 1 foot above the equilibrium position. Solution: From 0.1x + 0.4x + 2x = 0, x(0) = 1, x (0) = 0 we obtain x = e 2t [ cos4t 1 2 sin4t]. (b) [4 points] Express the equation of motion in the form given in x(t) = Ae λt sin( ω 2 λ 2 t + φ). Solution: x = 5 2 e 2t sin(4t ). (c) [4 points] Find the first time at which the mass passes through the equilibrium position heading upward. Solution: If x = 0 then 4t = 2π,3π,4π,..., so that the first time heading upward is t = seconds. 5

6 Question 5. [18 points] Find the steady-state charge the steady-state current in an LRCseries circuit when L = 1 h, R = 2 Ω, C = 0.25 f, E(t) = 50cos t V. Solution: Solving q +2q +4q = 0 we obtain q c = e t (cos 3t+sin 3t). The steady-state charge has the form q p = A cos t + B sin t. Substituting into the differential equation we find (3A + 2B)cos t + (3B 2A)sin t = 50cos t. Thus, A = 150/13 B = 100/13. The steady-state charge is the steady-state current is q p (t) = cos t sin t i p (t) = sin t cos t 6

7 Question 6. [30 points] Use the power series method to solve the given initial-value problem (x 1)y x y + y = 0, y(0) = 2, y (0) = 6 Solution: Substituting y = n=0 c n x n into the differential equation we have (x 1)y x y + y = n(n 1)c n x n 1 n(n 1)c n x n 2 nc n x n + c n x n = n=2 }{{} = 0. Thus k=n 1 n=2 }{{} k=n 2 n=1 }{{} k=n n=0 }{{} k=n (k + 1)kc k+1 x k (k + 2)(k + 1)c k+2 x k kc k x k + c k x k = 2c 2 + c 0 + k=0 k=0 [ (k + 2)(k + 1)c k+2 + (k + 1)kc k+1 (k 1)c k ]x k 2c 2 + c 0 = 0 (k + 2)(k + 1)c k+2 + (k + 1)kc k+1 (k 1)c k = 0 c 2 = 1 2 c 0 Choosing c 0 = 1 c 1 = 0 we find c k+2 = kc k+1 k + 2 (k 1)c k (k + 2)(k + 1), k = 1,2,3,... c 2 = 1 2, c 3 = 1 6, c 4 = 1 24, so on. For c 0 = 0 c 1 = 1 we obtain c 2 = c 3 = c 4 = = 0. Thus y = C 1 ( x x3 + ) +C 2 x y ( 1 = C 1 x + 2 x2 + ) +C 2 The initial conditions imply C 1 = 2 C 2 = 6, so y = 2 ( x x3 + ) + 6x = 6x 2e x. 7

8 Question 7. equation: [25 points] Verify that x = 0 is a regular singular point of the given differential 2x y y + 2y = 0. Use the method of Frobenius to obtain two linearly independent series solutions about x = 0. Form the general solution on (0, ). Solution: Substituting y = n=0 c n x n+r into the differential equation collecting terms, we obtain 2x y y + 2y = (2r 2 3r )c 0 x r 1 + which implies = 0, The indicial roots are r = 0 r = 3/2. For r = 0 the recurrence relation is k+r 1 [2(k + r 1)(k + r )c k (k + r )c k + 2c k 1 ]x 2r 2 3r = r (2r 3) = 0 (k + r )(2k + 2r 3)c k + 2c k 1 = 0. 2c k 1 c k = k(2k 3), k = 1,2,3,..., c 1 = 2c 0, c 2 = 2c 0, c 3 = 4 9 c 0, so on. For r = 3/2 the recurrence relation is 2c k 1 c k = (2k + 3)k, k = 1,2,3,..., c 1 = 2 5 c 0, c 2 = 2 35 c 0, c 3 = c 0, so on. The general solution on (0, ) is y = C 1 ( 1 + 2x 2x x3 + ) +C 2 x 3/2( x x x3 + ). 8

California State University Northridge MATH 280: Applied Differential Equations Midterm Exam 2

California State University Northridge MATH 280: Applied Differential Equations Midterm Exam 2 November 3, 203. Duration: 75 Minutes. Instructor: Jing Li Student Name: Student number: Take your time to