equações diferenciais ordinárias, Exercícios de Cálculo

Baixe equações diferenciais ordinárias e outras Exercícios em PDF para Cálculo, somente na Docsity! CHAPTER > fá 4 Higher Order Linear Equations 1. The differential equation is in standard form. Tts function g(t) = t, are continuous everywhere. Hence solutions are valid on the entire real line. oefhicients, as well as the ting the equation im standard form, the coeficients are rational functions with singularities at t = O and t = 1. Hence the solutions are valid on the intervals (-00,0), (0,1), and (1,00). 4. The coefficients are continuous everywhere, but the function g(t) = In t is defined and continuous only on the interval (0,00). Hence solutions are defined for positive reals. 5. Writing the equation in standard form, the coefficients are rational functions with a singularity at zo =1. Furthermore, pa(x) = tan «:/(x — 1) is undefined, and hence not continuous, at xp = (2k + 1)7/2, k = 0,1,2,.... Hence solutions are defined on any interval that does not contain ag or 6. Writing the equation in standard form, the coefhicients are rational functions with singularities at a = +2. Hence the solutions are valid on the intervals (—00, — 2) (-2,2), and (2,00). 137 138 Chapter 4. Higher Order Linear Equations 7. Evaluating the Wronskian of the three functions, W(fi, fo, f3) = —14. Hence the functions are linearly independent. 9. Evaluating the Wronskian of the four functions, W(f1, fo, f3, 1) = 0. Hence the functions are lincarly dependent. To find a linear relation among the functions, we need to find constants cy ca, not all zero, such that cifi(t) + cofo(t) + cafa(t) + cafa(t) = 0. Collecting the common terms, we obtain (o +20 ++ -c+eat+(-Sa+e+e)=0, results in three equations in four unknowi » cam solve the equations c» + 2c3 = 1, 2e1 — ca ca =2/7,0 =13/7, cs = 3/7. Hence 2fi(t) +13fa(t) — 3fa(t) = Tfa(t) = 0. Arbitrarily setting cu = —1, 1, to find that 10. Evaluating the Wronskian of the three functions, W(fi, fo, f3) = 156. Hence the functions are linearly independent. 11. Substitution verifies that the functions are solutions of the ODE. Furthermore, we have W(1,cos t,sin t)=1. 12. Substitution verifies that the functions are solutions of the ODE. Furthermore, we have W(1,t,cos t,sin t)=1. 14. Substitution verifies that the functions are solutions of the ODE. Furthermore, we have W(L,t,e-L,te-!) = et, 15. Substitution verifies that the functions are solutions of the ODE. Furthermore, we have W(1,2,2º) = 6a. 16. Substitution verifies that the functions are solutions of the ODE. Furthermore, we have W(a,42,1/2) = 6/x. 18. The operation of taking a derivative is lincar, and hence (en + ex) = crf + cogoP. Tt follows that (n) (n A Len + cayo] = exi” + caga é gn — Dr pi(erytoD + co DI +... + pa(esm + comp). Rearranging the terms, we obtain L [ey + c»32] = ci fy1] + c2L [5]. Since 3 and 4p are solutions, L [er + ca15] = 0. The rest follows by induction. 19.(a) Note that d*(t")/dt* = n(n = 1)...(n- k+ Dt fork=1,2,...,n. Hence Lt] =aon!+ a [n(n= 1)... t+. anant O ant”. 