Exercícios Selecionados Livro Calc2 Pt1-2021 2

Zill, D., Cullen, M. Differential equations with boundary-value problems 50 e CHAPTER 2. FIRST-ORDER DIFFERENTIAL EQUATIONS REMARKS (i) As we have just seen in Example 5, some simple functions do not possess an antiderivative that is an elementary function. Integrals of these kinds of functions are called nonelementary. For example, [% e~" dt and fsin x? dx are nonelementary integrals. We will run into this concept again in Section 2.3. (ii) In some of the preceding examples we saw that the constant in the one- parameter family of solutions for a first-order differential equation can be rela- beled when convenient. Also, it can easily happen that two individuals solving the same equation correctly arrive at dissimilar expressions for their answers. For example, by separation of variables we can show that one-parameter families of solutions for the DE (1 + y?) dx + (1 + x?) dy = Oare x+y arctan x + arctan y = c or aE 1 — xy As you work your way through the next several sections, bear in mind that fami- lies of solutions may be equivalent in the sense that one family may be obtained from another by either relabeling the constant or applying algebra and trigonom- etry. See Problems 27 and 28 in Exercises 2.2. EX E RC | S ES 2 . 2 . et aravels Answers to selected odd-numbered problems begin on page ANS- 1. In Problems(_22)solve the given differential equation by dy _ — _. ay > separation of variables. 21, dx I-y 22, (e+e x ~ dy . dy 5 In Problems @3-29 find an explicit solution of the given 1. — = sin 5x 2. — =(« + 1) ae dx dx initial-value problem. 3. dx + e*dy =0 4. dy — (y — 1)*dx =0 23. “ =402+ 1), x(7/4) = 1 t dy dy 5.x =4 6. — + 2xy? =0 2 “dx dx a4, & =X 1 yay =2 dx xr-1 dy _ 3x + 2y x dy _ -y —2x-y d 7 e y 8 eye te y 28. Po = y — xy, y(-1) = -1 x dx yt ‘) dy (2 + 2) dy _ _ 5 9. yInx— = | —— 10. — = |——_— 26. — + 2y=1, yO) =3 ve dy ( x dx \4x+5 dt ? 2 5 5 V3 11. csc ydx + sec*x dy = 0 27. V1 -— y dx — V1 -—xdy=0, yO) = a . 3 _— 12. sin 3x dx + 2y cos°3x dy = 0 28. (1 +.x4) dy + x(1 + 4y2) dx =0, y(1)=0 . (& + 1)%e7? + (eX + 1)33e7* _ 13. Ce Drew dx +(e Dien dy =0 In Problems 29 and 30 proceed as in Example 5 and find an 14. x(1 + y?)!2 dx = y(1 + x2)! dy explicit solution of the given initial-value problem. dy > dS d 29. —=ye*, y(4)=1 15. -=ks 16. “2 = Ko — 10) a 7 dr dt dy I dP dN 30. — = y’sinx’, y(-2) = 3 17. —=P-P 18. — + N = Nte*? dx” » ° dt dt 31. (a) Find a solution of the initial-value problem consisting 19 dy_ xy +3x-y-3 20 dy _xy +2y—x-2 of the differential equation in Example 3 and the ini- * dx xy — 2x + 4y —- 8 * dx xy —3y+x-3 tial conditions y(0) = 2, y(O) = —2, and y(t) = 1. 2.2. SEPARABLE VARIABLES e 51 (b) Find the solution of the differential equation in y3(1) = 2, and y4(—1) = 4. Graph each solution Example 4 when In c; is used as the constant of and compare with your sketches in part (a). Give integration on the /eft-hand side in the solution and the exact interval of definition for each solution. 