INSTRUCTOR S SOLUTIONS MANUAL SECTION 17.1 (PAGE 902)

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1 INSTRUCTOR S SOLUTIONS MANUAL SECTION 7 PAGE 90 CHAPTER 7 ORDINARY DIFFEREN- TIAL EQUATIONS Section 7 Classifing Differential Equations page 90 = 5: st order, linear, homogeneous d d + = : nd order, linear, nonhomogeneous d 3 = : st order, nonlinear d 4 + = sin : 3rd order, linear, nonhomogeneous 5 + sin = : nd order, linear, homogeneous = : nd order, nonlinear 7 d 3 dt 3 + t dt + t = t 3 : 3rd order, linear, nonhomogeneous 8 cos d + sin t = 0: st order, nonlinear, homogeneous dt e = 3 : 4th order, linear, homogeneous 0 + e = : nd order, nonlinear If = cos, then + = cos + cos = 0 If = sin, then + = sin + sin = 0 Thus = cos and = sin are both solutions of + = 0 This DE is linear and homogeneous, so an function of the form = A cos + B sin, where A and B are constants, is a solution also Therefore sin cos is a solution A =, B =, and sin + 3 = sin 3 cos + cos 3 sin is a solution, but sin is not since it cannot be represented in the form A cos + B sin If = e, then = e e = 0; if = e, then = e e = 0 Thus e and e are both solutions of = 0 Since = 0 is linear and homogeneous, an function of the form = Ae + Be is also a solution Thus cosh = e + e is a solution, but neither cos nor e is a solution 3 Given that = cosk is a solution of + k = 0, we suspect that = sink is also a solution This is easil verified since Since the DE is linear and homogeneous, = A + B = A cosk + B sink is a solution for an constants A and B It will satisf 3 = π/k = A cosπ + B sinπ = A 3 = π/k = Ak sinπ + Bk cosπ = Bk, provided A = 3 and B = 3/k The required solution is = 3cosk 3 k sink 4 Given that = e k is a solution of k = 0, we suspect that = e k is also a solution This is easil verified since k = k e k k e k = 0 Since the DE is linear and homogeneous, = A + B = Ae k + Be k is a solution for an constants A and B It will satisf 0 = = Ae k + Be k = = Ake k Bke k, provided A = e k /k and B = e k /k The required solution is = k ek k e k 5 B Eercise, = A cos + B sin is a solution of + = 0 for an choice of the constants A and B This solution will satisf 0 = π/ 0 = B A, 3 = π/4 = A + B, provided A = and B = The required solution is = cos + sin + k = k sink + k sink = 0 67

2 SECTION 7 PAGE 90 R A ADAMS: CALCULUS 6 = e r is a solution of the equation = 0if r e r re r e r = 0, that is, if r r = 0 This quadratic has two roots, r =, and r = Since the DE is linear and homogeneous, the function = Ae + Be is a solution for an constants A and B This solution satisfies = 0 = A + B, = 0 = A B, provided A = and B = 0 Thus, the required solution is = e 7 If = =, then = and = 0 Thus + = 0 + = B Eercise we know that = A cos + B sin satisfies the homogeneous DE + = 0 Therefore, b Theorem, = + = + A cos + B sin is a solution of + = This solution satisfies = π = π A, 0 = π = B, provided A = π and B = Thus the required solution is = + π cos + sin 8 If = = e, then = 0 and = 0 Thus = 0 + e = e B Eercise we know that = Ae + Be satisfies the homogeneous DE = 0 Therefore, b Theorem, = + = e + Ae + Be is a solution of = e This solution satisfies 3 d = + Let = v v + dv d = v + v dv d = v v3 v = + v + v + v v 3 dv = d v + ln v = ln +C + ln =C 4 ln =C d = + + v + dv d = + v + v dv d + v = tan v = ln +C = tan ln +C = tan ln +C Let = v 0 = = Ae + B e e, = = Ae B e, provided A = e + /e and B = ee / Thus the required solution is = e+ e+e + e e Section 7 page 907 d = + Solving First-Order Equations Let = v v + dv + v = d v dv d = + v v v = + v v v + v dv = d tan v ln + v = ln +C tan / ln + = ln +C tan / ln + = C 4 d = Let = v v + dv d = 3 + 3v 3 3v + v 3 dv d = + 3v 3v + v 3 v = v4 v3 + v 3 + v v dv d v 4 = Let u = v du = v dv 3 + u u du = ln +C 3 4 ln u + u 4 ln u =ln +C 3ln + ln = 4ln +C + 3 ln 4 4 = C ln + 4 = C + = C 68