4.2 141 28. First write the equation in standard form: m t+2 y t+ 1, 6 ut -g- É. moro” Sega" era! Let y(t) = t2v(t). Substitution into the given ODE results in Er at Do +3 Set w =v”. Then w is a solution of the first order differential equation This equation is linear, with integrating fa lution is w = c(t+ 3)/t?. Integrating tw ut) = ct+ ci + cat? + cat?. Finally, si ions, the third independent solution tor u(t) =t!/(t +3). The general so- t HI ret? + cot+ cs. Hem 2 and yo(t) = É +a. are given so- 1. The magnitude of 1 +i is R = 2 and the polar angle is 7/4. Hence the polar form is given by 1+i = 2 ei7/4, 3. The magnitude of —3 is R = 3 and the polar angle is 7. Hence —3 = 3€'7. 3ri/2 4. The magnitude of —i is R = 1 and the polar angle is 37/2. Hence —i = € 5. The magnitude of 3 —iis R=2 and the polar angle is 7/6 = 117/6. Hence the polar form is given by 3. 2ellni/6, R=v2 and the polar angle is 57/4. Hence the 2 eómila, 6. The magnitude of —1 — polar form is given by —1 — where m may be any , ely, we obtain the three Equivalently, the roots H(-1+iv3), cos(47/3) + 7. Writing the complex number in polar form, 1 integer. Thus 11/3 = e2mri/3, Setting m = 0, roots as 1/3 = 1, 1/8 = e2mi/3 qU/8 = qdmi/3, os(27/3) + i sin(27/3) = an also be n(47/3) = written as 1, A -1-iv3). 9. Writing the complex number in polar form, 1 = c2"7º, where m may be any integer. Thus 11/4 = emzi/4, Setting m = ly, we obtain the three roots as 1/4 = 1, 114 = eri/2, qU4 — . Equivalently, the cos(m/2) + s(7) + isin(m) = —1, roots can also be written as cos(37/2) + isin(37/2) = —i. 142 Chapter 4. Higher Order Linear Equations 10. In polar form, 2(cos 7/3 + i sin 7/3) = 2e'7/8+2m7 “in which m is any integer. Thus [2(cos 7/3 + i sin 7/3)]/2 = 21/2ein/6kmz, With m =0, one square root is given by 21/2ei7/6 = (V3 +i)/v2. With m=1, the other root is given by 91/2 ÁTr/6 — (-v3 — d/v2. 11. The characteristic equation is 13-12 - 1 41 =0. The roots are r = —1,1,1. One root is repeated, hence the general solution is y = cre! + coe! + este! 13. The characteristic equation is 1º - 22-17 +2=0, with roots r = —1,1, The roots are real and distinct, so the general solution is y = qe! + coe! + ese, 14. The are r = haracteristic equation can be written as r2(r? — 4r +4) = 0. The roots ,0,2,2. There are two repeated roots, and hence the general solution is given by y + cat + cao! + cqte!, 15. The chara that is, the six sixth roots of —1. They are e-7/6+m=i/8 m = 0,1 il (3 = D/2, (V3 + 0)/2, à, ci, (-V3 + 0)/2, (-V3 —i general solution is given by eristic equation is rº + 1 = 0. The roots are given by r = (—1)H6, ,5. Explic- /2. Hence the y = 2 [ey cos (t/2) + ca sin (t/2)] + cacos t+ +easin t+ e! [es cos (t/2) + cesim (t/2)]. 16. The characteristic equation can be written as (r2 — 1)(r2— 4) = 0. The roots are given by r = + 1,42. The roots are real and distinct, hence the general solution isy=ce!+ ce! +?! + cgel. 17. The characteri given byr=+1,es 2-1)3=0. The roots are the general solution is equation can be written as ch with multiplicity three. Hen yv= et +ete! +eRe! + cge! + este! + cotel. 18. The characteristic equation can be written as r2(r! — 1) = 0. The roots are given by r=0,0,+1,+i. The general solution is yv=a+et+ce!