2425_APAB_Unit_7_Notes_-_D.E.pdf

2425 APAB Unit 7 Notes - Di erential Equations 7.1 Modeling Situations with Di erential Equations 1. Unit 7: Di erential Equations What we're learning in Unit 7: 1. What is a di erential equation? 2. What is a slope eld and how can we use them? 3. Use sep...

2425 APAB Unit 7 Notes - Di erential Equations 7.1 Modeling Situations with Di erential Equations 1. Unit 7: Di erential Equations What we're learning in Unit 7: 1. What is a di erential equation? 2. What is a slope eld and how can we use them? 3. Use separation of variables to nd general solutions to a di erential equation. 4. Use initial conditions to nd particular solutions to di erential equations. 2. What is a Di erential Equation? a. y00 + 2y0 + 3 = 0 Second Order Di erential Equation d3 y b. dt3 +3=0 Third Order Di erential Equation c. f 0 (t) = 1 First Order Di erential Equation Solutions to di erential equations are FUNCTIONS! Solving di erential equations is very di cult so we will focus on one kind called separable di erential equations. For the more di cult Di Eqs, we'll use slope elds to "see" the solutions. page 1  1 2 3. How did Newton derive the equation for the height of a falling body? h (t) = 2 gt + v0 t + h0 2 a. He started with a di erential equation ddt2h = g. The height, h, of a falling body accelerates (second derivative) towards earth at g meters per second2. R R dt = g ! dv = ( g) dt ! v = gt + C b. Since dv. R 1 2 c. The object's position, h (t) = ( gt + C) dt = 2 gt + Ct + D. C and D are constants of integration. d. C and D can be found using initial conditions such as initial height, h0 and initial velocity, v0. 4. Write a Di erential Equation The RATE at which the velocity of an object is changing is 4 meters per second per second. Write a di erential equation describing this relationship. 5. The RATE at which a drink warms or cools is proportional to the di erence between the ambient temperature Ta and the current temperature of the drink, T. Which equation describes this relationship? A. dT dt = k Ta T B. T (t) = k (Ta T) C. dT dt = k (Ta T) D. T (t) = k Ta T 6. If the pressure P applied to a gas is increased while the gas is held at a constant temperature, then the volume V of the gas will decrease. The rate of change of the volume of gas with respect to the pressure is proportional to the reciprocal of the square of the pressure. Which of the following is a di erential equation that could describe this relationship? A. V = k P2 , where k is a positive constant. B. dV dP = k P2 , where k is a positive constant. C. dV dP = k P2 , where k is a negative constant. D. dP dV = k P2 , where k is a negative constant. page 2 2425 APAB Unit 7 Notes - Di erential Equations 7.2 Verifying Solutions for Di erential Equations 5y 7. dy dx = x Is y = 2x5 a solution to the di erential equation above? Show your work. 8. Of the following, which are solutions to the di erential equation y00 2y0 8y = 0 ? I. y = 2 sin (4x) II. y = 3e2x III. y = Ce4x A. I only B. II only C. III only D. II and III only page 3 2425 APAB Unit 7 Notes - Di erential Equations 7.3 Sketching Slope Fields 9. A Slope Field Helps Visualize Solutions to Di erential Equations This is a slope eld for the di erential equation dx dy = x + y. A short tangent line is drawn at coordinate points (x; y) having slope x + y. For example, look at the coordinate point (1,0). The short tangent line drawn there has a slope of 1 + 0 = 1. The curve drawn through the slope eld is a PARTICULAR solution. It is the member of the family of general solutions passing through the point (2, -1). Notice that the curve follows the slope of the tangent lines in the slope eld. page 4 2425 APAB Unit 7 Notes - Di erential Equations 10. This is an activity you can play with. https://www.geogebra.org/m/TSja3f3v 11. The slope eld to a certain di erential equation is shown above. Which of the following could be a solution to the di erential equation with initial condition