Giáo trình Toán cao cấp - Chương 13: Lý thuyết chuỗi

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MATH-EDUCARE Chu.o.ng 13 L´y thuyˆe´tchuˆo˜i 13.1 Chuˆo˜isˆo´ du.o.ng 178 . 13.1.1 C´acd.inh ngh˜ıaco ba’n 178 13.1.2 Chuˆo˜isˆo´ du.o.ng 179 13.2 Chuˆo˜ihˆo. itu. tuyˆe.tdˆo´i v`ahˆo.itu. khˆong tuyˆe.tdˆo´i 191 . 13.2.1 C´acd.inh ngh˜ıaco ba’n 191 13.2.2 Chuˆo˜idan dˆa´u v`adˆa´uhiˆe.u Leibnitz . . . . 192 13.3 Chuˆo˜il˜uy th`u.a 199 . 13.3.1 C´acd.inh ngh˜ıaco ba’n 199 . . 13.3.2 D- iˆe`ukiˆe.n khai triˆe’n v`aphuong ph´apkhai triˆe’n 201 13.4 Chuˆo˜iFourier 211 . 13.4.1 C´acd.inh ngh˜ıaco ba’n 211 . 13.4.2 Dˆa´uhiˆe.udu’ vˆe` su. hˆo.itu. cu’a chuˆo˜i Fourier 212 www.matheducare.com
MATH-EDUCARE 178 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i 13.1 Chuˆo˜isˆo´ du.o.ng . 13.1.1 C´acd.inh ngh˜ıaco ba’n . . Gia’ su’ cho d˜aysˆo´ (an). Biˆe’uth´ucda.ng ∞ a1 + a2 + ···+ an + ···= X an = X an (13.1) n=1 n>1 . . . duo. cgo.il`achuˆo˜isˆo´ (hay don gia’n l`achuˆo˜i). C´acsˆo´ a1, ,an, . . duo. cgo.il`ac´acsˆo´ ha. ng cu’achuˆo˜i, sˆo´ ha.ng an go.il`asˆo´ ha. ng tˆo’ng qu´at . . cu’a chuˆo˜i. Tˆo’ng n sˆo´ ha.ng d`ˆau tiˆencu’a chuˆo˜iduo. cgo.il`atˆo’ng riˆeng . . th´u n cu’a chuˆo˜i v`ak´yhiˆe.ul`asn,t´ucl`a sn = a1 + a2 + ···+ an. V`ısˆo´ sˆo´ ha.ng cu’achuˆo˜i l`avˆoha.n nˆenc´actˆo’ng riˆengcu’achuˆo˜ilˆa.p th`anhd˜ayvˆoha.n c´actˆo’ng riˆeng s1,s2, ,sn, . . D- .inh ngh˜ıa13.1.1. Chuˆo˜i (13.1) duo. cgo.il`achuˆo˜ihˆo. itu. nˆe´u d˜ay . . . . . c´actˆo’ng riˆeng(sn)cu’an´oc´ogi´oiha. nh˜uuha. n v`agi´oiha.nd´oduo. c . . go.il`atˆo’ng cu’achuˆo˜ihˆo.itu Nˆe´ud˜ay(sn) khˆongc´ogi´oiha.nh˜uuha.n th`ıchuˆo˜i (13.1) phˆank`y. D- .inh l´y13.1.1. Diˆe`ukiˆe. ncˆa`ndˆe ’ chuˆo˜i (13.1) hˆo. itu. l`asˆo´ ha. ng tˆo’ng . qu´atcu’an´odˆa`ndˆe´n 0 khi n →∞,t´ucl`a lim an =0. n→∞ . D.inh l´y13.1.1 chı’ l`adiˆe`ukiˆe.ncˆa`n ch´u khˆongl`adiˆe`ukiˆe.ndu’. . . Nhung t`u d´oc´othˆe’ r´ut ra diˆe`ukiˆe.ndu’ dˆe ’ chuˆo˜i phˆank`y: Nˆe´u lim an =06 th`ıchuˆo˜i P an phˆank`y. n→∞ n>1 . . . Chuˆo˜i P an thu duo. ct`u chuˆo˜i P an sau khi c˘a´tbo’ m sˆo´ ha.ng n>m+1 n>1 . . . . d`ˆa u tiˆenduo. cgo.il`aphˆa`nduth´u m cu’achuˆo˜i P an.Nˆe´u chuˆo˜i (13.1) n>1 . . hˆo.itu. th`ımo.i phˆa` nducu’an´od`ˆe uhˆo.itu.,v`amˆo.t phˆa` ndun`aod´o . . hˆo.itu. th`ıba’n thˆanchuˆo˜ic˜ung hˆo.itu Nˆe´u phˆa` nduth´u m cu’achuˆo˜i www.matheducare.com