3 INSTRUCTOR S SOLUTIONS MANUAL SECTION 7 PAGE d = + cos let = v v + dv d = v + cos v dv d = cos v sec v dv = d tan v = ln +ln C tan = ln C 6 = tan ln C d = e / v + dv d = v e v e v dv = d e v = ln +ln C e / = ln C = ln ln C let =v 7 We require d = + Thus 8 + = d = + C Since, 3 lies on the curve, = 4 + C Thus C = 8 and = 8, or = 4 d = + v + dv d = + v dv d = + v dv d + v = ln + v =ln +C Let = v + = C + = C Since, 3 lies on the curve, 4 = C Thus the curve has equation + = 4 9 If ξ = 0, η = 0, and d = a + b + c e + f+ g, then dη dξ = d = aξ bη c eξ f η g = aξ + bη + a 0 + b 0 + c eξ + f η + e 0 + f 0 + g aξ + bη = eξ + f η provided 0 and 0 are chosen such that a 0 + b 0 + c = 0, and e 0 + f 0 + g = 0 0 The sstem = 0, = 0 has solution 0 =, 0 = Thus, if ξ = and η =, where then dη dξ = ξ + η ξ η d = + 4 3, Let η = vξ v + ξ dv dξ = + v v ξ dv dξ = + v v v = + v v v dξ + v dv = ξ tan v ln + v = ln ξ +C 4 tan η ξ lnξ + η = C Hence the solution of the original equation is 4 tan ln + = C + d + + = 0 d + = 0 + = C e sin + d + e cos + = 0 de sin + + = 0 e sin + + = C 3 e + d + e = 0 d e = 0 e = C 4 + d + = 0 d + + = = C 69

4 SECTION 7 PAGE 907 R A ADAMS: CALCULUS 5 + d = 0 M = +, N = M N N = 3 indep of dµ µ = 3 d µ = d = 0 d ln = 0 ln = C = ln +C 6 e + ln + d + + ln + sin = 0 M = e + ln +, N = + ln + sin M = +, N = + ln + + sin M N N = N ln sin = dµ µ = d µ = d + e + ln + + ln + sin d e + ln + ln cos = 0 e + ln + ln cos = C 7 If µm, d + µn, is eact, then µm, = µn, µ M + µ M = µ N µ µ = M N M Thus M and N must be such that depends onl on N M M 8 + d = µ d µ = 0 M = µ µ N = + 63 µ For eactness we require + 4 µ = [ ]µ + 3 µ = + 3 µ µ = µ µ = + 3 d = 0 d + 3 = = C 9 Consider d + 3 e = 0 Here M =, N = 3 e, M Thus µ =, and N = µ = 3 µ = 3 d 3 + e = 0 d e = 0 e = C, or e = C 0 If µ is an integrating factor for Md+ N = 0, then µm = µn, µ M + µ M = µ N + µ N Thus M and N will have to be such that the right-hand side of the equation µ µ = N M N M depends onl on the product For cos + sin d or + we have M = cos +, sin N = M =, N = sin cos N M sin = + cos + M N = cos + + sin + M N N M = 630

5 INSTRUCTOR S SOLUTIONS MANUAL SECTION 73 PAGE 95 Thus, an integrating factor is given b µ t µt = t µt = t We multipl the original equation b / to make it eact: cos + sin d + = 0 sin d = 0 sin = C The solution is sin = C Section 73 Eistence, Uniqueness, and Numerical Methods page 95 A computer spreadsheet was used in Eercises The intermediate results appearing in the spreadsheet are not shown in these solutions We start with 0 =, 0 = 0, and calculate n+ = n + h, n+ = n + h n + n a For h = 0 we get 5 =, 5 = b For h = 0 we get 0 =, 0 = c For h = 005 we get 0 =, 0 = We start with 0 =, 0 = 0, and calculate n+ = n + h, u n+ = n + h n + n n+ = n + h n + n + n+ + u n+ a For h = 0 we get 5 =, 5 = b For h = 0 we get 0 =, 0 = 486 c For h = 005 we get 0 =, 0 = We start with 0 =, 0 = 0, and calculate n+ = n + h p n = n + n q n = n + h + n + h p n r n = n + h + n + h q n q n = n + h + n + hr n n+ = n + h 6 p n + q n + r n + s n a For h = 0 we get 5 =, 5 = b For h = 0 we get 0 =, 0 = c For h = 005 we get 0 =, 0 = We start with 0 = 0, 0 = 0, and calculate n+ = n + h, n+ = h n e n a For h = 0 we get 0 =, 0 = b For h = 0 we get 0 =, 0 = We start with 0 = 0, 0 = 0, and calculate n+ = n + h, u n+ = n + h n e n n+ = n + h ne n + n+ e u n+ a For h = 0 we get 0 =, 0 = b For h = 0 we get 0 =, 0 = We start with 0 = 0, 0 = 0, and calculate n+ = n + h p n = n e n q n = n + h e n+h/ p n r n = n + h e n+h/q n s n = n + he n+hr n n+ = n + h 6 p n + q n + r n + s n a For h = 0 we get 0 =, 0 = b For h = 0 we get 0 =, 0 = We start with 0 = 0, 0 = 0, and calculate n+ = n + h, n+ = n + h cos n a For h = 0 we get 5 =, 5 = b For h = 0 we get 0 =, 0 = c For h = 005 we get 0 =, 0 = We start with 0 = 0, 0 = 0, and calculate n+ = n + h, u n+ = n + h cos n n+ = n + h cos n + cos u n+ a For h = 0 we get 5 =, 5 = 0868 b For h = 0 we get 0 =, 0 = c For h = 005 we get 0 =, 0 =

