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

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1 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 read the entire paper before you begin to write, and 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, and 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, and 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 and 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! Problem Total (BONUS) Points Your Marks

2 Question. [8 points] Determine whether the given set of fuctions is linearly independent on the interval (, ). (a) [4 points] f (x) = x, f 2 (x) = x 2, f 3 (x) = 4x 3x 2 Solution: Since f 3 (x) = 4 f (x) 3 f 2 (x) = 4 f (x) 3 f 2 (x) f 3 (x) = 0, the set of functions f (x), f 2 (x), f 3 (x) is linearly dependent. (b) [4 points] f (x) = + x, f 2 (x) = x, f 3 (x) = x 2 Solution: Let c f (x) + c 2 f 2 (x) + c 3 f 3 (x) = 0 = c ( + x) + c 2 x + c 3 x 2 = 0 = c + (c + c 2 )x + c 3 x 2 = 0, then c = 0,c + c 2 = 0,c 3 = 0,= c = c 2 = c 3 = 0. Hence the set of functions f (x), f 2 (x), f 3 (x) is linearly independent. 2

3 Question 2. [24 points] Consider the differential equation x 3 y + x 2 y 2xy + 2y = 0 (a) [5 points] Verify that the fuctions of y = x, y 2 = x 2, y 3 = x differential equation. Solution: are solutions of the given Check y is the sol of x 3 y + x 2 y 2xy + 2y = 0 y = x = y =, y = 0, y = 0 Then x 3 y + x2 y 2xy + 2y = x + 2x = 0. Check y 2 is the solution of x 3 y + x 2 y 2xy + 2y = 0 y 2 = x 2 = y 2 = 2x, y 2 = 2, y 2 = 0 Then x 3 y 2 + x2 y 2 2xy 2 + 2y 2 = 0 + x 2 2 2x 2x + 2x 2 = 0. Check y 3 is the solution of x 3 y + x 2 y 2xy + 2y = 0 y 3 = x = y 3 = x 2, y 3 = 2x 3, y 3 = 6x 4 Then x 3 y 3 + x2 y 3 2xy 3 + 2y 3 = 0. (b) [7 points] Verify that y, y 2, y 3 in part (a) form a fundamental set of solutions for the given differential equaiton on the interval of (0, ). Solution: W (y, y 2, y 3 )(x) = y y 2 y 3 y y2 y3 y y2 y3 = Then y, y 2, y 3 form a fundamental set of solutions. x x 2 x 2x x x 3 = 6 x = 0, on (0, ) (c) [2 points] Using y, y 2 and y 3 in part (a), form the general solution of the given differential equation. Solution: The general solution is y(x) = c y + c 2 y 2 + c 3 y 3 = c x + c 2 x 2 + c 3 x. 3

4 Question 3. [24 points] Find the general solution of the given second-order differential equation. (a) [8 points] y 7y + 2y = 0 Solution: m 2 7m + 2 = 0 = (m 3)(m 4) = 0 = m = 3,m 2 = 4 = y = c e 3x + c 2 e 4x. (b) [8 points] y 6y + 9y = 0 Solution: m 2 6m + 9 = 0 = (m 3) 2 = 0 = m = m 2 = 3 = y = c e 3x + c 2 xe 3x. (c) [8 points] y 4y + 5y = 0 Solution: m 2 4m + 5 = 0 = m,2 = 2 ± i = y = e 2x (c cos x + c 2 sin x). 4

5 Question 4. [2 points] Find the general solution of the given third-order differential equation d 3 x dt 3 + d 2 x dt 2 2x = 0 Solution: m 3 + m 2 2 = 0 = m = is one of the roots, then using Long Division, we have Then m 2,3 = ± i. Hence, the general solution is m 3 + m 2 2 = (m )(m 2 + 2m + 2) x = c e t + e t (c cos t + c 2 sin t) 5

6 Question 5. [30 points] Solve the given differential equation by undetermined coefficients, y + 2y = 2x + 5 e 2x subject to the initial conditions y(0) = 0, y (0) = 2. Solution: The associated homogeneous equation is y + 2y = 0. Then the auxiliary equation is m 2 + 2m = 0 = m = 0andm = 2. Hence, the complementary function is y c = c e 2x + c 2. Now g (x) = 2x +5 e 2x = y p (x) = Ax+B +Ce 2x, however, there is duplications between y c and y p. Then the correct assumed particular solution is y p (x) = Ax 2 + Bx+Cxe 2x. Then y p (x) = 2Ax + B +Ce 2x 2Cxe 2x and y p (x) = 2A 4Ce 2x + 4Cxe 2x. Hence, y p (x)+2y p (x) = 2A +B +4Ax 2Ce 2x = 2x +5 e 2x = 2A +2B = 5,4A = 2, 2C = = A = 2,B = 2,C = 2 = y p(x) = 2 x2 + 2x + 2 xe 2x Then the general solution is y = c e 2x + c x2 + 2x + 2 xe 2x Using I.C. = 0 = c + c 2, 2 = 2c = c =,c 2 =. Therefore, the solution is y = e 2x + 2 x2 + 2x + 2 xe 2x 6