MATH-EDUCARE 13.1. Chuˆo˜isˆo´ du.o.ng 179 (13.1) hˆo.itu. v`atˆo’ng cu’an´ob˘a`ng Rm th`ı s = sm + Rm. Chuˆo˜ihˆo.itu. c´oc´act´ınhchˆa´t quan tro.ng l`a . . . (i) V´oisˆo´ m cˆo´ d.inh bˆa´tk`y chuˆo˜i (13.1) v`achuˆo˜i phˆa` nduth´u m . . cu’an´od`ˆong th`oihˆo.itu. ho˘a.cd`ˆong th`oi phˆank`y. (ii) Nˆe´u chuˆo˜i (13.1) hˆo.itu. th`ı Rm → 0 khi m →∞ (iii) Nˆe´u c´acchuˆo˜i P an v`a P bn hˆo.itu. v`a α, β l`ah˘a`ng sˆo´ th`ı n>1 n>1 X(αan + βbn)=α X an + β X bn. n>1 n>1 n>1 13.1.2 Chuˆo˜isˆo´ du.o.ng . . . . Chuˆo˜isˆo´ P an duo. cgo.i l`achuˆo˜isˆo´ duong nˆe´u an > 0 ∀ n ∈ N.Nˆe´u n>1 . . . . . . an > 0 ∀ n th`ıchuˆo˜iduo. cgo.il`achuˆo˜isˆo´ duong thu. csu. . . . Tiˆeuchuˆa’nhˆo. itu. . Chuˆo˜isˆo´ duong hˆo.itu. khi v`achı’ khi d˜aytˆo’ng riˆengcu’a n´obi. ch˘a.n trˆen. . . . . Nh`o diˆe`ukiˆe.n n`ay, ta c´othˆe’ thu duo. cnh˜ung dˆa´uhiˆe.udu’ sau dˆay: . Dˆa´uhiˆe.u so s´anhI. Gia’ su’ cho hai chuˆo˜isˆo´ A : X an,an > 0 ∀ n ∈ N v`a B : X bn,bn > 0 ∀ n ∈ N n>1 n>1 v`a an 6 bn ∀ n ∈ N. Khi d´o: (i) Nˆe´u chuˆo˜isˆo´ B hˆo.itu. th`ıchuˆo˜isˆo´ A hˆo.itu., (ii) Nˆe´u chuˆo˜isˆo´ A phˆank`yth`ıchuˆo˜isˆo´ B phˆank`y. . . Dˆa´uhiˆe.u so s´anhII. Gia’ su’ c´acchuˆo˜isˆo´ A v`a B l`anh˜ung chuˆo˜i . . . . an sˆo´ duong thu. csu. v`a ∃ lim = λ (r˜or`angl`a0 6 λ 6 +∞). Khi n→∞ bn d´o : . . . (i) Nˆe´u λ 0th`ıt`u su. hˆo.itu. cu’a chuˆo˜isˆo´ A k´eo theo su. hˆo.itu. cu’a chuˆo˜isˆo´ B www.matheducare.com
MATH-EDUCARE 180 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i . (iii) Nˆe´u0 0 sao cho a ∼ , n →∞th`ıchuˆo˜i a hˆoitunˆe´u p>1 n p P n . . n n>1 v`aphˆank`ynˆe´u p 6 1. . . . . C´acchuˆo˜ithu`ong duo. cd`ung dˆe ’ so s´anhl`a 1) Chuˆo˜icˆa´psˆo´ nhˆan P aqn, a =0hˆo6 .itu. khi 0 6 q 0 k`y khi q > 1. 1 2) Chuˆo˜i Dirichlet: hˆoitukhi α>1 v`aphˆank`ykhi α 6 1. P α . . n>1 n 1 Chuˆo˜i phˆank`y P go.il`achuˆo˜idiˆe`u h`oa. n>1 n . T`u dˆa´uhiˆe.u so s´anhI v`achuˆo˜i so s´anh1) ta r´utra: Dˆa´uhiˆe. u D’Alembert. Nˆe´u chuˆo˜i a1 + a2 + ···+ an + , an > 0 ∀ n c´o a lim n+1 = D n→∞ an th`ıchuˆo˜ihˆo.itu. khi D 1. Dˆa´uhiˆe. u Cauchy. Nˆe´u chuˆo˜i a1 + a2 + ···+ an + , an > 0 ∀ n c´o √ n lim an = C n→∞ th`ıchuˆo˜ihˆo.itu. khi C 1. . . . Trong tru`ong ho. p khi D = C = 1 th`ıca’ hai dˆa´uhiˆe.u n`ayd`ˆe u . khˆongcho cˆautra’ l`oi kh˘a’ng d.inh v`ıtˆo`nta.ichuˆo˜ihˆo.itu. lˆa˜nchuˆo˜i . phˆank`yv´oi D ho˘a.c C b˘a`ng 1. Dˆa´uhiˆe.ut´ıch phˆan. Nˆe´u h`am f(x) x´acd.inh ∀ x > 1 khˆongˆam v`agia’mth`ıchuˆo˜i P f(n)hˆo.itu. khi v`achı’ khi t´ıch phˆansuy rˆo.ng n>1 www.matheducare.com
MATH-EDUCARE 13.1. Chuˆo˜isˆo´ du.o.ng 181 ∞ Z f(x)dx hˆo.itu 0 1 T`u. dˆa´uhiˆeut´ıch phˆansuy ra chuˆo˜i hˆoitukhi α>1v`a . P α . . n>1 n 1 phˆank`ykhi 0 1 pn(n +1) n>7 . . Gia’i. 1) Su’ du.ng bˆa´td˘a’ng th´uchiˆe’n nhiˆen 1 1 > · pn(n +1) n +1 1 . . V`ıchuˆo˜i P l`aphˆa` ndusau sˆo´ ha.ng th´u nhˆa´tcu’achuˆo˜idiˆe`u n>1 n +1 h`oanˆenn´ophˆank`y. Do d´otheo dˆa´uhiˆe.u so s´anhI chuˆo˜id˜acho phˆank`y. 1 1 2) V`ıln n>2 ∀ n>7nˆen 7. nln n n2 1 Do chuˆo˜i Dirichlet hˆoitunˆensuy ra r˘a`ng chuˆo˜id˜acho hˆoi P 2 . . . n>7 n tu N . V´ı du. 2. Kha’o s´atsu. hˆo.itu. cu’a c´acchuˆo˜i: (n − 1)n √ 1) X , 2) X n2e− n. nn+1 n>1 n>1 . . Gia’i. 1) Ta viˆe´tsˆo´ ha.ng tˆo’ng qu´atcu’a c´acchuˆo˜idu´oida.ng: (n − 1)n 1 1 n = 1 − . nn+1 n n www.matheducare.com