4 In ci Is replaced by In c. Then solve the same 41. (a) Find an explicit solution of the initial-value problem initial-value problems in part (a). dy 2x+1 ; ; dy_ 4 = V2) = 1. G2) Find a solution of xy = y° — y that passes through dx 2y . . . x the indicated points. (b) Use a graphing utility to plot the graph of the solu- (a) (0, 1) (b) (0, 0) (c) ( 1) (d) (2, 1) tion in part (a). Use the graph to estimate the inter- val J of definition of the solution. (3) Find a singular solution of Problem 21. Of Problem 22. (c) Determine the exact interval / of definition by ana- Show that an implicit solution of lytical methods, 5 5 42. Repeat parts (a)—(c) of Problem 41 for the IVP consist- 2x sin’y dx — (x° + 10) cos ydy = 0 ing of the differential equation in Problem 7 and the ini- is given by In(x* + 10) + csc y = c. Find the constant tial condition y(0) = 0. solutions, if any, that were lost in the solution of the dif- . . ferential equation. Discussion Problems 43. Explain why the interval of definition of th licit Often a radical change in the form of the solution of a differen- (a) lL, ion “ y bs () r the onitinLveluc v blem in tial equation corresponds to a very small change in either the Exam ed is the open interval (—5, 5) P initial condition or the equation itself. In Problems 35-38 find P ; p : a an explicit solution of the given initial-value problem. Use a (b) Can any solution of the differential equation Cross graphing utility to plot the graph of each solution. Compare the ans! ee you ae that al + = oI an each solution curve in a neighborhood of (0, 1). implicit solution of the imtial-value problem F dy /dx = —x/y, y(1) = 0? 35. ak =(y- 1), y@)=1 (a) If a> 0, discuss the differences, if any, between the solutions of the initial-value problems consist- 36. dy _ (y — 1), y(0) = 1.01 ing of the differential equation dy/dx = x/y and dx each of the initial conditions y(a) = a, y(a) = —a, 37) = (y — 1° +001, yO) =1 CO GMO EE Vdx YD) O01, y(0) = (b) Does the initial-value problem dy/dx = x/y, dy y(0) = 0 have a solution? 38, dx (y- 1° — 0.01, y@)=1 (c) Solve dy/dx = x/y, y1) = 2 and give the exact interval J of definition of its solution. 39. Every autonomous first-order equation dy/dx = f(y) 45. In Problems 39 and 40 we saw that every autonomous is separable. Find exp hicit solutions Yi), Yas ¥ 300), first-order differential equation dy/dx =f(y) is and y 4) of the differentia I equation dy /dx=y—y separable. Does this fact help in the solution of the that satisfy, in turn, the initial conditions y,(0) = 2, d y,(0) = 4, y,(0) = —3, and y4(0) = —2. Use a graphing initial-value problem = =V1+y’sin’y, y(0) = 4? utility to plot the graphs of each solution. Compare these Di ketch. by h “4 lausible soluti f graphs with those predicted in Problem 19 of Exercises iscuss. Sketch, by hand, a plausible solution curve 0 2.1. Give the exact interval of definition for each solution. the problem. 40. (a) The autonomous first-order differential equation 46. Without the use of technology, how would you solve dy/dx = 1/(y— 3) has no. critical points. dy Nevertheless, place 3 on the phase line and obtain (Vx + x) dx =vVyty? a phase portrait of the equation. Compute d’y /dx” to determine where solution curves are concave up Carry out your ideas. and where they are concave down (see Problems 35 47. Find a function whose square plus the square of its and 36 in Exercises 2.1). Use the phase portrait derivative is 1. and concavity to sketch, by hand, some typical solution curves. 48. (a) The differential equation in Problem 27 is equiva- lent to the normal form (b) Find explicit solutions y;(x), yo(x), y3(x), and y4(x) 5 of the differential equation in part (a) that satisfy, dy _ jl—y in turn, the initial conditions y\(0) = 4, y2(0) = 2, dx 1-— x 60 e CHAPTER 2. FIRST-ORDER DIFFERENTIAL EQUATIONS USE OF COMPUTERS The computer algebra systems Mathematica and Maple are capable of producing implicit or explicit solutions for some kinds of differential equations using their dsolve commands.” REMARKS (®) In