6 SECTION 73 PAGE 95 R A ADAMS: CALCULUS 9 We start with 0 = 0, 0 = 0, and calculate n+ = n + h p n = cos n q n = cos n + h/p n r n = cos n + h/q n q n = cos n + hr n n+ = n + h 6 p n + q n + r n + s n a For h = 0 we get 5 =, 5 = b For h = 0 we get 0 =, 0 = c For h = 005 we get 0 =, 0 = We start with 0 = 0, 0 = 0, and calculate n+ = n + h, n+ = n + h cos n a For h = 0 we get 5 =, 5 = b For h = 0 we get 0 =, 0 = c For h = 005 we get 0 =, 0 = We start with 0 = 0, 0 = 0, and calculate n+ = n + h, u n+ = n + h cos n n+ = n + h cos n + cos n+ a For h = 0 we get 5 =, 5 = b For h = 0 we get 0 =, 0 = 0903 c For h = 005 we get 0 =, 0 = We start with 0 = 0, 0 = 0, and calculate n+ = n + h p n = cos n q n = cos n + h/ r n = cos n + h/ q n = cos n + h n+ = n + h 6 p n + q n + r n + s n a For h = 0 we get 5 =, 5 = b For h = 0 we get 0 =, 0 = c For h = 005 we get 0 =, 0 = = + t dt, d = = + 0 = = d = + C = + C C = 3 = 3/ = 3 4 u = + 3 t ut dt du d = 3 u, u = + 0 = du u = 3 d ln u = 3 + C 0 = ln = ln u = 3 + C C = 8 u = e For the problem = f, a = 0, the -step Runge- Kutta method with h = b a gives: 0 = a, 0 = 0, = 0 + h = b p 0 = f a, q 0 = f a + h a + b = f = r 0 s 0 = f a + h = f b = 0 + h 6 p 0 + q 0 + r 0 + s 0 = b a a + b f a + 4 f + f b, 6 which is the Simpson s Rule approimation to b a f d based on subintervals of length h/ 6 If φ0 = A 0 and φ kφ on an interval [0, X], where k > 0 and X > 0, then d d φ e k = ek φ ke k φ e k 0 Thus φ/e k is increasing on [0, X] Since its value at = 0isφ0 = A 0, therefore φ/e k A on [0, X], and φ Ae k there 7 a Suppose u = u, = +, and v = + v on [0, X], where u0 = 0 = v0 =, and X > 0 is such that v is defined on [0, X] In part b below, we will show that X <, and we assume this fact now Since all three functions are increasing on [0, X], we have u,, and v on [0, X] 63

7 INSTRUCTOR S SOLUTIONS MANUAL SECTION 74 PAGE 99 If φ = u, then φ0 = 0 and φ = + u u + u u φ on [0, X] B Eercise 6, φ 0on[0, X], and so u there Similarl, since X <, if φ = v, then φ0 = 0 and φ = + v v v + v φ on [0, X], so v there b The IVP u = u, u0 = has solution u =, obtained b separation of variables This solution is valid for < The IVP v = + v, v0 = has solution v = tan + π 4, also obtained b separation of variables It is valid onl for 3π/4 < <π/4 Observe that π/4 <, proving the assertion made about v in part a B the result of part a, the solution of the IVP = +, 0 =, increases on an interval [0, X] and as X from the left, where X is some number in the interval [π/4, ] c Here are some approimations to for values of near 09 obtained b the Runge-Kutta method with 0 = 0 and 0 = : For h = 005 n = 7 n = 085 n = 3739 n = 8 n = 090 n = n = 9 n = 095 n = For h = 00 n = 43 n = 086 n = n = 44 n = 088 n = n = 45 n = 090 n = n = 46 n = 09 n = n = 47 n = 094 n = For h = 00 n = 86 n = 086 n = n = 87 n = 087 n = n = 88 n = 088 n = n = 89 n = 089 n = n = 90 n = 090 n = n = 9 n = 09 n = n = 9 n = 09 n = n = 93 n = 093 n = n = 94 n = 094 n = The values are still in reasonable agreement at = 09, but the start to diverge quickl thereafter This suggests that X is slightl greater than 09 Section 74 Differential Equations of Second Order page 99 If = e, then 3 + = e 3 + = 0, so is a solution of the DE 3 + = 0 Let = e v Then = e v + v, = e v + v + v 3 + = e v + v + v 3v 3v + v = e v v satisfies 3 + = 0 provided w = v satisfies w w = 0 This equation has solution v = w = C e, so v = C e + C Thus the given DE has solution = e v = C e + C e If = e, then 6 = e = 0, so is a solution of the DE 6 = 0 Let = e v Then = e v v, = e v 4v + 4v 6 = e v 4v + 4v v + v 6v = e v 5v satisfies 6 = 0 provided w = v satisfies w 5w = 0 This equation has solution v = w = C /5e 5,sov = C e 5 + C Thus the given DE has solution = e v = C e 3 + C e 3 If = on 0,, then + = 0 + = 0, so is a solution of the DE + = 0 Let = v Then = v + v, = v + v + = 3 v + v + v + v v = v + 4v satisfies + = 0 provided w = v satisfies w + 4w = 0 This equation has solution v = w = 3C 4 obtained b separation of variables, so v = C 3 + C Thus the given DE has solution = v = C + C 633