MATH-EDUCARE 182 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i 1 n 1 1 Ta biˆe´tr˘a`ng lim 1 − = nˆen an ∼ . n→∞ n e n→∞ ne . 1 Nhung chuˆo˜i P phˆank`y,do d´o chuˆo˜id˜acho phˆank`y. n→∞ ne 2) R˜or`angl`adˆa´uhiˆeu D’Alembert v`aCauchy khˆonggia’i quyˆe´t . √ . . . − n − α duo. cvˆa´nd`ˆe vˆ`e su. hˆo.itu Ta nhˆa.nx´et r˘a`ng e =0(n 2 ) khi n →∞ (α>0). T`u. d´o 1 X a = X n a0 − 2 2 n>1 n>1 n √ 2 − n hˆo.itu. nˆe´u a0 > 6. Do vˆa.y theo dˆa´uhiˆe.u so s´anhI chuˆo˜i P n e n>1 hˆo.itu N . V´ı du. 3. Kha’o s´atsu. hˆo.itu. cu’a chuˆo˜i 2n + n2 (n!)2 1) X , 2) X · 3n + n (2n)! n>1 n>1 Gia’i. 1) Ta c´o: (n +1)2 n n+1 2 n 2+ 1+ a 2 +(n +1) 3 + n n n n+1 = × = 2 × 3 , a 3n+1 +(n +1) 2n + n2 n +1 n2 n 3+ 1+ 3n 2n v n2 u1+ √ 2 un n n u 2 an = u n · 3 t1+ 3n an+1 2 √ 2 . n T`u d´osuy ra lim = v`a lim an = . V`aca’ hai dˆa´uhiˆe.u n→∞ an 3 n→∞ 3 Cauchy, D’Alembert d`ˆe uchokˆe´t luˆa.n chuˆo˜ihˆo.itu 2) Ap´ du.ng dˆa´uhiˆe.u D’Alembert ta c´o: a (n +1)2 1 D = lim n+1 = lim = < 1. n→∞ an n→∞ (2n + 2)(2n +1) 4 Do d´o chuˆo˜id˜a c h o h ˆo.itu www.matheducare.com
MATH-EDUCARE 13.1. Chuˆo˜isˆo´ du.o.ng 183 . Nhˆa. nx´et. Nˆe´u ´apdu.ng bˆa´td˘a’ng th´uc n n n n 0. n2 +1 n lnp n n>1 n>2 2n Gia’i. 1) Ta c´o a = = f(n). Trong biˆe’uth´u.ccu’asˆo´ hang n n2 +1 . 2n tˆo’ng qu´atcu’a a = ta thay n bo’.ibiˆe´n liˆentuc x v`ach´u.ng to’ n n2 +1 . . . . . . . r˘a`ng h`am f(x)thuduo. cliˆen tu.cdondiˆe.u gia’m trˆennu’ a tru.cduong. Ta c´o: +∞ A Z 2x Z 2x 2 A dx = lim dx = lim ln(x +1) x2 +1 A→+∞ x2 +1 A→+∞ 1 1 1 = ln(+∞) − ln 2 = ∞. Do d´o chuˆo˜i 1) phˆank`y. 1 2) Nhu. trˆen,ta d˘a t f(x)= , p>0, x > 2. H`am f(x) tho’a . x lnp x +∞ dx m˜anmoidiˆe`ukiˆencu’adˆa´uhiˆeu t´ıch phˆan.V`ıt´ıch phˆan Z hˆoi . . . x lnp x . 2 tu. khi p>1 v`aphˆank`ykhi p 6 1 nˆen chuˆo˜id˜acho hˆo.itu. khi p>1 v`aphˆank`ykhi 0 <p6 1- N www.matheducare.com
MATH-EDUCARE 184 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i . n +2 V´ı du. 5. Ch´ung minh r˘a`ng chuˆo˜i P √ tho’a m˜andiˆe`ukiˆe.n n>1 (n +1) n . cˆa` nhˆo.itu. nhung chuˆo˜i phˆank`y. Gia’i. Ta c´o n +2 1 an = √ ∼ √ ⇒ lim an =0. (n +1) n (n→∞) n n→∞ Tiˆe´p theo ∀ k =1, 2, ,n ta c´o k +2 1 1 ak = √ > √ > √ (k +1) k k n v`ado d´o n 1 √ s = X a > n · √ = n → +∞ khi n →∞ n k n k=1 v`ado d´o chuˆo˜i phˆank`y. N ` ˆ BAI TA. P . Trong c´acb`aito´ansau dˆa y , b` ˘a ng c´ach kha’o s´atgi´oiha.ncu’atˆo’ng riˆeng,h˜ayx´aclˆa.p t´ınhhˆo.itu. (v`at´ınhtˆo’ng S) hay phˆank`ycu’achuˆo˜i 1 3 1. X .(DS. S = ) 3n−1 2 n>1 (−1)n 2 2. X .(DS. ) 2n 3 n>0 3. X(−1)n−1.(DS. Phˆank`y) n>1 1 4. X ln2n 2. (DS. ) 1 − ln2 2 n>0 1 137 5. X .(DS. ) n(n +5) 300 n>1 www.matheducare.com
