9 38 CHAPTER 2 First-Order Differential EquationsProblemsIn each of Problems 1 through 8, solve the given differential equation.x²1. y'==2.3.4.5.6.7.yy' + y² sinx = 0y' = cos(x) cos² (2y)xy' = (1 - y2) 1/2x-e-xal ala alaaladydxdydy x²dx=dy=||y + ey=1+ y²8.In each of Problems 9 through 16:a. Find the solution of the given initial value problem in explicitform.G b. Plot the graph of the solution.c. Determine (at least approximately) the interval in which thesolution is defined.9. y' = (1-2x) y²,y(0) = -1/610. y'=(1-2x)/y,y(1) = -211. xdx+ye dy = 0, y(0)-X= 112. dr/d0 = r²/0, r(1) = 213. y'=xy³ (1+x²)-¹/², y(0) = 114. y'= 2x/(1+2y), y(2)=0 alibadtto noituloer15. y' = (3x² - e*)/(2y-5), y(0) = 116. sin(2x) dx + cos(3y) dy = 0, y(π/2) = π/3o theThSome of the results requested in Problems 17 through 22 can beobtained either by solving the given equations analytically or byplotting numerically generated approximations to the solutions. Tryto form an opinion about the advantages and disadvantages of eachapproach.G 17. Solve the initial value problemdx yy' =1+3x²Зу2 — буy' ="3x²3y² - 4'and determine the interval in which the solution is valid.Hint: To find the interval of definition, look for points where theintegral curve has a vertical tangent.G 18. Solve the initial value problemy(0) = 1y(1) = 0G 19. Solve the initial value problemand determine the interval in which the solution is valid.Hint: To find the interval of definition, look for points where theintegral curve has a vertical tangent.y' = 2y² + xy², y(0) = 1and determine where the solution attains its minimum value.G 20. Solve the initial value problem2-et3 + 2yy' =and determine where the solution attains its maximum value.G21. Consider the initial value problemy' =y(0) = yo.G 22.y'ty(4- y)3a. Determine how the behavior of the solution as t increasesdepends on the initial value yo.b. Suppose that yo = 0.5. Find the time T at which the solutionfirst reaches the value 3.98.=Consider the initial value problem(.*)\ 11 sausood bail 23. Solve the equationty(4- y)1+ty(0) = 09a. Determine how the solution behaves as t → ∞.b. If yo = 2, find the time T at which the solution first reachesthe value 3.99.dydxc. Find the range of initial values for which the solution lies inthe interval 3.99

Question

9 38 CHAPTER 2 First-Order Differential EquationsProblemsIn each of Problems 1 through 8, solve the given differential equation.x²1. y'==2.3.4.5.6.7.yy' + y² sinx = 0y' = cos(x) cos² (2y)xy' = (1 - y2) 1/2x-e-xal ala alaaladydxdydy x²dx=dy=||y + ey=1+ y²8.In each of Problems 9 through 16:a. Find the solution of the given initial value problem in explicitform.G b. Plot the graph of the solution.c. Determine (at least approximately) the interval in which thesolution is defined.9. y' = (1-2x) y²,y(0) = -1/610. y'=(1-2x)/y,y(1) = -211. xdx+ye dy = 0, y(0)-X= 112. dr/d0 = r²/0, r(1) = 213. y'=xy³ (1+x²)-¹/², y(0) = 114. y'= 2x/(1+2y), y(2)=0 alibadtto noituloer15. y' = (3x² - e*)/(2y-5), y(0) = 116. sin(2x) dx + cos(3y) dy = 0, y(π/2) = π/3o theThSome of the results requested in Problems 17 through 22 can beobtained either by solving the given equations analytically or byplotting numerically generated approximations to the solutions. Tryto form an opinion about the advantages and disadvantages of eachapproach.G 17. Solve the initial value problemdx yy' =1+3x²Зу2 — буy' =&#34;3x²3y² - 4'and determine the interval in which the solution is valid.Hint: To find the interval of definition, look for points where theintegral curve has a vertical tangent.G 18. Solve the initial value problemy(0) = 1y(1) = 0G 19. Solve the initial value problemand determine the interval in which the solution is valid.Hint: To find the interval of definition, look for points where theintegral curve has a vertical tangent.y' = 2y² + xy², y(0) = 1and determine where the solution attains its minimum value.G 20. Solve the initial value problem2-et3 + 2yy' =and determine where the solution attains its maximum value.G21. Consider the initial value problemy' =y(0) = yo.G 22.y'ty(4- y)3a. Determine how the behavior of the solution as t increasesdepends on the initial value yo.b. Suppose that yo = 0.5. Find the time T at which the solutionfirst reaches the value 3.98.=Consider the initial value problem(.*)\ 11 sausood bail 23. Solve the equationty(4- y)1+ty(0) = 09a. Determine how the solution behaves as t → ∞.b. If yo = 2, find the time T at which the solution first reachesthe value 3.99.dydxc. Find the range of initial values for which the solution lies inthe interval 3.99 < y < 4.01 by the time t = 2.=y(0) = yo > 0.aoiuloa toetanoo & 2nday+b 1 bas of sulny amos totcy + dTOT (S) nailsups9where a, b, c, and d are constants.24. Use separation of variables to solve the differential equationdQdt=r(a+bQ), Q(0) = 20,igoin 608where a, b, r, and Qo are constants. Determine how the solutionbehaves as t→∞garbingen yd bajariliant odHomogeneous Equations. If the right-hand side of the equationdy/dx = f(x, y) can be expressed as a function of the ratio y/xonly, then the equation is said to be homogeneous.¹ Such equationsLOREcan always be transformed into separable equations by a change ofthe dependent variable. Problem 25 illustrates how to solve first-orderhomogeneous equations.The word “homogeneous&#34; has different meanings in different mathematicalcontexts. The homogeneous equations considered here have nothing to do withthe homogeneous equations that will occur in Chapter 3 and elsewhere.Vosling meneillups sign000nitityd moitampoili lisilqzsbibliqmi apbuloz 19. a. y = 12 +12 as t → ∞b. t = 10.065778820.21. yoYo29. See Problem 2.30. See Problem 4.1.2.3.y =4.5.6.7.65cos 2t+Section 2.2, page 3833y² - 2x³ = c; y = 0-1y-¹ + cos x = cify # 0;6465y = kx8. y = ± √x² +c9. a. y = 1/(x²-x-6)c. -2 < x < 33sin 2t -e-t/4; y oscillateyo = -5/2= -16/3; y →→∞o as to for yo = -16/378865DE-also y = 0; everywhere2 tan(2y) - 2x - sin(2x) = c if cos(2y) # 0; also±(2n + 1)π/4 for any integer n; everywherey = sin(ln |x| + c) if x # 0 and ly| < 1; also y = ±1y² - x² +2(e-ex) = c;y+e³ #03y+y³ - x³ = c; everywhereJE 1500

Accepted Answer

We have to solve the  question  number (9).

y' = (1-2x) y², y(0) = -1/