general, a linear DE of any order is said to be homogeneous when g(x) = 0 in (6) of Section 1.1. For example, the linear second-order DE y” — 2y' + 6y = 0 is homogeneous. As can be seen in this example and in the special case (3) of this section, the trivial solution y = 0 is always a solution of a homogeneous linear DE. (ii) Occasionally, a first-order differential equation is not linear in one variable but is linear in the other variable. For example, the differential equation el dx x+y? is not linear in the variable y. But its reciprocal dx +¥ dx P —=%x or —-x= dy 7 dy " is recognized as linear in the variable x. You should verify that the integrating factor e/(~ 4’ = e-” and integration by parts yield the explicit solution x = —y? — 2y — 2+ ce? for the second equation. This expression is, then, an implicit solution of the first equation. (iii) Mathematicians have adopted as their own certain words from engineer- ing, which they found appropriately descriptive. The word transient, used earlier, is one of these terms. In future discussions the words input and output will occasionally pop up. The function f in (2) is called the input or driving function; a solution y(x) of the differential equation for a given input is called the output or response. (iv) The term special functions mentioned in conjunction with the error func- tion also applies to the sine integral function and the Fresnel sine integral introduced in Problems 49 and 50 in Exercises 2.3. “Special Functions” is actually a well-defined branch of mathematics. More special functions are studied in Section 6.3. *Certain commands have the same spelling, but in Mathematica commands begin with a capital letter (Dsolve), whereas in Maple the same command begins with a lower case letter (dsolve). When discussing such common syntax, we compromise and write, for example, dsolve. See the Student Resource and Solutions Manual for the complete input commands used to solve a linear first-order DE. EXERCISES 2.3 || Nn e a re S Answers to selected odd-numbered problems begin on page ANS-2. In Problems(_24.ind the general solution of the given dif- , 2 2 , _ 3 . ; . . . 5. y' + 3x*y=x 6. y’ + 2xy =x ferential equation. Give the largest interval J over which the general solution is defined. Determine whether there are any 7. xy’ +xy=1 8. y’ = 2y + +5 transient terms in the general solution. d d y _ 2G: y = 9. x— —y=x'sinx 10. x— + 2y =3 dy dy dx dx 1. — =5y 2.— +2y=0 dx dx dy dy d d HW. x7 + dy =x 12. (lta oo xyaxta? x x 3, 2+ y =e 4.324 2y=4 dx dx 13. x°y’ + x(x + 2)y = e* 2.3. LINEAR EQUATIONS e 61 , —X a: d 14. xy’ + (1 + xy =e sin 2x 33. a. + 2xy = f(x), yO) = 2, where x 15. ydx — 4x + y®) dy = 0 y (x + y”) dy fx O<x<1 16. y dx = (ye — 2x) dy fa) = 0, x21 17 dy + (sinx)y = 1 d » COSK TT sinOy 34. (1 + x2) a + 2xy = f(x), (0) = 0, where x d 18. cos?x sin x + (cos*x)y = 1 x O<x<1 x fx) = - dy —x, x21 19. (x + 1) a + (x + 2)y = 2xe 35. Proceed in a manner analogous to Example 6 to solve the dy initial-value problem y’ + P(x)y = 4x, y(O) = 3, where 20. (x + 2? = =5 — 8y — 4xy dx _ | 2, Os=x=l, P(x) = d —2/x, > 1. 21. + rsec 8 = cos 8 (x x Use a graphing utility to graph the continuous function d graphing y to grap P 22. + 2P = P + At ~ 2 y(@). dy (36) Consider the initial-value problem y’ + e*y = f(x), 23. x ax + Bx + ly =e* y(0) = 1. Express the solution of the IVP for x > O asa nonelementary integral when f(x) = 1. What is the so- 24. (x2 — 1) dy +2y =(x + 1p lution when f(x) = 0? When f(x) = e*? dx 37. Express the solution of the initial-value problem In Problems@5—30) solve the given initial-value problem. y’ — 2xy = 1, y(1) = 1, in terms of erf(x). Give the largest interval J over which the solution is defined. 