MATH-EDUCARE 13.1. Chuˆo˜isˆo´ du.o.ng 185 1 1 6. X , α > 0. (DS. ) (α + n)(α + n +1) α +1 n>1 1 25 7. X .(DS. ) n2 − 4 48 n>3 2n +1 8. X .(DS. 1) n2(n +1)2 n>1 √ √ √ √ 9. X( 3 n +2− 1 3 n +1+ 3 n). (DS. 1 − 3 2) n>1 1 73 10. X .(DS. ) n(n + 3)(n +6) 1080 n>1 . Su’ du.ng diˆe`ukiˆe.ncˆa` n2)dˆe ’ x´acd.inh xem c´acchuˆo˜i sau dˆaychuˆo˜i n`aophˆank`y. 11. X(−1)n−1.(DS. Phˆank`y) n>1 2n − 1 12. X .(DS. Phˆank`y) 3n +2 n>1 13. Xpn 0, 001. (DS. Phˆank`y) n>1 1 . 14. X√ .(DS. Dˆa´uhiˆe.ucˆa` n khˆongcho cˆautra’ l`oi) n>1 2n 2n 15. X .(DS. Dˆa´uhiˆeucˆa` n khˆongcho cˆautra’ l`o.i) 3n . n>1 1 16. X √ .(DS. Phˆank`y) n 0, 3 n>1 1 √ 17. X n .(DS. Phˆank`y) n>1 n! 1 18. Xn2 sin .(DS. Phˆank`y) n2 + n +1 n>1 www.matheducare.com
MATH-EDUCARE 186 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i 1 n2 1+ 19. X n .(DS. Phˆank`y) en n>1 2n2 +1 n2 20. X .(DS. Phˆank`y) 2n2 +3 n>1 n+ 1 n n 21. X .(DS. Phˆank`y) 1 n n>1 n + n n +2 22. X √ .(DS. Dˆa´uhiˆeucˆa` n khˆongcho cˆautra’ l`o.i) (n +1) n . n>1 1 23. X(n + 1)arctg .(DS. Phˆank`y) n +2 n>1 Trong c´acb`aito´ansau dˆay, h˜ayd`ungdˆa´uhiˆe.u so s´anhdˆe ’ kha’o . s´atsu. hˆo.itu. cu’a c´acchuˆo˜id˜a c h o 1 24 X√ .(DS. Phˆank`y) n n>1 1 25. X .(DS. Hˆoitu). Chı’ dˆa˜n. nn > 2n ∀ n > 3. nn . . n>1 1 26. X .(DS. Phˆank`y). Chı’ dˆa˜n. So s´anhv´o.i chuˆo˜idiˆe`u h`oa. ln n n>1 1 27. X .(DS. Hˆoitu) n3n−1 . . n>1 1 28. X√ .(DS. Phˆank`y) 3 n +1 n>1 1 29. X .(DS. Hˆoitu) 2n +1 . . n>1 n 30. X .(DS. Hˆoitu) (n + 2)2n . . n>1 www.matheducare.com
MATH-EDUCARE 13.1. Chuˆo˜isˆo´ du.o.ng 187 1 31. X .(DS. Hˆoitu) 2 . . n>1 p(n + 2)(n +1) 5n2 − 3n +10 32. X .(DS. Hˆoitu) 3n5 +2n +17 . . n>1 5+3(−1)n 33. X .(DS. Hˆoitu). Chı’ dˆa˜n. 2 6 5+3(−1)n 6 8. 2n+3 . . n>1 ln n 34. X .(DS. Phˆank`y). Chı’ dˆa˜n. ln n>1 ∀ n>2. n n>1 ln n 35. X .(DS. Hˆoitu) n2 . . n>1 . . α . Chı’ dˆa˜n. Su’ du.ng hˆe. th´uclnn 0v`an du’ l´on. ln n 36. X √ .(DS. Phˆank`y) 3 n n>1 n5 37. X √ .(DS. Hˆo.itu.) 5 n n>1 1 1 38. X√ sin .(DS. Hˆoitu) n n . . n>1 n4 +4n2 +1 39. X .(DS. Hˆoitu) 2n . . n>1 √ √ 40. Xn2( n a − n+1 a), a>0. (DS. Phˆank`y ∀ a =1)6 n>1 √ √ n n+1 41. X( 2 − 2). (DS. Hˆo.itu.) n>1 1 42. X , a>0. (DS. Hˆoitukhi a>1. Phˆank`ykhi0 1 πn 43. X sin √ .(DS. Hˆoitu) n2 n + n +1 . . n>1 . Trong c´acb`aito´ansau dˆay, h˜ayx´acd.inh nh˜ung gi´atri. cu’a tham sˆo´ p dˆe ’ chuˆo˜id˜acho hˆo.itu. ho˘a.c phˆank`y: www.matheducare.com