y (0) = 0? x2 1 A. y = e2x B. y = x+1 C. y = ln (x + 1) D. y = x+1 page 5 2425 APAB Unit 7 Notes - Di erential Equations 12. The slope eld for a di erential equation is shown above. Which of the following statements about a solution y = f (x) must be false? A. The graph of the particular solution that satis es f (1) = 1 is concave down on the interval (0,2). B. The graph of the particular solution that satis es f ( 1) = 1 is concave up on the interval (-1, 0). C. The graph of the particular solution that satis es f (0) = 0 is concave up on the interval (-1, 1). D. The graph of the particular solution that satis es f (0) = 0 has a point of in ection at x = 0. page 6 2425 APAB Unit 7 Notes - Di erential Equations 13. a. On the axes provided below, sketch the slope eld of the di erential equation dy dx = y+1 x2 at the six points indicated. b. Sketch the particular solution y = f (x) passing through the point (1,0). c. Use separation of variables to nd the particular solution passing through the point (1,0). page 7 2425 APAB Unit 7 Notes - Di erential Equations 7.4 Reasoning With Slope Fields 14. The slope eld for a di erential equation is shown above. Which of the following could be a solution to the di erential equation with the initial condition y (0) = 1? p 1 A. y = cos (x) B. y = 1 x2 C. y = 1 x2 D. y = 1+x2 page 8 2425 APAB Unit 7 Notes - Di erential Equations 15. FL The slope eld for a di erential equation is shown above. Which of the following could be a solution, y = f (x), to the di erential equation passing through the point (0,0)? 2x2 A. y = 1 cos (x) B. y = sin (x) C. y = x2 D. y = x2 +1 x 2 16. Consider the di erential equation dy dx = y for all y 6= 0. Describe all points in the xy plane for which dy dx = 2. page 9 2425 APAB Unit 7 Notes - Di erential Equations 17. No Calculator The slope eld for a certain di erential equation is shown above. Which of the following could be a solution to the di erential equation? 1 A. y = ex B. y = x C. y = ln (x) D. y = x2 page 10 2425 APAB Unit 7 Notes - Di erential Equations 7.6 Finding General Solutions Using Separation of Variables Note: 7.5 is a BC topic not covered in this class. 18. Separable Di erential Equations A SEPARABLE di erential equation can be written in the form g (y) dy = f (x) dx These di erential equations are separable: a. dy dx = x y ! ydy = xdx b. dy dx = yex ! 1y dy = ex dx These di erential equations are NOT separable: a. dy dx =x y b. dy dx = exy 19. Example (General Solution) Find the general solution g0 (x) = y sin (x) 20. Find the general solution. 3x2 +2 dy dx = 5 cos(y) cos(x)esin(x) 21. Find the general solution to the di erential equation dy dx = cos(y). Show your work and solve explicitly for y. page 11 2425 APAB Unit 7 Notes - Di erential Equations 22. Find all solutions to the di erential equation dy dx =y y sin (x). for all y > 0. cos(x) 1 A. y = Cex B. y = p x+cos(x)+C C. y = Cex+cos(x) D. y = (x + cos (x) + C)2 7.7 Finding Particular Solutions Using Initial Conditions and Separation of Variables 23. To nd a particular solution, 1. nd the general solution 2. apply the initial condition or initial value to nd C. 3. solve explicitly for y, if necessary. 24. Example Consider the di erential equation dy dx = y. x+1 a. Find the general solution. Show your work. b. Find the particular solution satisfying y (0) = 1. Show your work. **Try solving this problem in two ways. 1. Leave part (a) unsimpli ed and solve for C, leaving part (b) unsimpli ed. 2. Simplify part (a) and solve for C, leaving part (b) simpli ed (solved explicitly for y). 25. Example (Particular Solution) 4 If f (3) = 2 and f 0 (x) = x3 Find f (1) page 12 2425 APAB Unit 7 Notes - Di erential Equations 26. Example (Implicit General and Particular Solution) (cos (y) + 2) dx dy = 2x and y (1) = 0 Find all values of x when y =. 