25. xy’ ty =e", yl) =2 Discussion Problems 26 dx _ x = 2, (1) =5 38. Reread the discussion following Example 2. Construct a ‘7 dy mY linear first-order differential equation for which all di nonconstant solutions approach the horizontal asymp- 27. Lt Ria E, i(0) = ig, tote y = 4asx—>~™, L, R, E, and ig constants 39. Reread Example 3 and then discuss, with reference to Theorem 1.2.1, the existence and uniqueness of a 28. aT =k(T—T,,); T(O) = Ty solution of the initial-value problem consisting of dt xy’ — 4y = x°e* and the given initial condition. k, T,,, and Tp constants ° (a) yO) =0 — (b) y(0) = yon Yo > 0 29. (x + pay sing y(1) = 10 (c) yo) = yo, Xo > 0, yo > 0 dx , , 5 40. Reread Example 4 and then find the general solution of 30. y’ + (tan x)y = cos*x, y(0) = —1 the differential equation on the interval (—3, 3). In ProblemsG1 -34)proceed as in Example 6 to solve the 41. Reread the discussion following Example 5. Construct a given initial-value problem. Use a graphing utility to graph linear first-order differential equation for which all solu- the continuous function y(x). tions are asymptotic to the line y = 3x —5 asx >. dy 42. Reread Example 6 and then discuss why it is technically 31. dx + 2y = f(x), y(O) = 0, where incorrect to say that the function in (13) is a “solution” of the IVP on the interval [0, ©). 1, 0Sx53 a SQ) = 0 x>3 43. (a) Construct a linear first-order differential equation of , the form xy’ + ao(x)y = g(x) for which y, = c/x? d and yy = x. Give an interval on which 32. 7” + y = f(x), y(O) = 1, where y = x° + c/x’ is the general solution of the DE. x (b) Give an initial condition y(xo) = yo for the DE fo) = {! O0Sx=1 found in part (a) so that the solution of the IVP —i, x>1 is y=x>—1/x>. Repeat if the solution is 2.4 EXACT EQUATIONS 67 sometimes possible to find an integrating factor p(x, y) so that after multiplying, the left-hand side of p(x, y)M(x, y) dx + w(x, y)N(x, y) dy = 0 (8) is an exact differential. In an attempt to find 2, we turn to the criterion (4) for exact- ness. Equation (8) is exact if and only if (uM), = (wN),, where the subscripts denote partial derivatives. By the Product Rule of differentiation the last equation is the same as uM, + wyM = WN, + Nor jxN — pyM = (My — N,)p (9) Although M, N, M,, and N, are known functions of x and y, the difficulty here in determining the unknown p(x, y) from (9) is that we must solve a partial differential equation. Since we are not prepared to do that, we make a simplifying assumption. Suppose y is a function of one variable; for example, say that uw depends only on x. In this case, uw» = du /dx and pry = 0, so (9) can be written as du M,—N, Pay 10 dx wy tt (10) We are still at an impasse if the quotient (M, — N,)/N depends on both x and y. However, if after all obvious algebraic simplifications are made, the quotient (My, — N,) /N turns out to depend solely on the variable x, then (10) is a first-order ordinary differential equation. We can finally determine ps because (10) is separa- ble as well as linear. It follows from either Section 2.2 or Section 2.3 that p(x) = ef(%-Nd/N)dx Ty like manner, it follows from (9) that if 4 depends only on the variable y, then du N,-—M, — = ———u. (1) dy M In this case, if (NV, — M,)/M is a function of y only, then we can solve (11) for p. We summarize the results