MATH-EDUCARE 188 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i π p 44. X sin , p>0. (DS. Hˆoitunˆe´u p>1, phˆank`ynˆe´u p 6 1) n . . n>1 π 45. Xtgp , p>0. (DS. Hˆoitukhi p>1, phˆank`ykhi p 6 1) n +2 . . n>1 1 1 46. X sin · tg , p>0, q>0. np nq n>1 (DS. Hˆo.itu. khi p + q>1, phˆank`ykhi p + q 6 1) 1 47. X1 − cos , p>0. np n>1 1 1 (DS. Hˆoitukhi p> , phˆank`ykhi p 6 ) . . 2 2 √ √ 2n +1 48. X( n +1− n)p ln . 2n +3 n>1 (DS. Hˆo.itu. khi p>0, phˆank`ykhi p 6 0) . Trong c´acb`aito´ansau dˆay, h˜aykha’o s´atsu. hˆo.itu. cu’a chuˆo˜id˜a . cho nh`o dˆa´uhiˆe.udu’ D’Alembert n 49. X .(DS. Hˆoitu) 2n . . n>1 2n−1 50. X .(DS. Hˆoitu) nn . . n>1 2n−1 51. X .(DS. Hˆoitu) (n − 1)! . . n>1 n! 52. X .(DS. Phˆank`y) 2n +1 n>1 4nn! 53. X .(DS. Phˆank`y) nn n>1 3n 54. X .(DS. Phˆank`y) n2n n>1 www.matheducare.com
MATH-EDUCARE 13.1. Chuˆo˜isˆo´ du.o.ng 189 1 · 3 ···(2n − 1) 55. X .(DS. Hˆoitu) 3nn! . . n>1 π 56. Xn2 sin .(DS. Hˆoitu) 2n . . n>1 n(n +1) 57. X .(DS. Hˆoitu) 3n . . n>1 73n 58. X .(DS. Hˆoitu) (2n − 5)! . . n>1 (n + 1)! 59. X .(DS. Hˆoitu) 2nn! . . n>1 (2n − 1)!! 60. X .(DS. Phˆank`y) n! n>1 n!(2n + 1)! 61. X .(DS. Hˆoitu) (3n)! . . n>1 π nn sin n 62. X 2 .(DS. Phˆank`y) n! n>1 nn 63. X .(DS. Hˆoitu) n!3n . . n>1 n!an 64. X , a =6 e, a>0. (DS. Hˆoitukhi a e) nn . . n>1 . Trong c´acb`aito´ansau dˆay, h˜aykha’o s´atsu. hˆo.itu. cu’a chuˆo˜id˜a . cho nh`o dˆa´uhiˆe.udu’ Cauchy n n 65. X .(DS. Hˆoitu) 2n +1 . . n>1 1 n 66. Xarc sin .(DS. hˆoitu) n . . n>1 1 n +1 n2 67. X .(DS. Hˆoitu) 3n n . . n>1 www.matheducare.com
MATH-EDUCARE 190 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i 3n +2 n 68. Xn5 .(DS. Hˆoitu) 4n +3 . . n>1 3n n n +2 n2 69. X .(DS. Phˆank`y) n +5 n +3 n>1 n! 70. X √ .(DS. Phˆank`y) n n n>1 n n√ Chı’ dˆan. Su’. dung cˆongth´u.c Stirling n! ∼ 2πn, n →∞ . e n − 1 n(n−1) 71. X .(DS. Hˆoitu) n +1 . . n>1 n2 +3 n3+1 72. X .(DS. Hˆoitu) n2 +4 . . n>1 n n2 73. X3n .(DS. Phˆank`y) n +1 n>1 √ 3n +2 74. Xarctgn √ .(DS. Phˆank`y) n +1 n>10 an n 75. X , a>0. n +2 n>1 (DS. Hˆo.itu. khi 0 1) nα ∀ 76. X n/2 , α>0. (DS. Hˆo.itu. α) n>1 ln(n +1) 5+(−1)n 77. X .(DS. Hˆoitu) 4n+1 . . n>1 78. X2(−1)n+n.(DS. Phˆank`y) n>1 (−1)n−n 79. X2 .(DS. Hˆo.itu.) n>1 [5 − (−1)n]n 80. X .(DS. Phˆank`y) n24n n>1 www.matheducare.com
MATH-EDUCARE 13.2. Chuˆo˜ihˆo. itu. tuyˆe.tdˆo´iv`ahˆo. itu. khˆongtuyˆe.tdˆo´i 191 [5 + (−1)n]n 81. X .(DS. Hˆoitu) n27n . . n>1 [3 + (−1)n] 82. X .(DS. Hˆoitu) 3n . . n>1 √ n4[ 5+(−1)n]n 83. X .(DS. Hˆoitu) 4n . . n>1 2+(−1)n 84. X · (DS. Hˆoitu) 5+(−1)n+1 . . n>1 13.2 Chuˆo˜ihˆo.itu. tuyˆe.tdˆo´iv`ahˆo. itu. khˆongtuyˆe.tdˆo´i . 