+cae! + es cos t+cesim t. 19. The characteristic equation can be written as r(r! - 31º 4312 -3r+2)=0. Examining the coefficients, it follows that 3º +32 -3r4+2=(r—1D(r-2)(12 +41). Hence the roots are r y=c+ce+cet+c, 20. The characteristic equation can be written as r(rê — 8) = 0, with roots r = 0, Zemails, 1,2. That is, r =0,2,-1 + ivV3. Hence the general solution is cos V3t + cusin VB]. m= y=a+ce! re 4.2 143 i 21. The characteristic equation can be written as (11 + 4)? = 0. The roots of the equation r!+4=0arer=1 &i,-1i. Each of these roots has multiplicity two. The general solution is y=e'fecost+esint]+te os t+ casin +] + +e! [cs cos t+ cesin t] + te"! [ey cos t+ cssin t]. 22. The characteristic equation can be written as (r2+1)2=0. The roots are given by r = <i, each with multiplicity two. The general solution is cost + cosin t+t[cz = os t+ casin t]. 24. The characteristic equation is 7º 4+5r24+6r+2=0. Examining the coeffi- i we find that 7º + 5r2 + 6r +2= (r+1)(r? +4r + 2). Hence the roots are »ddasr=-1,-2 +v2. The general solution is (23 (-2-vB deduc: y= ce! + ce + ce 25. The characteristic equation is 1812 + 2112 + 14r +4=0. By examining the first and last coefficients, we find that 1813 + 2112 + 14r +45 (2º + 1)(9r2 + 6r + 4). Hence the roots are r = —1/2, (-1 + V31)/3. The general solution of the ODE is given by y -H2 + e-H/8 fe; INB) + essin(t/V3)]. 26. The characteristic equation is 71 — 713 + 6r24+30r — 36=0. By examining the first and last coefficients, we find that rt Tr3+6r2+30r—36=(r—3)(r+2)(12 — 6r +6). The roots are r = —2,3,3 + V3.. The general solution is y= qe t+ cet + cg VB 4 cy VI, 28. The characteristic equation is r! + 6rê + 1772 + 22r + 14 = 0. It can be shown that 1! + 6r3 + 1772 +22r + 14 = (12 +27 +2)(r2 + 4r + 7). Hence the roots are r=-1 +i,-2 + iv3. The general solution is y=e"[ecost+cosint] + e? |egcosv3t+ casin v3t] . 146 Chapter 4. Higher Order Linear Equations The solution is c: = —18,c3 = 8. Therefore the solution of the initial value problem is y(t) = 8 — 18€!/8 + 8e-t2, 36. The general solution is derived in Problem 28 as u(t) = eles cos t + cosin t] + e [es cosv3t + cg sin v3t] . Invoking the initial conditions, we obtain the system of equations ate=1 -“ateo-23+V3ea=-2 20 +c3- 4/3 4=0 2e1 + 203 + 1003 +93 4 =3 = 8/13, with solution cy = 21/13, c» = —38/13, = 173/39. 38. (b) W(e!,e-*,cos t,sin t) = —8. (e) W(cosh t, sinh t, cos t, sin t) = 4. 40. Suppose that ce"! + oe! +... + cnc"! =0, and each of the r; are real and different. Multiplying this equation by et, ey + dA + eneltu= rt = 0. Differentiation results in ca(ra — rt + ca(ra — ret = 0, 4.2 147 ra—r)t Now multiplying the latter equation by e , and differentiating, we obtain es(ry — ro)(r3 — rd TD + cara — r9)(rn — me = 0, Following the above steps in a similar manner, it follows that en(ra = Tn=a) e(ra — ret = 0, Since these equations hold for all t, and all the ry are different, we have cn = 0. Hence get prot +. teme! =0, —o0<t<o00. The same procedure can now be repeated, successively, to show that aq=0=...=c=0. 41.