for the differential equation M(x, y) dx + N(x, y) dy = 0. (12) ° If, — N,) /N is a function of x alone, then an integrating factor for (12) is esc ——— dx Mx) =e NO, (13) ° If(N, — My) /M is a function of y alone, then an integrating factor for (12) is fuga u(y) = e . (14) | EXAMPLE 4 A Nonexact DE Made Exact The nonlinear first-order differential equation xy dx + (2x? + 3y? — 20) dy = 0 is not exact. With the identifications M = xy, N = 2x? + 3y* — 20, we find the partial derivatives M, = x and N, = 4x. The first quotient from (13) gets us nowhere, since M, — Nx x — 4x —3x N 2x2 + 3y?- 20 9 2x? + 3y? — 20 depends on x and y. However, (14) yields a quotient that depends only on y: N.-~M, 4x—x 3x _ 3 M = xy oxy oy’ 68 e CHAPTER 2. FIRST-ORDER DIFFERENTIAL EQUATIONS The integrating factor is then e/°4”” = e3"Y = e'™” = y?, After we multiply the given DE by p(y) = y°, the resulting equation is xy* dx + (2x’y3 + 3y° — 20y) dy = 0. You should verify that the last equation is now exact as well as show, using the method of this section, that a family of solutions is 4x?y+ + }y°—5yt=c. & REMARKS (i) When testing an equation for exactness, make sure it is of the precise form M(x, y) dx + N(x, y) dy =0. Sometimes a differential equation is written G(x, y) dx =H(x, y) dy. In this case, first rewrite it as G(x, y) dx — H(x, y) dy = 0 and then identify M(x, y) = G(x, y) and N(x, y) = —H(a, y) before using (4). (ii) In some texts on differential equations the study of exact equations precedes that of linear DEs. Then the method for finding integrating factors just discussed can be used to derive an integrating factor for y’' + P@)y = f(@). By rewriting the last equation in the differential form (P@)y — f(x)) dx + dy = 0, we see that M, — N, P wo From (13) we arrive at the already familiar integrating factor e/?*, used in Section 2.3. exercises 2.4 EXatas€iINCKAatasenm In Problems({_20)determine whether the given differential 12. 3x2y + &) dx + (x3 + xe — 2y) dy =0 equation is exact. If it is exact, solve it. 1. (2x — 1) dx + By + 7) dy =0 B. xP = mer y + 6x" x 2. (2x + y) dx — (x + 6y) dy = 0 3. (5x + 4y) dx + (4x — 8y7) dy = 0 14. (1-2 +r)@ey=3-4 4. (sin y — ysin x) dx + (cosx + xcosy — y)dy=0 y * “ 2 2 = 5. (2xy — 3) dx + (2xy + 4) dy =0 15. (vy - 1 JS a xyr=o \ a 1 + 9x?/ dy 6. (2 — A+ cos 3x) 4+ 3 — a + 3y sin 3x = 0 x dx x 16. (Sy — 2x)y' — 2y =0 2_ 2 2 = 7. Q — y") dx + GO" — 2xy) dy = 0 17. (tan x — sin x sin y) dx + cos x cos y dy = 0 8. ( +Inx + *) dx = (1 — Inx) dy 18. (2y sinx cos x — y + 2ye*”) dx x = (x — sin? x — 4xye*” 9. (x — y> + y? sin x) dx = (xy? + 2y cos x) dy ( ~ sin’ rye) dy 10. (x3 + y3) dx + 3xy* dy =0 19. (4 y — 152° — y) dt + (4 + 3y? — dy =0 UU. (yin -eax+ (2 +xm )a =0 20 (t+ 4-52 )ar+( reatoa =0 ty y y y y . t r 2 4 y? ye 2 4 y? y= 2.4 EXACT EQUATIONS e 69 In Problem(21-26olve the given initial-value problem. (b) Show that the initial conditions yO) = —2 and 21. (x+y) dx + Qxy +x7-1)dy=0, y(l)=1 y(1) = 1 determine the same implicit solution. 22. (e + y)dx + (2 +x +4 ye”) dy=0, y(0)=1 (c) Find explicit solutions yy(x) and yo(x) of the dif- ferential equation in part (a) such that y,(0) = —2 23. (4y + 2t — 5) dt + (6y + 4t—-— 1)dy=0, y(-1) =2 and y2(1) = 1. Use a graphing utility to graph y;(x) 5 5 and y2(x). 