Test1_23F_a_sol

Test I A Fall Semester 2023 MATH 308 Differential Equations Dr. Minchul Kang Texas A&M University | Department of Mathematics | 335B Blocker Building | College Station | TX 77840 | grcorg@tamu.edu Direction □ Do not turn to the next page until you are told to do so. □ Print your name and UIN below as well as on the Scantron form. □ Please turn off all mobile phones, calculators, and other electronic device, and stow those away in your bags. □ Your desk may only contain pens, pencils, erasers, and student IDs during the test. □ In part I (Multiple choice questions), mark your answer on the scantron from using No. 2 pencil. □ In part II (Short answer questions), ensure that your final response is typed clearly and legibly in the provided answer box. You don’t have to show your work. □ In part III (Work out questions), show your work to earn credits for work out problems: merely having final answers does not suffice. □ There are 8 pages in this test set. Name: , UIN: Last Name First Name THE AGGIE HONOR CODE "An Aggie does not lie, cheat, or steal or tolerate those who do.” Signature: Test I A| MATH 308 Differential Equations [1/8]

Part I. Multiple choice (4 points each) Print your answer in the box provided. Problem 1. Find differential equation of the direction fields. A . y ′ = y x * B . y ′ = x y C . y ′ = − y x D . y ′ = − x y E . None of above [Answer] Problem 2. Find the order of x ( y ′′′ ) 2 + x 2 ( y ′′ ) 3 + x 3 ( y ′ ) 4 + x 4 y 5 = 0 A . 2 B . 3 * C . 4 D . 5 [Answer] Problem 3. Which of the following is a separable dif- ferential equation? A . y 2 ln x − x 2 y = xy ′ B . y ln x − xy 2 = xy ′ C . y ln x − x 2 y = xy ′ * D . y ln x − xy + x = xy ′ E . None of above [Answer] Problem 4. Which of followings is an exact differential equation? A . (3 x − 2 y ) dx + (3 y + 2 x ) dy = 0 B . (2 x − 3 y ) dx + (3 x − 2 y ) dy = 0 C . (3 x − 2 y ) dx + (2 y − 3 x ) dy = 0 D . (2 x − 3 y ) dx + (2 y − 3 x ) dy = 0 * E . None of above [Answer] Test I A| MATH 308 Differential Equations [2/8]

Part II. Short answer (8 points each) Print your answer in the box provided. You don’t have to show your work. Choose the equation category from □ 2nd order linear w/ constant coefficients □ Bernoulli’s equation □ Cauchy-Euler equation □ Exact (w/ or w/o integration factor) □ Homogeneous degree zero □ Integrable □ Linear 1st order □ Linear composite □ Separable Choose the solution method from □ Characteristic equation □ Direct integration □ Integrating factor □ Partial integrations □ Separation of variables □ u = ax + by + c substitution □ u = y k substitution □ y = ux substitution Problem 9. Identify the equation category, outline so- lution methods, and solve the equation. y ′ = 2 e − y x − 2 e − y , y (1) = 0 Category: Separable Method: Separation of variables Solution: e y = x 2 − 2 x + 2 Problem 10. Identify the equation category, outline solution methods, and solve the equation. ty ′ = − 3 y + 6 t 3 , y (1) = 2 . Category: 1st order linear Method: Integrating factor Solution: y = t 3 + t − 3 Test I A| MATH 308 Differential Equations [4/8]

Problem 11. Identify the equation category, outline solution methods, and solve the equation. x (1 − sin y ) dy = (cos x − cos y − y ) dx . Category: Exact Method: Partial integration Solution: x cos y − sin x + xy = C Problem 12. Identify the equation category, outline solution methods, and solve the equation. y ′′ − 4 y ′ + 4 y = 0 , y (0) = 1 , y ′ (0) = 1 Category: 2nd order linear w/ constant coefficients Method: Characteristic equation Solution: y = e 2 t − te 2 t Problem 13. Identify the equation category, outline solution methods, and solve the equation. y ′′ + 2 y ′ + 2 y = 0 Category: 2nd order linear w/ constant coefficients Method: Characteristic equation Solution: y = e − t ( C 1 cos t + C 2 sin t ) Test I A| MATH 308 Differential Equations [5/8]

Step 3: Show that y 1 and y 2 form the complete set of solution (i.e. independent). [Show your work here.] W ( y 1 , y 2 ) = x x ln x 1 ln x + 1 = ln x + 1 − x ln x = 1 > 0 Problem 15. Following suggested steps to determine the general solution to y ′′ + 2 y ′ − 3 y = 4 e t Step 1: Solve the associated homogeneous equation to find y c [Show your work here.] y ′′ + 2 y ′ − 3 y = 0 λ 2 + 2 λ − 3 = 0 ( λ + 3)( λ − 1) = 0 y c = C 1 e t + C 2 e − 3 t Test I A| MATH 308 Differential Equations [7/8]

Step 2: Find y p by the method of undetermined coeffi- cients [Show your work here.] y p ( t ) = Ate t ( ∵ e t is part of y c ) y ′ p ( t ) = A (1 + t ) e t y ′′ p ( t ) = A (2 + t ) e t Plugging into the differential equation, A (2 + t ) e t + 2 A (1 + t ) e t − 3 Ate t = 4 e t A (1 + 2 − 3) te t + A (2 + 2) e t = 4 e t 4 Ae t = 4 e t A = 1 ∴ y p ( t ) = te t Step 3: Find the general solution to y ′′ + 2 y ′ − 3 y = 4 e t [Show your work here.] y c = C 1 e t + C 2 e − 3 t + te t Test I A| MATH 308 Differential Equations [8/8]