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y' = (1-2x) y², y(0) = -1/

y' = (1-2x) y², y(0) = -1/

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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38 CHAPTER 2 First-Order Differential Equations Problems In each of Problems 1 through 8, solve the given differential equation. x² 1. y'= = 2. 3. 4. 5. 6. 7. y y' + y² sinx = 0 y' = cos(x) cos² (2y) xy' = (1 - y2) 1/2 x-e-x al ala alaala dy dx dy dy x² dx = dy = || y + ey = 1+ y² 8. In each of Problems 9 through 16: a. Find the solution of the given initial value problem in explicit form. G b. Plot the graph of the solution. c. Determine (at least approximately) the interval in which the solution is defined. 9. y' = (1-2x) y², y(0) = -1/6 10. y'=(1-2x)/y, y(1) = -2 11. xdx+ye dy = 0, y(0) -X = 1 12. dr/d0 = r²/0, r(1) = 2 13. y'=xy³ (1+x²)-¹/², y(0) = 1 14. y'= 2x/(1+2y), y(2)=0 alibadtto noituloer 15. y' = (3x² - e*)/(2y-5), y(0) = 1 16. sin(2x) dx + cos(3y) dy = 0, y(π/2) = π/3 o the Th Some of the results requested in Problems 17 through 22 can be obtained either by solving the given equations analytically or by plotting numerically generated approximations to the solutions. Try to form an opinion about the advantages and disadvantages of each approach. G 17. Solve the initial value problem dx y y' = 1+3x² Зу2 — бу y' = " 3x² 3y² - 4' and determine the interval in which the solution is valid. Hint: To find the interval of definition, look for points where the integral curve has a vertical tangent. G 18. Solve the initial value problem y(0) = 1 y(1) = 0 G 19. Solve the initial value problem and determine the interval in which the solution is valid. Hint: To find the interval of definition, look for points where the integral curve has a vertical tangent. y' = 2y² + xy², y(0) = 1 and determine where the solution attains its minimum value. G 20. Solve the initial value problem 2-et 3 + 2y y' = and determine where the solution attains its maximum value. G21. Consider the initial value problem y' = y(0) = yo. G 22. y' ty(4- y) 3 a. Determine how the behavior of the solution as t increases depends on the initial value yo. b. Suppose that yo = 0.5. Find the time T at which the solution first reaches the value 3.98. = Consider the initial value problem (.*)\ 11 sausood bail 23. Solve the equation ty(4- y) 1+t y(0) = 0 9 a. Determine how the solution behaves as t → ∞. b. If yo = 2, find the time T at which the solution first reaches the value 3.99. dy dx c. Find the range of initial values for which the solution lies in the interval 3.99 < y < 4.01 by the time t = 2. = y(0) = yo > 0. aoiuloa toetanoo & 2nd ay+b 1 bas of sulny amos tot cy + d TOT (S) nailsups 9 where a, b, c, and d are constants. 24. Use separation of variables to solve the differential equation dQ dt =r(a+bQ), Q(0) = 20, igoin 608 where a, b, r, and Qo are constants. Determine how the solution behaves as t→∞ garbingen yd bajariliant od Homogeneous Equations. If the right-hand side of the equation dy/dx = f(x, y) can be expressed as a function of the ratio y/x only, then the equation is said to be homogeneous.¹ Such equations LORE can always be transformed into separable equations by a change of the dependent variable. Problem 25 illustrates how to solve first-order homogeneous equations. The word “homogeneous" has different meanings in different mathematical contexts. The homogeneous equations considered here have nothing to do with the homogeneous equations that will occur in Chapter 3 and elsewhere. Vosling me neillups sign 000 nitit yd moitampo ili lisilqzs bibliqmi apbuloz