3y? —t je t 24, | ——— ]— + = = 0, 1)=1 ( y> dt 2y* yQ) 25. (y? cos x — 3x7y — 2x) dx Discussion Problems + Qysinx — x3 +Iny)dy=0, y(0) = (2y sinx ~ x ny) dy yO) =e 40. Consider the concept of an integrating factor used in 1 dy ; Problems 29—38. Are the two equations M dx + Ndy = 0 26. l+y + cosx — 2xy dx y(y + sin x), yO) = 1 and uM dx + uN dy = 0 necessarily equivalent in the sense that a solution of one is also a solution of the other? In Problems 27 and 28 find the value of k so that the given Discuss. differential equation is exact. 41. Reread Example 3 and then discuss why we can con- (y? + kxy* — 2x) dx + Gxy? + 20x?y3) dy = 0 clude that the interval of definition of the explicit solution of the IVP (the blue curve in Figure 2.4.1) is (6xy> + cos y) dx + (2kx*y* — x sin y) dy = 0 (-1, 1). In Problems 29 and 30 verify that the given differential equa- 42. Discuss how the functions M(x, y) and N(x, y) can be tion is not exact. Multiply the given differential equation found so that each differential equation is exact. Carry by the indicated integrating factor w(x, y) and verify that the out your ideas. new equation is exact. Solve. 1 (—xy sin x + 2y cos x) dx + 2x cos x dy = 0; (a) M(x, y) dx + (se + 2xy + ‘) dy = 0 M(x, y) = xy -/2,02 4 —* _ (x? + xy — y") dx + (y? + 2xy — x?) dy = 0; (b) (: y + Jax + Nevidy=0 L(x, y) = (+ y) (vide eqs. (13) e (14) no texto acima.) 43. Differential equations are sometimes solved by In ProblemsG1—39)solve the given differential equation by having a clever idea. Here is a little exercise in finding, as in Example 4, an appropriate integrating factor. cleverness: Although the differential equation 5 _ (x — Vx" + y’) dx + y dy = 0 is not exact, show how 31. (2y* + 3x) dx + 2xy dy =0 the rearrangement (x dx + y dy) /Vx? + y? = dx and 32. yx + y+ 1) dx + (x + 2y) dy =0 the observation 5 d(x? + y?) = x dx + y dy can lead to a solution. 33. 6xy dx + (4y + 9x’) dy =0 True or False: Every separable first-order equation 2 dy/dx = g(x)h()y) is exact. 34. cos x dx + (: +2) sinx dy = 0 »/ seyaty) y 35. (10 — 6y +e **) dx —2dy=0 . Mathematical Model 36. (y? + xy?) dx + (Sy? — xy + y? sin y) dy = 0 (yi Fay) dx + Oy" ay Fy ») dy 45. Falling Chain A portion of a uniform chain of length In Problems 37 and 38 solve the given initial-value problem 8 ft is loosely coiled around a peg at the edge of a high : . we : horizontal platform, and the remaining portion of the by finding, as in Example 4, an appropriate integrating factor. : chain hangs at rest over the edge of the platform. See " x dx + (x*y + 4y)dy=0, y(4) =0 Figure 2.4.2. Suppose that the length of the overhang- 2 2 _ _ ing chain is 3 ft, that the chain weighs 2 Ib/ft, and that (i ty S)dx=(y + xy)dy, yO) =1 the positive direction is downward. Starting at t = 0 (a) Show that a one-parameter family of solutions of seconds, the weight of the overhanging portion causes the equation the chain on the table to uncoil smoothly and to fall to the floor. If x(t) denotes the length of the chain over- (4xy + 3x2) dx + 2y + 2x?) dy =0 hanging the table at time ¢ > 0, then v = dx/dt is its velocity. When all resistive forces are ignored, it can isx + 2x?y + yy? =c. be shown that a mathematical model relating v to x is 74 e CHAPTER 2. FIRST-ORDER DIFFERENTIAL EQUATIONS EXERCISES 2.5 O I Y ) O1 e Nn aS: selected odd-numbered problems begin on page ANS-2. Each DE in Problems 1-14 is homogeneous. Each DE in Problems 23-30 is of the form given in (5). In ProblemsG—1 solve the given differential equation by In Problems 23-28 solve the given differential equation by using an appropriate substitution. using an appropriate substitution. 