8 SECTION 74 PAGE 99 R A ADAMS: CALCULUS 4 If = on 0,, then = = 0, so is a solution of the DE = 0 Let = v Then Therefore = / cos is a solution of the Bessel equation + + = 0 4 Now let = / cos v Then = v + v, = v + 4v + v = 4 v v + v 3 3 v 6 v + 4 v = 3 v + v satisfies = 0 provided w = v satisfies w + w = 0 This equation has solution v = w = C / obtained b separation of variables, so v = C ln + C Thus the given DE has solution = v = C ln + C 5 If =, then = and = 0 Thus = 0 Now let = v Then = v + v, = v + v Substituting these epressions into the differential equation we get v + 3 v v v 3 v v + v + v = 0 3 v 3 v = 0, or v v = 0, which has solution v = C + C e Hence the general solution of the given differential equation is 6 If = / cos, then Thus = C + C e = 3/ cos / sin = 3 4 5/ cos + 3/ sin / cos = 3 4 / cos + / sin 3/ cos = 0 / cos / sin + 3/ cos 4 / cos = 3/ cos v / sin v + / cos v = 3 4 5/ cos v + 3/ sin v 3/ cos v / cos v / sin v + / cos v If we substitute these epressions into the equation, man terms cancel out and we are left with the equation cos v sin v = 0 Substituting u = v, we rewrite this equation in the form cos du = sin u d du u = tan d ln u =ln sec +C 0 Thus v = u = C sec, from which we obtain v = C tan + C Thus the general solution of the Bessel equation is = / cos v = C / sin + C / cos 7 If = and = where satisfies + a + a 0 = f, then = and = a 0 a + f Thus d 0 0 = + d a 0 a f 8 If satisfies n + a n n + +a + a 0 = f, then let =, =, 3 =, n = n Therefore =, = 3, n = n, and n = a 0 a a 3 a n n + f, 634

9 INSTRUCTOR S SOLUTIONS MANUAL SECTION 75 PAGE 93 and we have d d n + = f 9 If = C e λ v, then a 0 a a a n n Section 75 Linear Differential Equations with Constant Coefficients page = 0 Auiliar: r 3 4r + 3r = 0 rr r 3 = 0 r = 0,, 3 General solution: = C + C e t + C 3 e 3t 4 + = 0 Auiliar: r 4 r + = 0 r = 0 r =,,, General solution: = C e t + C te t + C 3 e t + C 4 te t 0 = C λe λ v = C e λ Av = A provided λ and v satisf Av = λv λ 3 λ = 6 5λ + λ = λ 5λ + 4 = λ λ 4 = 0 if λ = orλ = 4 Let A = 3 If λ = and Av = v, then v A = = 3 v v v v + v = 0 Thus we ma take v = v = If λ = 4 and Av = 4v, then v v A = = 4 v 3 v = 0 v Thus we ma take v = v = v B the result of Eercise 9, = e v and = e 4 v are solutions of the homogeneous linear sstem = A Therefore the general solution of the sstem is that is = C e v + C e 4 v, = C e = C e + C e 4 = C e + C e 4 + C e 4, or = 0 Auiliar: r 4 + r + = 0 r + = 0 r = i, i, i, i General solution: = C cos t + C sin t + C 3 t cos t + C 4 t sin t = 0 Auiliar: r 4 + 4r 3 + 6r + 4r + = 0 r + 4 = 0 r =,,, General solution: = e t C + C t + C 3 t + C 4 t 3 5 If = e t, then 4 = e t = 0 The auiliar equation for the DE is r 3 r 4 = 0, for which we alrea know that r = is a root Dividing the left side b r, we obtain the quotient r + r + Hence the other two auiliar roots are ± i General solution: = C e t + C e t cos t + C 3 e t sin t 6 Au eqn: r r r 4 = 0 r + r r r + = 0 r =,,,,,,, The general solution is = e t C + C t + C 3 t + C 4 t 3 + e t C 5 + C 6 t + e t C 7 + C 8 t 7 + = 0 au: rr r + = 0 r r + = 0 r = 0, r =, Thus = A + B ln 8 3 = 0 rr r 3 = 0 r r 3 = 0 r 3r + = 0 r = and r = 3 Thus, = A + B 3 635