2 27. Consider the di erential equation dx dy = xy 1. Let y = f (x) be the particular solution to the given di erential equation with the initial condition f (2) = 3. a. On the axes provided, sketch the slope eld for the di erential equation at the six points (0, 0), (0, 1), (0,2), (2,0), (2,1), and (2, 2). b.. Write an equation for the line tangent to the graph of f (x) at x = 2. Use your equation to estimate f (2:1). c. Find the particular solution y = f (x) to the given di erential equation with the initial condition f (2) = 3. a. page 13 2425 APAB Unit 7 Notes - Di erential Equations 7.8 Exponential Models With Di erential Equations 28. Model Exponential Growth and Decay with Di erential Equations A population grows at a rate proportional to its size. dP dt = kP where k is a constant of proportionality (growth factor). Show that the general solution to this di erential equation is P (t) = P (0) ekt , where P (0) is the initial population. 29. Example (Exponential Model) The number of barrels of oil a certain company exports annually increases at a rate that is proportional at any time to the number of barrels they export at that time. The company exported 5.4 million barrels annually initially, and it exported 10.8 million barrels annually after 6 years. How many million barrels of oil did the company export annually after 10 years? Round to the nearest integer. 30. Example (Exponential Models) Find the particular solution to each di erential equation. a. dy dx = 2y and y = 4 when x = 0 b. g0 (x) = 7g (x) and g (0) = 12 c. h0 (x) = h (x) and h (1) = 2 c. page 14 2425 APAB Unit 7 Notes - Di erential Equations Unit 7 Sample Test Questions- Calculator Active 31. At 12pm, t = 0 minutes, a parking lot contains 200 cars. Between 12pm and 1pm, 0 t 60 minutes, cars enter the lot at the rate E (t) = 1:5e 0:1t cars per minute. For 0 t 60 minutes, cars leave the lot at the rate L (t) = 0:2 arctan 4t + 20t cars per minute. a. To the nearest whole number, nd the total number of cars in the parking lot at 1pm (t = 60 minutes). Write the expression that generates your answer. Indicate units of measure. b. Between 12 pm and 1 pm (0 t 60 minutes), what is the maximum number of cars in the lot? Leave your answer as a whole number, show your work, and justify your answer. page 15 2425 APAB Unit 7 Notes - Di erential Equations  x 2 32. Consider the di erential equation dy dx = y. Let y = f (x) be the particular solution satisfying the initial condition f (1) = 2. a. On the axes provided, sketch a slope eld for the given di erential equation at the nine points indicated. b. Write an equation of the line tangent to the graph of y = f (x) at x = 1. Use this tangent line to approximate the value of y (1:1). c. Use separation of variables to nd the particular solution y = f (x) passing through the point 1,2). Show your work. Please solve explicitly for y. 33. During optimal conditions, the rate of change of the population of a certain organism is proportional to the population at time t, in hours. At time t = 0 hours, the population is 100. At time t = 24 hours, the population is 1000. At what time t is the population 500? Round your answer to three decimals and show your work. Hint: write an expression for P (t) that includes the constants P (0) and k. Then use the two data points given to nd the values of these constants. page 16 2425 APAB Unit 7 Notes - Di erential Equations 34. The slope eld for a certain di erential equation is shown above. Which of the following statements about a solution y = f (x) to the di erential equation is true? A. The graph of the particular solution that satis es f (0) = 1 is concave up on the interval [-2,2]. B. The graph of the particular solution that satis es f (0) = 1 is concave down on the interval [-2,2]. C. The graph of the particular solution that satis es f (0) = 1 is decreasing on the interval [0,2]. D. The graph of the particular solution that satis es f (0) = 1 is increasing on the interval [-2,0]. page 17 2425 APAB Unit 7 Notes - Di erential Equations 35. The slope eld for a certain di erential equation is shown