1. @- y)dx+xdy=0 2. (x + y)dx+xdy =0 3, = G+ y 41) yg, Lo xy dx dx x+y 3. xdx + (y — 2x) dy =0 4. ydx =2(x+ y) dy d d 5. (y? + yx) dx — x2 dy =0 25. x = tan’*(x + y) 26. mn = sin(x + y) 6. (y*? + yx) dx + x* dy =0 d d 27, 24 Vy Oe F3 28,2 = 14+ e> 7 dy = ya * dx dx “dx ytx ; dy x +3y In Problems 29 and 30 solve the given initial-value problem. , - d dx 3xty 29. = = cos(x + y), (0) = 7/4 x 9, ~ydx+(x+ V = y dx (x xy) dy 0 w dy 3x +2y Cpe dy “dx 3x+2y+2’ y 10. x= yt Vie — yy, x>0 x In Problems 19) solve the given initial-value problem. Discussion Problems dy 6) Explain why it is always possible to express any homoge- 11. xy? dx =y-—x, yd) =2 neous differential equation M(x, y) dx + M(x, y) dy = Oin the form dx 12. (x? + 2y?)— = xy, y(-D=1 dy (2) dy —=F\=-]. dx x y/x _ y/x = = 13. (x + ye’) dx — xe" dy =0, yd) =0 You might start by proving that 14. +x(Unx-Iny-1 = l= ydx + x(nx~Iny~ I dy = 0, yl) =e M(x, y) =x*M(,y/x) and N(x, y) = x*N(L, y/2). Each DE in Problems 15-22 is a Bernoulli equation. @) Put the homogeneous differential equation In Problems 15-20 solve the given differential equation by (5x? — 2y’) dx — xy dy =0 i iat bstitution. Using an appropriate supsmumon into the form given in Problem 31. 15. +2 +y= i 16. dy ~y=e'y 33. (a) Determine two singular solutions of the DE in dx y" dx Problem 10. d d (b) If the initial condition y(5) = 0 is as prescribed in 17. oY _ y(xy3 — 1) 18. 7~2 — (1+ xy = xy? Problem 10, then what is the largest interval J over dx dx which the solution is defined? Use a graphing util- d d ity to graph the solution curve for the IVP. y J 19. P—+y=t 20. 30. + P) = = 2n(y? - 1 dt ” » 0. 3 ) dt x0 ) 34. In Example 3 the solution y(x) becomes unbounded as x — +, Nevertheless, y(x) is asymptotic to a curve as In Problems 21 and 22 solve the given initial-value problem. +> ~ and to a different curve as x —> °. What are the equations of these curves? d 21. x? aa 2xy = 3y*, yd) = 5 35. The differential equation dy/dx = P(x) + O(x)y + R(x)y” dx . as : is known as Riccati’s equation. dy (a) A Riccati equation can be solved by a succession 2 2X 3/2 — — 22. y Tx +y 1, yO) =4 of two substitutions provided that we know a EXERCISES 1.1 (PAGE 1 di d CISES 1.1 (PAGE 10) 15. Lo + Ri= E(t) 17. m— = mg — kv 1. linear, second order 3. linear, fourth order dt dt : : : 2 2 R2 5. nonlinear, second order 7. linear, third order 19. m ax kx 21. dr +8 9 9. linear in x but nonlinear in y dt’ dt? r? 15. domain of function is [—2, ©); largest interval of dA dx definition for solution is (—2, %) 23. dt kK(M—A),k>0 25. dt tkx=rk>0 A 17. domain of function is the set of real numbers except > ew x = 2 and x = —2; largest intervals of definition for 27. dy = ox t Va ty Ee solution are (—%, —2), (—2, 2), or (2, ©) dx y oS ‘—] 19. X = — defined on (—, In 2) or on (In 2, ©) 5 e—2 CHAPTER 1 IN REVIEW (PAGE 32) ; 27. m= —2 29. m= 2,m=3 31. m=0,m=—1 d 33. y=2 35. no constant solutions 1. a = 10y 3. y"+ ky =0 $ x Lu 5. y"—2y'+y=0 7. (a), (d) a EXERCISES 1.2 (PAGE 17) 9. (b) 11. (b) O 1. y =1/C - 4e*) 1. y= er and y= c2e*, c, and c2 constants a 3. y= 1/7 = 1); (1, %) Fy a) The demain is the set of all real numb a 5. y = 1/(x2 + 1); (—%, ©) . (a) he domain is the set of all real numbers. = 7. x= —cost+ 8sint (b) either (=, 0) or (0, *) ea 19. For x9 = —1 the interval is (—~, 0), and for x9 = 2 the Ss 9x = “3 cost +isint I. y =se"—fe™ interval is (0, %). 5 —x7, x <0 1 13. y=5e*! 15. y=0,y =x? 21. (ce) y= {3 ~¢ 23. (—, 2) Q 17. half-planes defined by either y > 0 or y < 0 > o= a 19. half-planes defined by either x > 0 or x <0 25. (0, ~) 27. y= je" —fSe*— 2x a 21. the regions defined by y > 2, y< —2,or -2<y<2 29. y= Re 3 4 Sextl — 2x. Lu 23. any region not containing (0, 0) 5 31. yo = —3,y1 = 0 m 25. yes ee 27. no dP rm . 