Transcribed Image Text:38 CHAPTER 2 First-Order Differential Equations Problems In each of Problems 1 through 8, solve the given differential equation. x² 1. y'= = 2. 3. 4. 5. 6. 7. y y' + y² sinx = 0 y' = cos(x) cos² (2y) xy' = (1 - y2) 1/2 x-e-x al ala alaala dy dx dy dy x² dx = dy = || y + ey = 1+ y² 8. In each of Problems 9 through 16: a. Find the solution of the given initial value problem in explicit form. G b. Plot the graph of the solution. c. Determine (at least approximately) the interval in which the solution is defined. 9. y' = (1-2x) y², y(0) = -1/6 10. y'=(1-2x)/y, y(1) = -2 11. xdx+ye dy = 0, y(0) -X = 1 12. dr/d0 = r²/0, r(1) = 2 13. y'=xy³ (1+x²)-¹/², y(0) = 1 14. y'= 2x/(1+2y), y(2)=0 alibadtto noituloer 15. y' = (3x² - e*)/(2y-5), y(0) = 1 16. sin(2x) dx + cos(3y) dy = 0, y(π/2) = π/3 o the Th Some of the results requested in Problems 17 through 22 can be obtained either by solving the given equations analytically or by plotting numerically generated approximations to the solutions. Try to form an opinion about the advantages and disadvantages of each approach. G 17. Solve the initial value problem dx y y' = 1+3x² Зу2 — бу y' = " 3x² 3y² - 4' and determine the interval in which the solution is valid. Hint: To find the interval of definition, look for points where the integral curve has a vertical tangent. G 18. Solve the initial value problem y(0) = 1 y(1) = 0 G 19. Solve the initial value problem and determine the interval in which the solution is valid. Hint: To find the interval of definition, look for points where the integral curve has a vertical tangent. y' = 2y² + xy², y(0) = 1 and determine where the solution attains its minimum value. G 20. Solve the initial value problem 2-et 3 + 2y y' = and determine where the solution attains its maximum value. G21. Consider the initial value problem y' = y(0) = yo. G 22. y' ty(4- y) 3 a. Determine how the behavior of the solution as t increases depends on the initial value yo. b. Suppose that yo = 0.5. Find the time T at which the solution first reaches the value 3.98. = Consider the initial value problem (.*)\ 11 sausood bail 23. Solve the equation ty(4- y) 1+t y(0) = 0 9 a. Determine how the solution behaves as t → ∞. b. If yo = 2, find the time T at which the solution first reaches the value 3.99. dy dx c. Find the range of initial values for which the solution lies in the interval 3.99 < y < 4.01 by the time t = 2. = y(0) = yo > 0. aoiuloa toetanoo & 2nd ay+b 1 bas of sulny amos tot cy + d TOT (S) nailsups 9 where a, b, c, and d are constants. 24. Use separation of variables to solve the differential equation dQ dt =r(a+bQ), Q(0) = 20, igoin 608 where a, b, r, and Qo are constants. Determine how the solution behaves as t→∞ garbingen yd bajariliant od Homogeneous Equations. If the right-hand side of the equation dy/dx = f(x, y) can be expressed as a function of the ratio y/x only, then the equation is said to be homogeneous.¹ Such equations LORE can always be transformed into separable equations by a change of the dependent variable. Problem 25 illustrates how to solve first-order homogeneous equations. The word “homogeneous" has different meanings in different mathematical contexts. The homogeneous equations considered here have nothing to do with the homogeneous equations that will occur in Chapter 3 and elsewhere. Vosling me neillups sign 000 nitit yd moitampo ili lisilqzs bibliqmi apbuloz
19. a. y = 12 + 12 as t → ∞ b. t = 10.065778 8 20. 21. yo Yo 29. See Problem 2. 30. See Problem 4. 1. 2. 3. y = 4. 5. 6. 7. 65 cos 2t+ Section 2.2, page 38 3 3y² - 2x³ = c; y = 0 -1 y-¹ + cos x = cify # 0; 64 65 y = kx 8. y = ± √x² +c 9. a. y = 1/(x²-x-6) c. -2 < x < 3 3 sin 2t - e-t/4 ; y oscillate yo = -5/2 = -16/3; y →→∞o as to for yo = -16/3 788 65 DE- also y = 0; everywhere 2 tan(2y) - 2x - sin(2x) = c if cos(2y) # 0; also ±(2n + 1)π/4 for any integer n; everywhere y = sin(ln |x| + c) if x # 0 and ly| < 1; also y = ±1 y² - x² +2(e-ex) = c; y+e³ #0 3y+y³ - x³ = c; everywhere JE 1500
Transcribed Image Text:19. a. y = 12 + 12 as t → ∞ b. t = 10.065778 8 20. 21. yo Yo 29. See Problem 2. 30. See Problem 4. 1. 2. 3. y = 4. 5. 6. 7. 65 cos 2t+ Section 2.2, page 38 3 3y² - 2x³ = c; y = 0 -1 y-¹ + cos x = cify # 0; 64 65 y = kx 8. y = ± √x² +c 9. a. y = 1/(x²-x-6) c. -2 < x < 3 3 sin 2t - e-t/4 ; y oscillate yo = -5/2 = -16/3; y →→∞o as to for yo = -16/3 788 65 DE- also y = 0; everywhere 2 tan(2y) - 2x - sin(2x) = c if cos(2y) # 0; also ±(2n + 1)π/4 for any integer n; everywhere y = sin(ln |x| + c) if x # 0 and ly| < 1; also y = ±1 y² - x² +2(e-ex) = c; y+e³ #0 3y+y³ - x³ = c; everywhere JE 1500
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