7 Question 6. [30 points] Solve the given differential equation by variation of parameters 3y 6y + 6y = e x sec x Solution:. For 3y 6y + 6y = 0 we have 3m 2 6m + 6 = 0 3(m 2 2m + 2) = 0. So m,m 2 = 2 ± = 2 ± 2i 2 = ± i and α = = β. So,y c = e x (c cosx+c 2 sin x). Hence y = e x cos x and y 2 = e x sin x. 2. We have w(y, y 2 ) = y y 2 y y2 = e x cos x e x sin x e x cos x e x sin x e x sin x + e x cos x = e x cos x(e x sin x + e x cos x) e x sin x(e x cos x e x sin x) = e 2x cos x sin x + e 2x cos 2 x e 2x sin x cos x + e 2x sin 2 x = e 2x (cos 2 x + sin 2 x) = e 2x. 3. 3y 6y + 6y = e x sec x y 2y + 2y = 3 ex sec x. f (x) = 3 ex sec x. 4. We have So w = 0 y 2 f (x) y2 = 0 e x sin x 3 ex sec x e x sin x + e x cos x = 3 e2x sin x sec x w 2 = y 0 y f (x) = e x cos x 0 e x cos x e x sin x 3 ex sec x = 3 e2x cos x sec x. u = w w = u 2 = w 2 w = 3 e2x sin x sec x e 2x = 3 sin x sec x = 3 sin x cos x = 3 3 e2x cos x sec x e 2x = 3 cos x sec x = 3. tan x. Therefore, u = 3 tan xdx = sin x sin x 3 cos x dx = 3 cos x dx = 3 cos x d(cos x) = ln cos x. 3 u 2 = 3 dx = 3 x. 5. y p = u y + u 2 y 2 = 3 ex cos x ln cos x + 3 xex sin x. 6. y = e x (c cosx+c 2 sin x) + 3 ex cos x ln cos x + 3 xex sin x. 7

8 Question 7. [20 pionts] A 00-volt electromotive force is applied to an RC-series circuit in which the resistance is 200 ohms and the capacitance is 0 4 farad. (a) [6 points] Find the charge q(t) on the capacitor if q(0) = 0. Solution: Assume that R dq dt + C q = E(t) where R = 200 ohms, C = 0 4 farad, and E(t) = 00. Then 200 dq dt + 04 q = 00 = 2 dq dq + 00q = = dt dt + 50q = 2 = q = 00 + ce 50t Using I.C., we get c =. Hence the solution is 00 q = e 50t. (b) [4 points] Find the current i(t). Solution: i = dq dt = 00 ( 50)e 50t = 2 e 50t. 8

9 Question 8. [BONUS: 24 points] Find the general solution of x 4 y + x 3 y 4x 2 y = given that y = x 2 is a solution of the associated homogeneous equation. [HINT: Reduction of Order & Variation of Parameters] Solution: We are given that y = x 2 is a sol. of x 4 y + x 3 y 4x 2 y =. To find a second solution, we use reduction of order. Let y = x 2 u(x). Then the product rule gives y = x 2 u + 2xu and y = x 2 u + 4xu + 2n. So x 4 y + x 3 y 4x 2 y = x 5 (xu + 5u ) = 0 Letting w = u, this becomes xw + 5w = 0 Separating variables and integrating we have dw dt = 5 dx and ln w = 5ln x +C x Thus, w = x 5 and u = 4 x 4. A second solution is then y 2 = x 2 x 4 =, and the general solution of the homogeneous x 2 DE is y c = c x 2 + c 2. x 2 To find a particular solution, y p, we use variation of parameters. The Wronskian is W = 4 x. Identifying f (x) =, we obtain x 4 u = 4 x 5 = u = 6 x 4 u2 = 4 x = u 2 = 4 ln x So y p = 6 x 4 x 2 4 (ln x)x 2 = 6 x 2 4 x 2 ln x. The general solution is y = c x 2 + c 2 x 2 6 x2 ln x. 4x2 9

California State University Northridge MATH 280: Applied Differential Equations Midterm Exam 1

California State University Northridge MATH 280: Applied Differential Equations Midterm Exam 1 October 9, 2013. Duration: 75 Minutes. Instructor: Jing Li Student Name: Student number: Take your time to