(a) Recall the derivative formula d” Pu (mn) do du nm dv (u : + 1) & a tee n) qo qa Let u=(r—r1)º and v = (r). Note that Cle-nPI= ss) sono and a ars r-m)]=s!. Therefore mo qa (lr = ra)ºa(r)] dr” rem 0 only if n<s, since it is assumed that q(r) £ 0. (b) Differential operators commute, so that DE dr ta € Likew o dt 0 er d* atas = qua) = ql e) Tt follows that, O pt ris 7 [e] = [ter]. From Eq. (i), we have o 57 ler mm] =L[ er). Based on the product formula in Part (: =0 r=m Vim 27 fer! 2(m)] if j<s. Therefore L [ti em] =0ifj<s. 148 Chapter 4. Higher Order Linear Equations 43 2. The general solution of the homogencous equation is Ve = cre! + coe! + cs cos t+ casin t. Let g(t) = 3t and go(t) = cost. By inspection, we find that Yi(t) = —3t. Since g2(t) is à solution of the homogeneous equation, set Ya(t) = E(Acos t+ Bin t). Substitution into the given ODE and comparing the coefficients of similar term resultsin 4 = 0 and B = —1/4. Hence the general solution of the nonhomogencous problem is u(t) = velt) — 3t — Es . 3. The characteristic equation corresponding to the homogeneous problem can be written as (r + 1)(r2 4+1) = 0. The solution of the homogeneous equation is ye = cre! +cocost+casin t. Let gi(t) = e! and go(t) = 4t. Since gi(t) is a solution of the homogeneous equation, set Yi(t) = Ate-!. Substitution into the ODE results in 4=1/2. Now let Yo(t) = Bt + C. We find that B = -C = 4. Hence the general solution of the nonhomogeneous problem is y(t) = ye(t) + te!/2+ 4(t-1). 4. The characteristic equation corresponding to the homogeneous problem can be written as r(r+ 1)(r— 1) = 0. The solution of the homogeneous equation is Ve=ci+ce'+cge!. Since g(t) = 2 sin t is not a solution of the homogencous problem, we can set Y(t) = A cost + Bsin t. Substitution into the ODE results in 4= land B=0. Thus the general solution is y(t) = ci + coe + cge! + cost. 6. The characteris written as (12 + 1) equation corresponding to the homogencous problem can be =0. It follows that Ye = ci cos t+ casin t+ t(es cos t + ca sin 1). Since g(t) is not a solution of the homogeneous problem Csin 2t. Substitution into the ODE results in A = general solution is y(t) = yelt) + 3 + d cos 2t. set Y(t) = A + Bcos t+ B=1/9,C=0. Thus the 7. The characteristic equation corresponding to the homogencous problem can be written as +1) =0. Thus the homogencous solution is m=a+rotrat+ce! + e [escos(V3 t/2) + cs sin(V3 t/2)] . Note the g(t) = t is a solution of the homogenous problem. Consider a particular solution of the form Y(t) = t3(At + B). Substitution into the ODE gives us that A=1/24 and B=0. Thus the general solution is y(t) = ye(t) + t!/24. 8. The written as equation corresponding to the homogencous problem can be 0. Hence the homogencous solution is - v=atot+ret + cu 4.3 151 (and its derivative Ys(t) = (Dt + E)e is independent of the homogeneous solution. Therefore . Conclude that the particular solution has the form Y(t) = At cos 2t + Bt sin 28 + CÊ + (Dt + E)e!. ic equation can be written as r2(r2 4+2r+2)=0, with roots y two, and r=—1 +. This means that the homoge- 1 +cat+cs t+csetsint. The function gr(t) = 3e! + 2te”!, and all of its derivative: independent of the homogeneous solu- tion. Therefore set Yi(t) = Ac! + (Bt + C)e!. Now ga(t) = “sin tis a solution of the homogeneous equation, associated with the complex roots. We need to set Y(t)=tDe"'cost+ Ee-"sin t). It follows that the particular solution has the form Y(g)=4d+(Bt+C)e! +H(De'cost+ Ee sin t). 