10 SECTION 75 PAGE 93 R A ADAMS: CALCULUS 9 + = 0 au: rr + r = 0 r =± = A + B 0 Consider + 5 = 0 Since a =, b =, and c = 5, therefore b a < 4ac Then k = a b/a = and ω = 4 Thus, the general solution is = A cosln + B sinln + = 0 au: rr + r = 0 r = 0, 0 Thus = A + B ln Given that + + = 0 Since a =, b =, c = therefore b a < 4ac Then k = a b/a = 0 and ω = Thus, the general solution is = A cosln + B sinln = 0 Tring = r leads to the auiliar equation rr r + r = 0 r 3 3r + 3r = 0 r 3 = 0 r =,, Thus = is a solution To find the general solution, tr = v Then = v + v, = v + v, = v + 3v Now 3 + = 4 v v + v + v v = v + 3v + v, and is a solution of the given equation if v = w is a solution of w + 3w + w = 0 This equation has auiliar equation rr + 3r + = 0, that is r + = 0, so its solutions are v = w = C + C 3 ln v = C + C ln + C 3 ln The general solution of the given equation is, therefore, = C + C ln + C 3 ln For a particular solution p of the given equation tr = A This satisfies the given equation if A = / Thus the general solution of the given equation is = + C e + C e + = The complementar function is h = C e + C e,as shown in Eercise For a particular solution tr = A + B Then = A and = 0, so satisfies the given equation if = A A + B = A B A We require A B = 0 and A =, so A = / and B = /4 The general solution of the given equation is = C e + C e 3 + = e The complementar function is h = C e + C e,as shown in Eercise For a particular solution tr = Ae Then = Ae and = Ae,so satisfies the given equation if e = e A A A = Ae We require A = / The general solution of the given equation is = e + C e + C e 4 + = e The complementar function is h = C e + C e,as shown in Eercise For a particular solution tr = Ae Then = Ae +, = Ae +, Section 76 Nonhomogeneous Linear Equations page 99 + = The auiliar equation for + = 0is r + r = 0, which has roots r = and r = Thus the complementar function is h = C e + C e so satisfies the given equation if e = Ae = 3Ae We require A = /3 The general solution of the given equation is = 3 e + C e + C 3 e 636

11 INSTRUCTOR S SOLUTIONS MANUAL SECTION 76 PAGE = The homogeneous equation has auiliar equation r + r + 5 = 0 with roots r = ± i Thus the complementar function is h = C e cos + C e sin For a particular solution, tr = A + B + C Then = A + B and = A We have = = A + 4A + B + 5A + 5B + 5C Thus we require 5A =, 4A + 5B = 0, and A + B + 5C = 0 This gives A = /5, B = 4/5, and C = /5 The given equation has general solution = e C cos + C sin = e The homogeneous equation has auiliar equation r + 4r + 4 = 0 with roots r =, Thus the complementar function is h = C e + C e For a particular solution, tr = A e Then = e A A and = e A 8A+4A We have e = = e A 8A + 4A + 8A 8A + 4A = Ae Thus we require A = / The given equation has general solution = e + C + C = The complementar function is = C cos + C sin For the given equation, tr = A + B + C Then = + 4 = A + 4A + 4B + 4C Thus A + 4C = 0, 4A =, 4B = 0, and we have A = 4, B = 0, and C = The given equation has 8 general solution = C cos + C sin 7 6 = e The homogeneous equation has auiliar equation r r 6 = 0 with roots r = and r = 3 Thus the complementar function is h = C e + C e 3 For a particular solution, tr = Ae Then = e A A and = e 4A + 4A We have e = 6 = e 4A + 4A A + A 6A = 5Ae Thus we require A = /5 The given equation has general solution = 5 e + C e + C e = e sin The homogeneous equation has auiliar equation r + r + = 0 with roots r = ± i Thus the complementar function is h = C e cos + C e sin For a particular solution, tr = Ae cos + Be sin Then = A + Be cos + B Ae sin = Be cos Ae sin This satisfies the nonhomogeneous DE if e sin = + + = e cos B + A + B + A + e sin A + B A + B = e cos 4A + 4B + e sin 4B 4A Thus we require A + B = 0 and 4B A =, that is, B = A = /8 The given equation has general solution = e 8 sin cos + e C cos + C sin = e sin The complementar function is the same as in Eercise 9, but for a particular solution we tr = Ae cos + Be sin = e cos A A + B + e sin B B A = e cos B B A + e sin A A B 637

12 SECTION 76 PAGE 99 R A ADAMS: CALCULUS This satisfies the nonhomogeneous DE if e sin = + + = Be cos Ae sin Thus we require B = 0 and A = / The given equation has general solution = e cos + e C cos + C sin + = e The homogeneous equation has auiliar equation r + r = 0 with roots r = 0 and r = Thus the complementar function is h = C + C e For a particular solution, tr = A + B + Ce Then = A + B + e C C = B + e C + C This satisfies the nonhomogeneous DE if e = + = A + B + B Ce Thus we require A + B = 4, B =, and C =, that is, A =, B =, C = The given equation has general solution = + e + C + C e + + = e The homogeneous equation has auiliar equation r + r + = 0 with roots r = and r = Thus the complementar function is h = C e + C e For a particular solution, tr = e A + B 3 Then = e A + 3B A B 3 = e A + 6B 4A 6B A + B 3 This satisfies the nonhomogeneous DE if e = + + = e A + 6B Thus we require A = 0 and B = /6 The given equation has general solution = 6 3 e + C e + C e 3 + = e The complementar function is h = C e + C e For a particular solution use p = e u + e u, where the coefficients u and u satisf Thus e u + e u = e u = 3 e u = 3 e Thus p = 3 e 6 e e u + e u = 0 solution of the given equation is u = 3 e u = 6 e = e The general = e + C e + C e 4 + = e The complementar function is h = C e + C e For a particular solution use p = e u + e u, where the coefficients u and u satisf Thus e u + e u = e e u + e u = 0 u = 3 e3 u = 9 e3 u = 3 u = 3 Thus p = 9 e + 3 e The general solution of the given equation is = 9 e + 3 e + C e + C e = 3 e + C e + C 3 e 5 + = If = A, then = A and = A Thus = + = A + A A = 3A, 638