13.2.1 C´acd.inh ngh˜ıaco ba’n . Chuˆo˜iv´oi c´acsˆo´ ha.ng c´odˆa´u kh´acnhau a1 + a2 + ···+ an + ···= X an (13.2) n>1 . . . . duo. cgo.il`achuˆo˜ihˆo. itu. tuyˆe.tdˆo´i nˆe´u chuˆo˜isˆo´ duong |a1| + |a2| + ···+ |an| + ···= X |an| (13.3) n>1 . . hˆo.itu Chuˆo˜i (13.2) duo. cgo.il`achuˆo˜i hˆo. itu. c´odiˆe`ukiˆe.n (khˆongtuyˆe. t dˆo´i) nˆe´u n´ohˆo.itu. c`onchuˆo˜i (13.3) phˆank`y. . . D- .inh l´y13.2.1. Mo. i chuˆo˜ihˆo. itu. tuyˆe.tdˆo´id`ˆe uhˆo. itu. ,t´uc l`asu. hˆo. i . tu. cu’a chuˆo˜i (13.3) k´eo theo su. hˆo. itu. cu’a chuˆo˜i (13.2). Chuˆo˜ihˆo.itu. c´odiˆe`ukiˆe.n c´ot´ınhchˆa´trˆa´td˘a.cbiˆe.t l`a:nˆe´u chuˆo˜i . (13.2) hˆo.itu. c´odiˆe`ukiˆe.n th`ıv´oisˆo´ A ⊂ R bˆa´tk`y luˆonluˆonc´othˆe’ . . ho´anvi. c´acsˆo´ ha.ng cu’a chuˆo˜id´o d ˆe ’ chuˆo˜ithuduo. c c´otˆo’ng b˘a`ng A. www.matheducare.com
MATH-EDUCARE 192 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i 13.2.2 Chuˆo˜idan dˆa´u v`adˆa´uhiˆe.u Leibnitz Chuˆo˜ida.ng n−1 n−1 X(−1) an = a1 − a2 + a3 − a4 + ···+(−1) an + , n>1 an > 0 ∀ n ∈ N (13.4) . . duo. cgo.il`achuˆo˜idan dˆa´u. Dˆa´uhiˆe. u Leibnitz. Nˆe´u lim an =0v`aan > an+1 > 0 ∀ n ∈ N th`ı n→∞ chuˆo˜idan dˆa´u (13.4) hˆo.itu. v`a |S − Sn| 6 an+1 (13.5) . trong d´o S l`atˆo’ng cu’a chuˆo˜i (13.4), Sn l`atˆo’ng riˆength´u n cu’a n´o. . . Nhu vˆa.ydˆe ’ kha’o s´atsu. hˆo.itu. cu’a chuˆo˜idan dˆa´u ta cˆa` nkiˆe’m tra hai diˆe`ukiˆe.n i) an > an+1 > 0 ∀ n ∈ N, ii) lim an =0. n→∞ . . Hˆe. th´uc (13.5) ch´ung to’ r˘a`ng sai sˆo´ g˘a.p pha’i khi thay tˆo’ng S cu’a . chuˆo˜idan dˆa´uhˆo.itu. bo’ itˆo’ng cu’amˆo.tsˆo´ sˆo´ ha.ng d`ˆa u tiˆencu’an´ol`a . . . . khˆongvuo. t qu´agi´atri. tuyˆe.tdˆo´icu’asˆo´ ha.ng th´u nhˆa´tcu’a chuˆo˜idu bi. c˘a´tbo’. . . Dˆe ’ x´aclˆa. psu. hˆo. itu. cu’achuˆo˜iv´oi c´acsˆo´ ha.ng c´odˆa´u kh´acnhau ta . . . c´othˆe’ su’ du.ng c´acdˆa´uhiˆe.uhˆo.itu. cu’achuˆo˜iduong v`ad.inh l´y13.1.1. . . Nˆe´u chuˆo˜i P |an| phˆank`yth`ısu. hˆo.itu. cu’a chuˆo˜i P an tro’ th`anh n>1 n>1 . . . . . . vˆa´nd`ˆe dˆe ’ mo’ ngoa. itr`u tru`ong ho. psu’ du.ng dˆa´uhiˆe.u D’Alembert v`a . dˆa´uhiˆe.u Cauchy v`ıc´acdˆa´uhiˆe.u n`ayx´aclˆa.psu. phˆank`ycu’a chuˆo˜i . . . chı’ du. a trˆensu. ph´av˜o diˆe`ukiˆe.ncˆa` n. Nhˆa. nx´et. Chuˆo˜idan dˆa´u tho’a m˜andˆa´uhiˆe.u Leibnitz go.i l`achuˆo˜i Leibnitz. CAC´ V´IDU. www.matheducare.com
MATH-EDUCARE 13.2. Chuˆo˜ihˆo. itu. tuyˆe.tdˆo´iv`ahˆo. itu. khˆongtuyˆe.tdˆo´i 193 n−1 . (−1) V´ı du. 1. Kha’o s´atsu. hˆo.itu. v`ad˘a.c t´ınhhˆo.itu. cu’achuˆo˜i P √ . n>1 n 1 Gia’i. D˜aysˆo´ √ do.ndiˆeu gia’mdˆa` ndˆe´n 0 khi n →∞.Dod´o n . theo dˆa´uhiˆe.u Leibnitz n´ohˆo.itu Dˆe ’ kha’os´atd˘a.c t´ınhhˆo.itu. (tuyˆe.t . . 