19. Differentiating y = u(t)u(t), suc »ssively, we have ro! ' y'=ulv+uv y"=ulv+2u'v! + uv” n UM =3> (Dutoiuo j=0 =e, vlD) = adet, So forany p= 1,2 p gy?) = et »” (5) ai ur), j=0 Setting v(t Tt follows that n Pp L[etu) = 3) Jan) (os ulr=5) (+). J j=0 p=0 It is evident that the right hand side of Eq. (*) is of the form est [ho ul + ku D 4 knmu! + kn ] . Hence the operator equation L [e“tu] = (bot + bit" 1 4...4 Dat + Dm ) can be written as hou + ut D+ kau! + knu= =bot" + bt + bmt+ bm. The coefficients k 0,1,...,n can be determined by collecting the like terms in the double summation in Eq. (+). For example, ko is the coefficient of ul”). The only term that contains ul”) is when p= and j=0. Hence ko=ag. On 152 Chapter 4. Higher Order Linear Equations the other hand, ky is the coefficient of u(t). The inner terms with u, given by a”u (when j = p), for each p= n =D anpo”. p=0 summation in (+) contains ;n. Hence t 21.(a) Clearly, e2 is a solution of y' — 2y =0, and te! is a solution of the dif ferential equation y” + 2y'+y=0. The latter ODE has characteristic equation (r+1)2=0. Hence (D-2) Be] =3(D-2) [2] =0 and (D+ 1) [te Furthermore, we have (D-2(D +12 [te] =(D-2)[0)=0, and (D-2(D +17 [82] =(D+1)4D — 2) [38] = (D+ 1)2[0] = 0. (b) Based on part (a) (D=-2(D+1P2 [(D -29(D+1)Y] =(D-2(D+1)2 [8 - te] =0, since the operators are linear. The implied operations are associative and commu- tative. Hence (D=-9(D+1)Y =0. The operator equation corresponds to the solution of a linear homogeneous ODE with characteristic equation (r — 2)4(r + 1) = 0. The roots are r = 2, with mul- tiplicity 4and r = —1, with multiplicity 3. It follows that the given homogeneous solution is YO = pet + gt + eae + ese! + este! + ee, which is a linear combination of seven independent solutions. 22. (15) Observe that (D — 1) [e!] = O and (D2 41) [sin £) = 0. Hence the operator H(D)=(D — 1)(D2 +1) is an annihilator of e! + sin t. The operator correspond- ing to the left hand side of the given ODE is (D? — 1)2. It follows that (D+DAD-1)(D2+1)Y =0. The resulting ODE is homogeneous, with solution Y(t) = ce! +este! + cae! + cute! + cste! + cocos t+ crsint. After examining the homogencous solution of Problem 15, and eliminating dupli- cate terms, we have Y(b)= 4.4 153 > s 22. (16) We find that D [4] =0, (D— 1)? [te] = 0, and (D? + 4) [sin 24] = 0. The operator H(D) = D(D — 1)%(D? + 4)is an annihilator of t? + te! + sin 2t. The op- erator corresponding to the left hand side of the ODE is D?(D? 44). It follows that DD -VAD +4)2y =0. The resulting ODE is homogencous, with solution YM)= ca +et+ ct + ce! + este! + cocos 2t + cy sin 2t+ c; cos 2t + cotsin 2t. After examining the homogencous solution of Problem 16, and eliminating dupli- cate terms, we have Y(t) = est? + cye! + este! + catcos 2t + cotsin 2t. 22. (18) Observe that (D — 1) [] = 0, (D+ 1)2[te!] = 0. The function e-sin t is a solution of a second order ODE with characteristic roots r = —1 + i. It follows that (D? + 2D + 2) [€*sin t] = 0. Therefore the operator H(D)=(D-1(D+1)(D2 +2D +2) is an annihilator of 3e! + 2te-! + e!sin t. The operator corresponding to the left hand side of the given ODE is DD? + 2D + 2). It follows that DD -1(D+1)(D? +2D+2)y =0. The resulting ODE is homogencous, with solution tet+ +e(egcost+ersint)+te “(cscost+cosint). Y(t)=ca+ot+ce+ce!