13 INSTRUCTOR S SOLUTIONS MANUAL SECTION 76 PAGE 99 so A = /3 A particular solution of the given equation is = /3 The auiliar equation for the homogeneous equation + = 0is4rr + r = 0, or r = 0, which has solutions r =± Thus the general solution of the given equation is = 3 + C + C 6 + = r has a solution of the form = A r provided r = ± If this is the case, then r = A r rr + r = A r r Thus A = /r and a particular solution of the DE is = r r 7 + = Tr = A ln Then = Aln + and = A/ We have = A + Aln + A ln = A Thus A = / The complementar function was obtained in Eercise 5 The given equation has general solution = ln + C + C 8 + = Tr = u + u, where u and u satisf u + u = 0, u u = Solving these equations for u and u, we get u =, u = Thus u = ln and u = A particular solution is 4 = ln 4 The term /4 can be absorbed into the term C in the complementar function, so the general solution is = ln + C + C = 3 Since and e are independent solutions of the corresponding homogeneous equation, we can write a solution of the given equation in the form = u + e u, where u and u are chosen to satisf u + e u = 0, u + + e u = Solving these equations for u and u, we get u = and u = e Thus u = and u = e The particular solution is = Since is a solution of the homogeneous equation, we can absorb that term into the complementar function and write the general solution of the given DE as = + C + C e = 3/ 4 A particular solution can be obtained in the form = / cos u + / sin u, where u and u satisf / cos u + / sin u = 0 3/ cos / sin u 3/ sin / cos u = / We can simplif these equations b dividing the first b /, and adding the first to times the second, then dividing the result b / The resulting equations are cos u + sin u = 0 sin u + cos u =, which have solutions u = sin, u = cos, so that u = cos and u = sin Thus a particular solution of the given equation is = / cos + / sin = / The general solution is = / + C cos + C sin 639

14 SECTION 76 PAGE 99 R A ADAMS: CALCULUS Section 77 Series Solutions of Differential Equations page 933 = Tr = = = a n n nn a n n n= n + n + a n+ n 0 = = n + n + a n+ n a n n+ = n + n + a n+ n a n n n= = a + 6a 3 ] + [n + n + a n+ a n n n= a n Thus a = a 3 = 0, and a n+ = for n n + n + Given a 0 and a we have a 4 = a a 8 = a 4 a 4n = = 7 8 = a a n 4n a 0 4 n n! 3 7 4n a 5 = a 4 5 a 9 = a 5 a 4n+ = 8 9 = a n4n + a = 4 n n! 5 9 4n + a 4n+3 = a 4n+ = =a 3 = a = 0 The solution is = a a + a 4n 4 n n! 3 7 4n 4n+ 4 n n! 5 9 4n + = Tr a n n Then = na n n = na n n n= = nn a n n = n + n + a n+ n Thus we have 0 = = n + n + a n+ n a n n+ = n + n + a n+ n a n n = a + ] [n + n + a n+ a n n a n Thus a = 0 and a n+ = for n n + n + Given a 0 and a,wehave a 3 = a 0 3 a 6 = a = a = 4 a 0 6! a 9 = a = 4 7 a 0 9! a 3n = 4 3n a 0 3n! a 4 = a 3 4 = a 4! a 7 = a = 5 a 7! a 3n+ = 5 3n a 3n +! 0 = a = a 5 = a 8 = =a 3n+ Thus the general solution of the given equation is = a a 4 3n 3n 3n! 5 3n 3n+ 3n +! 640