33. — = k(P — 200 + 10t ”n 29. (a) y=cx dt ( ) oc (b) any rectangular region not touching the y-axis ° (c) No, the function is not differentiable at x = 0. ua 31. (b) y= 1/(1 — x) on (—%, 1); EXERCISES 2.1 (PAGE 41) ac y = —1/(x + 1) on (—1, ©); 21. 0 is asymptotically stable (attractor); 3 is unstable = (c) y = O0on(—~%, %) (repeller). 3 23. 2 is semi-stable. xt 25. —2 is unstable (repeller); 0 is semi-stable; 2 is EXERCISES 1.3 (PAGE 27) asymptotically stable (attractor). dP dP 27. —1 is asymptotically stable (attractor); 0 is unstable 1. dt =kPt+r; Ut =kP-r (repeller). dP 39. 0< Po <h/k 3. dt = k,P _— ky P? 41. Ving/k dx 7. — = kx(1000 — x) dt EXERCISES 2.2 (PAGE 50) dA 1 1 1 4-3 9. Tt T9904 = 8 AW = 50 1. y = —;cos5x + c¢ 3. y=7e “+e _— 4 _ —2y 3x dA 7 dh 5. y=cx 7. 3e % =2e* +e 11. — + ——A=6 B.S = -2wva 9, 13 _— 13 _1\,2 dt 600—f dt 450 230 Inx — 5x =3y + 2y + Inly| +c ANS-1 ANS-2 e ANSWERS FOR SELECTED ODD-NUMBERED PROBLEMS 1. 4cosy =2x + sin2x +c EXERCISES 2.4 (PAGE 68) 13. (e+ 1) 7+2(e?+ 1) '=c le -—xt+3y+7y=c 3. 32° + 4xy — 2yt=c 15. S = cek" 17. P= Tw 5. x°y?-3x+4y=c 7. not exact ; ; 9. xy + y?cosx — 32x? = — y = sin (4 19. (y+ 3)’e c(x + 4 e 21. y sin ($x + c) 11. not exact 3 e (tix) 13. xy — 2xe* + 2e*- 2x7 =c 23. x = tan(4r — 3m) 25. y = —— 15. ey? — tan”! 3x=c x . , VA ; Prera 17. —In|cos x| + cos xsiny = c¢ 27,.y=5xt+V1-x 29. y = e/4 19. fy —58-ty+y3=c 3 — gtr l 143 2 2yna4 31. (a) y= 2. y = —2.y = 23TH 21. 39° + x’y + xy y= e N 2 2 VE ce 33. y= —l andy = 1 are singular solutions of Problem 21; 23. Aty + ’ o r 3 y=8 uu _ 25. y°sinx —x°y—x° +ylny-y=0 y = 0 of Problem 22 07. k= 10 29, x22 _ 5. 35. y=1 Lk= . xy" COSxX =C a , ; , 31. xy? +22 =c 33. 3x°yi + yt=c UO 37. y= 1 + 96 tan (jp) 35. —2ye* + Ree +x=c e 41. (a) y= -—Vx Fx - 1 (c) (—20, -1— 1/5) 37, e (x2 + 4) = 20 n _ 2) 2 = 49, y(x) = (4h/L yx +a 39. (c) y (x) = 2 — Ve =P +4 a yo(x) = — x? + Viet +4 ° EXERCISES 2.3 (PAGE 60) 9 = L. y = ce, (—2, 20) 45. (a) v(x) = 8, 5 -~ 5 (b) 12.7 ftls x = 3. y= ier + ce *, (—%, ©); ce * is transient Fa 5. y= j + ce, (—©, %); ce is transient S 4 i . ; EXERCISES 2.5 (PAGE 74) Ss 7. y=x ‘Inx + cx‘, (0, ©); solution is transient Z 9. y=cx — x cos x, (0, ©) 1. y + xIn|x| = cx 2 1. y =5x — ix + cx 4, 0, ©); cx ‘4 is transient 3. (x — y)Injx — y| = y + c(x — y) oO 13. y= tx 7e" + cx ?e™, (0, ©); cx 7e” is transient 5. x + yIn|x| = cy QO 15. x= 2y6 + cy4, (0, ) 7. In(x? + y?) +2 tan”'(y/x) =e , , 5 17. y = sinx + ccos x, (—7/2, p /2) 9. 4x = y(in|y| —c) 11. y + 3x In|x| = 8x Wy 19. (x + De*y = x? +c, (—1, %); solution is transient 13. In|x| = e ‘ 1 15. y a 1+ cx n 21. (sec 6 + tanu)r = 0 — cos@ +c, (—p /2, 7/2) 17. y% =x+3 + ce* 19, e = ct oc 23. y= e>* + cx~!e73*, (0, ©); solution is transient 21. y3= —2y-1 4 9 4-6 Oo 25. y=x le*+ (2-e)x!, (0, %) ° ° an 23. y= —x—1 +4 tan(x + c) ce 27.i- E +4 (i _ few , (—%, ©) 25. 2y —2x+sin2a¢+ y)=c = R R 27. 4(y — 2x + 3) = (x + c)? 3 29, (x + Ly =xInx — x + 21, (0, ~) 29. —cot(x + y) t+ esc(x + y) =x + V2-1 2 i. lq -e%), 0<x<3 35. (b) y= = + (—1x + ex)! 31. y= 41 « 5 x x(e° — le, x>3 1 3 -¥ 33. y = ( F296", Osx! EXERCISES 2.6 (PAGE 79) ° i 3),-" yxySl (re + se", 1. y> = 2.9800, ya = 3.1151 35. y = 2x — 1+ 4e7*, O<x=1 > yO on v0 ee y=e 7) |4a2inx + (1 + 4e2)x2, x>1 7 y ~ 0.5630 - ~ 0.5565 37. y =e"! + } Vire® (erf(x) — erf(1)) 9. ys = 1.2194, vio = 1.2696 13. Euler: yjo = 3.8191, yoo = 5.9363 47. E(t) = Eye V/RC RK4: yo = 42.9931, y29 = 84.0132