above. Which of the following could be a solution to the di erential equation with the initial condition y (0) = 2 ? x 2 2 2 A. y = x2 +1 B. y = 2ex C. y = 2ex D. y = 2e0:5x 36. Of the following, which are solutions to the di erential equation y00 + 4y = 0 ? I. y = sin (2x) II. y = e2x III. y = 1 e 2x A. I, II, and III B. I and III only C. II only D. I only page 18 2425 APAB Unit 7 Notes - Di erential Equations  Problem-Attic format version 4.4.693 c 2011–2024 EducAide Software _ Licensed for use by Will Catterson Terms of Use at www.problem-attic.com 2425 APAB Unit 7 Notes - Di erential Equations 3/23/2025 1. 13. Answer: Answer: a. Sketch drawn in class. Points: 1 b. 2. Answer: Points: 1 3. Answer: Points: 1 4. Answer: dv dt =4 Points: 1 5. Answer: C Points: 1 R R dx 1 dy = ! ln jy + 1j = +C !y= 6. y+1 1 x2 x Answer: C Ce x 1 Points: 1 y (1) = Ce 1 1=0!C=e 7. Particular solution: y = e1 1 x 1 Answer: yes Points: 1 y = 2x5 ! dy dx = 10x4 14. 5y 2x5 Answer: D dy dx = x =5 x = 10x4 Points: 1 Points: 1 15. 8. Answer: A Answer: C Points: 1 Points: 1 16. 9. Answer: dy = x 2 = 2 dx y Answer: Points: 1 2y = x 2!y=1 x 2 for all y 6= 0. Points: 1 10. 17. Answer: Answer: C Points: 1 Points: 1 11. 18. Answer: B Answer: Points: 1 Points: 1 12. 19. Answer: C Answer: dy = sin (x) dx ! ln jyj = cos (x) + C ! Points: 1 y jyj = e cos(x)+C = e cos(x) eC jg (x)j = Ce cos(x) Points: 1  Teacher's Key Page 2 20. 26. 3x2 +2 Answer: dy dx = 5 cos(y) Answer: Find general solution: R R 3x2 +2 sin (y) + 2y = x2 + C Implicit general cos (y) dy = 5 dx ! sin (y) = 51 x3 + 2x + solution C sin (0) + 2 0 = 12 + C ! C = 1 1 1 y = sin x3 + 2x + C 5 sin (y) + 2y = x2 1 Implicit particular Points: 1 solution p 21. sin ( ) + 2 = x2 1!x= 2 +1 cos(x)esin(x) Answer: dy dx = cos(y) ! cos (y) dy = cos (x) esin(x) dx Points: 1 sin (y) = esin(x) + C (U-substitution) 27. y = sin 1 esin(x) +C Answer: Points: 1 22. Answer: C Points: 1 23. Answer: Points: 1 24. R R Answer: dx = y ! dy x+1 ydy = (x + 1) dx ! 12 y2 = 1 2 2x + x + C ! y2 = x2 + 2x + C p a. 21 y2 = 12 x2 +x+C OR y = x2 + 2x + C b. dy = 32 =9 dx (2;3) 2 1 1 2 1 2 1 b. 2 (1) = 2 (0) + 0 + C ! C = 2 y 3 = 9 (x 2) unsimpli ed f (2:1) y (2:1) = 3 + 9 (2:1 1) = 3:9 1 2 1 2 1 2y = 2x +x+ 2 OR p 1 = 02 + 2 0 + C ! C = 1 simpli ed - solved explicitly for y. p y = x2 + 2x + 1 Points: 1 25. 2 Answer: f (x) = 4x 2 + C ! f (x) = 2 x2 + C General Solution 2 16 f (3) = +C = 2!C = 9 (3)2 2 16 Points: 1 f (x) = x2 9 Particular Solution 16 34 28. f (1) = 2 = R dP R 9 9 Answer: = k dt ! ln jPj = kt + C ! jPj = ekt+C Alternatively; P R3 0 1. Assuming population is always 1 f (x) Rdx = f (3) f (1) ! f (1) = positive, we can drop the absolute value. 3 4 f (3) 1 x3 dx P (t) = ekt eC ! P (t) = Cekt f (1) = 2 16 = 34 2. Notice that P(0) = C. So, C is the 9 9 initial population. Points: 1 P (t) = P (0) ekt Points: 1  Teacher's Key Page 3 29. 36. Answer: P (t) = P (0) ekt Answer: D P (0) = 5:4 and P (6) 10:8 = 5:4ek 6 ! Points: 1 k =0.115524530093 P (10) = 5:4e0:115525(10) 17 million barrels Points: 1 30. Answer: a. y = 4e2t b. g (x) = 12e7x c. h (x) = 2e e x = 2e1 x Points: 1 31. Answer: R 60 a. 200 + 0 (E (t) L (t)) dt = 109 cars in the lot at 1 pm. b. E (t) L (t) = 0 ! t = 8:418 minutes. Since R (t) E (t) changes from positive to negative, the number of cars in the lot reaches a maximum at time t = 8.418 minutes (roughly 12:08 pm). At this time, there are 205 cars in the lot. Points: 1 32. Answer: a. Slope eld b. y 2 = 21 (x 0) y (1:1) = 1 1 2 2 (1:1 1) = 20 = 0:05 x 2 R R c. dy dx = y ! (y) dy = (x 2) dx ! y2 x2 2 = 22x + C. Using the initial condition (1,2), C = 72. The particular solution passing p through (1,2) is f (x) = x2 4x + 7 Points: 1 33. Answer: Exponential growth: population grows at a rate proportion to its size. P (t) = 100ekt ! P (24) = 1000 = 100ek 24 ! k = ln(10) 24 ln(10) P (t) = 100e 24 t The population reaches 500 at time t = 16:775 hours Points: 1 34. Answer: A Points: 1 35. Answer: D Points: 1