15 INSTRUCTOR S SOLUTIONS MANUAL SECTION 77 PAGE Let + + = 0 0 = 0 = = a n n = na n n n= = nn a n n = n + n + a n+ n Substituting these epressions into the differential equation, we get n + n + a n+ n + na n n + a n n = 0, a + + so [n + n + a n+ + n + a n ] n = 0 It follows that a =, a n+ = a n, n =,, 3, n + Since a 0 = 0 =, and a = 0 =, we have Thus, 0 = + + = n + n + a n+ n + na n n + a n n = a + a 0 + ] [n + n + a n+ + n + a n n Since coefficients of all powers of must vanish, therefore a + a 0 = 0 and, for n, that is, n + n + a n+ + n + a n = 0, a n+ = a n n + If 0 =, then a 0 =, a =, a 4 =!, a 6 = 3 3!, a 8 = 4 4!, If 0 = 0, then a = a 3 = a 5 = = 0 Hence, = = 5 + sin = 0, 0 =, 0 = 0 Tr n n n! n = a 0 + a + a + a a a a 0 = a = a 4 = 3 a 6 = 3 5 a 8 = The patterns here are obvious: a = n a n = 3 5 n = n n n! n! Thus = n [ n n! n 4 If = a 3 = a 5 = 4 a 7 = 4 6 a 9 = n! a n+ = n n n! ] + n+ n n! a n n, then = na n n and = nn a n n = n + n + a n+ n n= Then a 0 = and a = 0 We have = a + 6a 3 + a 4 + 0a sin = a + a a a = + a 3 + a a 4 6 a Hence we must have a = 0, 6a 3 + = 0, a 4 = 0, 0a 5 + a 6 = 0, That is, a = 0, a 4 = 0, a 3 = 6, a 5 = The solution is 0 = = 0, 0 = 0, 0 = Tr = a n n 64

16 SECTION 77 PAGE 933 R A ADAMS: CALCULUS Then a 0 = 0 and a = We have = = na n n nn a n n n= 0 = + 9 = n + n + a n+ n nn a n n na n n + 9 a n n n= = a + 9a 0 + 6a 3 + 8a ] + [n + n + a n+ n 9a n n n= Thus a + 9a 0 = 0, 6a 3 + 8a = 0, and Therefore we have a n+ = n 9a n n + n + a = a 4 = a 6 = =0 a 3 = 4 3, a 5 = 0 = a 7 = a 9 = The initial-value problem has solution = = 0 Since = 0 is a regular singular point of this equation, tr = a n n+µ a 0 = = = Then we have n + µa n n+µ n + µn + µ a n n+µ 0 = [ ] = 3n + µ n + µ a n n+µ + a n n+µ = 3µ µ µ [ ] + 3n + µ n + µ a n + a n n+µ a n Thus 3µ µ = 0 and a n = 3n + µ n + µ for n There are two cases: µ = 0 and µ = /3 a n CASE I µ = 0 Then a n = n3n Since a 0 = we have a =, a = 5 a 3 = a n = One series solution is = n n! 5 3n CASE II µ = 3 Then n n n! 5 3n a n a n = 3 n + = a n 3 n + n3n + 3 Since a 0 = wehave a = 4, a = 4 7 a 3 = a n = n n! 4 7 3n + A second series solution is = /3 n n + n! 4 7 3n = 0 Since = 0 is a regular singular point of this equation, tr = a n n+µ a 0 = = = n + µa n n+µ n + µn + µ a n n+µ 64

17 INSTRUCTOR S SOLUTIONS MANUAL REVIEW EXERCISES 7 PAGE Then we have 0 = + + [ ] = n + µn + µ + n + µ a n n+µ + a n n+µ+ = n + µ a n n+µ + a n n+µ n= = µ µ + + µ a µ ] + [n + µ a n + a n n+µ n= Thus µ = 0, a = 0, and a n = a n n for n It follows that 0 = a = a 3 = a 5 =, and, since a 0 =, One series solution is a =, a 4 = 4, n a n = 4 n = n n n! = + n n n n! Review Eercises 7 page 934 d = = d ln = + C = Ce d = e sin e = sin d e = cos + C = lnc cos d = + d = d d e = e d = e e = e d = e 4 e + C = 4 + Ce d = + v + dv d = + v v dv d = + v v v dv v = d let = v v = v v lnv = ln + ln C = ln C = C = C d = + + d + = 0 eact d + = 0 + = C + e = d + e + e d + + e = 0 eact d + e + e = 0 + e + e = C d dt = dp dt let p = /dt dt = p dp p = dt p = C t dt = p = C t = dt C t = ln t C +C 8 d dt + 5 dt + = 0 Au: r + 5r + = 0 r = /, = C e t/ + C e t = 0 Au: 4r 4r + 5 = 0 r + 4 = 0 r = ± i = C e / cos + C e / sin 0 + = 0 Au: rr + = 0 r r + = 0 r = ± i = C / cos ln + C / sin ln 643