+e After examining the homogencous solution of Problem 18, and eliminating dupli- cate terms, we have Y()= +ee! +este! + te cacost+ cosint). 2. The characteristic equation is r(r2 — 1) = 0. Hence the homogencous solution is yelt) = ci + coe! + cge”!. The Wronskian is evaluated as W(1,e!,e!) = 2. Now compute the three determinants oe mi(g=|0 é Le 10 wa()=|o 0 0 1 156 Chapter 4. Higher Order Linear Equations 9. Based on Problem 4, u/(t) = sec t, us(t) = —1, us(t) = — tan t. The particular solution can be expressed as Y(t) = [um (t)] + cos t [uo(t)] + sin t [us(t)]. That is, Y()=In see(t) + tan(t)| — t cos t + sin t Im [cos(t)] . Hence the general solution of the initial value problem is u(t) = ci + cocos t+ cssin é + In |sec(t) + tan(H)] — t cos t+ sin t In |cos(f)]. Invoking the initial conditions, we require that + =2, c=1,-c=-2. Therefore u(t) = 2.cos t+ sin t+ In |see(t) + tan(t)] — t cos t + sin t In cos(f)] 10. From Problem 6, y(t)=cicost+casint+c; Im order to satisfy the initial conditions, we require that c; = 2,» +c3 -3/4- co — 3e3 = 1. Therefore 204 = — u(t) = 2cos t+ [Tsint — Tt cos t+ 4t sin t 12. From Problem 8, the general solution of the initial value problem is y(t) = ci + coe' + ce! + In ese(t) + cot(t)| + et Ft c(s)ds + J 4.4 157 (7/2), y'(m/2) = ye(m/2), and sm of equations se, to = 7/2. Observe that y(7/2) = u!'(7/2). Therefore we obtain the s; atear +eeT2=2 ge eg =1 oe + ee = Hence the solution of the initial value problem is 4 ut) =3- 7/24 m |ese(t) + cot(t)| +[ cosh(t — s to 13. First write the equation as y" + ay" - 27-2y' +217%y = 2x. The Wron- skian is evaluated as W(x,a?2,1/x) = 6/1. Now compute the three determinants 0 a? m 0 2 = 13 2 0 Wa(a)=[1 0 =9/a, 01 EM a? = % 02 snce u! (x) = —: Hence u/(x) = tion can be expres a a to É s, + a o e. o S s, + sim q = = 7 s e + = G t+ casinh t. Now the four additional determinants are given by Wi(t) in t, Wo(t) = 2 cos t, Walt) = —2 sinh t, Wa(t) = 2 cosh t. If follows that w(t) = g(t) sin(t)/2, us(t) = —g(t) cos(t)/2, 158 Chapter 4. Higher Order Linear Equations us(t) = —g(t) sinh(t)/2, ua(t) = g(t) cosh(t)/2. Therefore the particular solution can be expressed as os(t) | sin(t Y(g = cor ) [ g(s) sin(s) ds — em ) [ g(s) cos(s) ds— 1 , - cost) [ g(s) sinh(s) ds + SBMtD guto) f g(s) cosh(s) ds. to Using the appropriate identities, the integrals can be combined to obtain Y()= 5| gls) sinh(t — s) ds — 5/, g(s) sin(t — s) ds. to to 17. First write the equation : as sur =B4"y “+612y' - Gay = g(u) /42. 3 is a solution of the fomegendos equation. The Wronskian of this fundamental set of solutions is W 3 2x3, The three additional determinants are given by Wi(x) = at, Wo(x) 2, 3, 2, Hence uí(x) = 9(1)/222, us(x) = —g(x) /x?, us(a) = 9(12)/20!. Now vular solution can be expressed as Y(2) EA Mato a [a Dara f 10 ap 202 q? = + &) g(t)dt can be shown that ye the part