1 dˆo´i hay khˆongtuyˆe.tdˆo´i) ta x´etchuˆo˜iduong P √ . Chuˆo˜i n`ayphˆan n>1 n k`y. Do vˆa.y chuˆo˜id˜acho hˆo.itu. c´odiˆe`ukiˆe.n. N . V´ı du. 2. Kha’o s´atsu. hˆo.itu. v`ad˘a.c t´ınhhˆo.itu. cu’a chuˆo˜i ln2 n X(−1)n−1 · n n>1 ln2 n Gia’i. Dˆe ’ kha’o s´atd´angdiˆeucu’a d˜ay ta x´eth`am ϕ(x)= . n ln2 x ln x . R˜or`angl`a lim ϕ(x)=0v`aϕ0(x)= (2 − ln x). T`u. d´osuy x x→∞ x2 ln2 n ra khi x>e2 th`ı ϕ0(x) e.V`ıvˆa.y chuˆo˜id˜acho hˆo.itu Dˆ˜e d`angthˆa´y 2 . . ln n r˘a`ng chuˆo˜isˆo´ duong P phˆank`ynˆenchuˆo˜idan dˆa´ud˜achohˆo.i n>1 n tu. c´odiˆe`ukiˆe.n. N . . V´ı du. 3. C˜ung ho’inhutrˆenv´oi chuˆo˜i cos nα X · 2n n>1 Gia’i. Dˆayl`achuˆo˜idˆo’idˆa´u. X´etchuˆo˜idu.o.ng | cos nα| X (*) 2n n>1 | cos αn| 1 V`ı 6 ∀ n ∈ N nˆentheo dˆa´uhiˆeu so s´anhchuˆo˜i (*) hˆoitu 2n 2n . . . v`ado vˆa.y chuˆo˜id˜acho hˆo.itu. tuyˆe.tdˆo´i. N www.matheducare.com
MATH-EDUCARE 194 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i . . V´ı du. 4. C˜ung ho’inhutrˆendˆo´iv´oi chuˆo˜i (−1)n X · n(n +1) n>1 1 Gia’i. Dˆ˜e d`angthˆa´yr˘a`ng d˜ay do.ndiˆeu gia’mdˆa` ndˆe´n0 n(n +1) . . khi n →∞.Dod´otheo dˆa´uhiˆe.u Leibnitz n´ohˆo.itu Tax´et su. hˆo.itu. . . 1 cu’achuˆo˜iduong P . Chuˆo˜i n`ayhˆo.itu.,ch˘a’ng ha.n theo dˆa´u n>1 n(n +1) hiˆe.u t´ıch phˆan ∞ A 1 dx + A Z dx Z 2 x = lim 2 = lim ln =ln2. x(x +1) A→∞ 1 1 A→∞ x +11 x + − 1 1 2 4 Do d´o chuˆo˜id˜a c h o h ˆo.itu. tuyˆe.tdˆo´i. N 1 V´ı du 5. Cˆa` nlˆa´y bao nhiˆeusˆo´ hang cu’achuˆo˜i (−1)n−1 dˆe ’ tˆo’ng . . P 2 n>1 n cu’ach´ung sai kh´acv´o.itˆo’ng cu’achuˆo˜id˜acho khˆongqu´a0,01 ? 0,001 ? Gia’i. 1+ Chuˆo˜id˜a cho l`achuˆo˜i Leibnitz. Do d´ophˆa` ndu. cu’an´o tho’a m˜andiˆe`ukiˆe.n 1 |R | 10. n (n +1)2 . . . . Nhu vˆa.ydˆe ’ t´ınhtˆo’ng cu’achuˆo˜iv´oi sai sˆo´ khˆongvuo. t qu´a0,01 ta chı’ . . cˆa` n t´ınhtˆo’ng mu`oisˆo´ ha.ng d`ˆaul`adu’. + . . . 2 Dˆe ’ t´ınhtˆo’ng cu’a chuˆo˜iv´oi sai sˆo´ khˆongvuo. t qu´a0,001 ta cˆa` n t´ınhtˆo’ng 31 sˆo´ ha.ng d`ˆaul`adu’ (ta.i sao ?) N www.matheducare.com
MATH-EDUCARE 13.2. Chuˆo˜ihˆo. itu. tuyˆe.tdˆo´iv`ahˆo. itu. khˆongtuyˆe.tdˆo´i 195 Nhˆa. nx´et. Ta thˆa´yr˘a`ng chuˆo˜i Leibnitz l`acˆongcu. t´ınhto´antiˆe.n 1 . . ˜ . . ’ ’ ’ ’ ˜ hon so v´oichuˆoiduong. Ch˘ang ha.ndˆe t´ınhtˆong cua chuˆoi P 2 n>1 n . . . . v´oi sai sˆo´ khˆongvuo. t qu´a0,001 ta cˆa` n pha’ilˆa´y 1001 sˆo´ ha.ng m´oidu’. Thˆa.tvˆa.y ta c´othˆe’ ´apdu.ng dˆa´uhiˆe.u t´ıch phˆan.Ta c´o ∞ ∞ Z Z f(x)dx 1000, n . t´ucl`aR1001 1 ta c´o n +1 1 1 = + n3 n2 n3 www.matheducare.com