18 REVIEW EXERCISES 7 PAGE 934 R A ADAMS: CALCULUS t d dt t dt + 5 = 0 Au: rr r + 5 = 0 3 r + 4 = 0 r = ± i = C t cosln t + C t sinln t d 3 dt d + 6 = 0 dt dt Au: r 3 + 8r + 6r = 0 rr + 4 = 0 r = 0, 4, 4 = C + C e 4t + C 3 te 4t d d 5 d + 6 = e + e 3 Au: r 5r + 6 = 0 r =, 3 Complementar function: = C e + C e 3 Particular solution: = Ae + Be 3 = Ae + B + 3e 3 = Ae + B6 + 9e 3 Thus A =, B = 4, C = 6 The general solution is = C e + C e 6 d d = 3 The corresponding homogeneous equation has auiliar equation rr = 0, with roots r = and r =, so the complementar function is = C + C / A particular solution of the nonhomogeneous equation can have the form = A 3 Substituting this into the DE gives 6A 3 A 3 = 3, so that A = /4 The general solution is = C + C 4 e + e 3 = Ae Be = Ae + Be 3 Thus A = / and B = The general solution is = e + e 3 + C e + C e 3 d d = e d Same complementar function as in Eercise 3: C e + C e 3 For a particular solution we tr = A + Be Substituting this into the given DE leads to e = A Be Ae, d =, = = d 3 = 3 + C = 8 + C C = 7 3 = 3 7 = 3 7 /3 d =, = = d = C = + C C = = + = + d =, 0 = Let = v Then + 5 so that we need A = / and B = A = The general solution is = + e + C e + C e 3 d d + d + = Au: r + r + = 0 has solutions r =, Complementar function: = C e + C e Particular solution: tr = A + B + C Then v + dv d = v + v dv d = v v3 v = + v + v + v v 3 dv = d ln v =ln +ln C v = v = lncv = lnc C = e /, 0 = C = = A + A + B + A + B + C = e /, or = e / 644

19 INSTRUCTOR S SOLUTIONS MANUAL REVIEW EXERCISES 7 PAGE cos = cos, π = d d e sin = e sin + cos = cos e sin d d e sin = e sin + C = + Ce sin = + Ce 0 C = = e sin = 0, 0 =, 0 = Au: r + 3r + = 0 r =, = Ae + Be = A + B = Ae Be = A B Thus B = 3, A = 4 The solution is = 4e 3e π = 0, = 0, = π Au: r + r + + π = 0 r = ± πi = Ae cosπ + Be sinπ = e cosπ A + Bπ+ e sinπ B Aπ Thus Ae = 0 and A Bπe = π, so that A = 0 and B = e The solution is = e sinπ = 0, = e 5, = 0 Au: r + 0r + 5 = 0 r = 5, 5 = Ae 5 + Be 5 = 5Ae 5 + B 5e 5 We require e 5 = A + Be 5 and 0 = e 5 5A 4B Thus A + B = and 5A = 4B, so that B = 5 and A = 4 The solution is = 4e 5 + 5e = 0, e = e, e = 0 Au: rr 3r + 4 = 0, or r = 0, so that r =, 5 = A + B ln = A + B ln + B We require e = Ae + Be and 0 = Ae + 3Be Thus A + B = and A = 3B, so that A = 3 and B = The solution is = 3 ln, valid for > 0 d dt + 4 = 8et, 0 =, 0 = Complementar function: = C cost + C sint Particular solution: = Ae t, provided 4A + 4A = 8, that is, A = Thus = e t + C cost + C sint = e t C sint + C cost We require = 0 = + C and = 0 = + C Thus C = 0 and C = The solution is = e t sint 6 d d + 5 d 3 = 6 + 7e/, 0 = 0, 0 = Au: r + 5r 3 = 0 r = /, 3 Complementar function: = C e / + C e 3 Particular solution: = A + Be / = Be / + = Be / + 4 We need Be / A = 6 + 7e / This is satisfied if A = and B = The general solution of the DE is = + e / + C e / + C e 3 Now the initial conditions impl that 0 = 0 = + C + C = 0 = + C 3C, which give C = /7, C = /7 Thus the IVP has solution = + e / + 7 e/ + e 3 7 [ + Ae sin + cos ] d + [e cos + B sin ] = 0 is Md+ N We have M = + Ae cos sin N = e cos + B sin + e cos These epressions are equal and the DE is eact if A = and B = If so, the left side of the DE is dφ,, where φ, = e sin + cos The general solution is e sin + cos = C d + = 0 Multipl b n : n + 3 d + n+ = 0 is eact provided 6 n = n + n, that is, provided n = 5 In this case the left side is dφ, where φ, =

20 REVIEW EXERCISES 7 PAGE 934 R A ADAMS: CALCULUS The general solution of the given DE is = C 9 + cot + + cot = 0 If =, then = and = 0, so the DE is clearl satisfied b To find a second, independent solution, tr = v Then = v + v, and = v + v Substituting these epressions into the given DE, we obtain v + 3 v v + v + cot + v + cot = 0 3 v 3 v cot = 0, or, putting w = v, w = cot w, that is, dw cos d = w sin ln w = ln sin + ln C v = w = C sin v = C C cos A second solution of the DE is cos, and the general solution is = C + C cos 30 + cot + + cot = 3 sin Look for a particular solution of the form = u + cos u, where u + cos u = 0 u + cos sin u = sin Divide the first equation b and subtract from the second equation to get sin u = sin Thus u = and u = The first equation now gives u = cos, so that u = sin The general solution of the DE is = sin cos + C + C cos 3 Suppose = f, and 0 = 0, where f, is continuous on the whole -plane and satisfies f, K there B the Fundamental Theorem of Calculus, we have Therefore, 0 = 0 = 0 t dt = 0 0 K 0 f t, t dt Thus is bounded above and below b the lines = 0 ± K 0, and cannot have a vertical asmptote anwhere Remark: we don t seem to have needed the continuit of f/, onl the continuit of f to enable the use of the Fundamental Theorem 646