MATH-EDUCARE 196 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i 1 ˜ ∀ ˜ v`ado chuˆoi P α hˆo.itu. α>1nˆenchuˆoid˜acho hˆo.itu n>1 n . Bˆaygi`o x´et chuˆo˜i ( ). R˜or`angsˆo´ ha.ng tˆo’ng qu´atcu’a ( ) khˆong dˆa` ndˆe´n 0 khi n →∞,dod´o chuˆo˜i ( ) phˆank`y. N ` ˆ BAI TA. P . . Su’ du.ng dˆa´uhiˆe.u Leibnitz dˆe ’ ch´ung minh c´acchuˆo˜i sau dˆa y h ˆo.i tu. c´odiˆe`ukiˆe.n (−1)n+1 (−1)n ln n 1. X√ 6. X n2 − 4n +1 n n>4 n>1 (−1)n+1n9 2n +1 2. X√ 7. X(−1)n+1 20 3 n(n +1) n>1 n +4n +1 n>1 n (−1) n n π 3. X √ (−1) cos (n +1)3 n +2 8. X n n>1 n √ n>1 (−1)n n √ 4. X n n n +20 9. X(−1) ( 2 − 1) n>1 n>1 n 1 − 5. X(−1) √ n n 1 1 4 10. X(−1) √ n n +1 100 n n>1 n>1 . Kha’o s´atsu. hˆo.itu. v`ad˘a.c t´ınhhˆo.itu. cu’a c´acchuˆo˜i 2n +1 n 11. X(−1)n .(DS. Hˆoitutuyˆetdˆo´i) 3n − 2 . . . n>1 3n +1 5n+2 12. X(−1)n .(DS. Phˆank`y) 3n − 2 n>1 2+(−1)n 13. X(−1)n .(DS. Phˆank`y) n n>1 √ (−1)n−1 n 14. X sin .(DS. Hˆoitutuyˆetdˆo´i) n n +1 . . . n>1 www.matheducare.com
MATH-EDUCARE 13.2. Chuˆo˜ihˆo. itu. tuyˆe.tdˆo´iv`ahˆo. itu. khˆongtuyˆe.tdˆo´i 197 ln(n +1) 15. X(−1)n+1arctg .(DS. Hˆoitutuyˆetdˆo´i) (n +1)2 . . . n>1 √ . . Chı’ dˆa˜n. Su’ du.ng bˆa´td˘a’ng th´uc ln(n +1) 2. n+1 1 16. X(−1) .(DS. Hˆo.itu. c´odiˆe`ukiˆe.n) n − ln3 n n>1 Trong c´acb`aito´ansau dˆayh˜ayx´acd.inh gi´atri. cu’a tham sˆo´ p dˆe ’ chuˆo˜isˆo´ hˆo.itu. tuyˆe.tdˆo´i ho˘a.chˆo.itu. c´odiˆe`ukiˆe.n (−1)n−1 17. X , p>0. (2n − 1)p n>1 (DS. Hˆo.itu. tuyˆe.tdˆo´i khi p>1; hˆo.itu. c´odiˆe`ukiˆe.nkhi0 0. n n n>1 2 2 (DS. Hˆoitutuyˆetdˆo´i khi p> ;hˆoituc´odiˆe`ukiˆen khi 0 0. 2 n>1 n n +3 2 2 (DS. Hˆoitutuyˆetdˆo´i khi p> ;hˆoituc´odiˆe`ukiˆen khi 0 0. n n +1 n>1 1 1 (DS. Hˆoitutuyˆetdˆo´i khi p> ;hˆoituc´odiˆe`ukiˆen khi 0 1 1 22. X(−1)n+1 .(DS. Hˆoitutuyˆetdˆo´i) (2n − 1)3 . . . n>1 www.matheducare.com
MATH-EDUCARE 198 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i (2n + 1)!! 23. X(−1)n−1 .(DS. Hˆoitutuyˆetdˆo´i) 2 · 5 · 8 ···(3n − 1) . . . n>1 n+1 π 24. X(−1) 1 − cos √ .(DS. Hˆo.itu. c´odiˆe`ukiˆe.n) n n>1 π (−1)n sin 25. X n .(DS. Hˆoitutuyˆetdˆo´i) n . . . n>1 (−1)n 26. X√ .(DS. Hˆo.itu. c´odiˆe`ukiˆe.n) n +2 n>1 (−1)n 27. X √ .(DS. Phˆank`y) n n n>1 (−1)n+1 28. X .(DS. Hˆoituc´odiˆe`ukiˆen) n − ln n . . . n>1 (−1)n−1 29. X . (n +1)a2n n>1 (DS. Hˆo.itu. tuyˆe.tdˆo´i khi |a| > 1, hˆo.itu. c´odiˆe`ukiˆe.n khi |a| =1, phˆank`ykhi |a| 1 1 n (−1)n+12+ 31. X n .(DS. Hˆoitutuyˆetdˆo´i) 5n . . . n>1 π 32. X(−1)ntg .(DS. Hˆoitutuyˆetdˆo´i) 3n . . . n>1 Trong c´acb`aito´ansau dˆay, h˜ayt`ımsˆo´ sˆo´ ha.ng cu’a chuˆo˜id˜acho cˆa` nlˆa´ydˆe ’ tˆo’ng cu’ach´ung v`atˆo’ng cu’achuˆo˜itu.o.ng ´u.ng sai kh´acnhau . . . . . . mˆo.tda.iluo. ng khˆongvuo. t qu´asˆo´ δ cho tru´oc 1 33. X(−1)n−1 , δ =0, 01. (DS. No =7) 2n2 n>1 www.matheducare.com
MATH-EDUCARE 13.3. Chuˆo˜il˜uy th`u.a 199 cos(nπ) 34. X , δ =0, 001. (DS. No =5) n! n>1 (−1)n−1 35. X√ , δ =10−6.(DS. No =106) 2 n>1 n +1 cos nπ 36. X , δ =10−6.(DS. No = 15) 2n(n +1) n>1 (−1)n2n 37. X , δ =0, 1?; δ =0, 01? (DS. No =2,No =3) (4n + 1)5n n>1 (−1)n 38. X , δ =0, 1; δ =0, 001? (DS. No =4,No =6) n! n>1 13.3 Chuˆo˜il˜uy th`u.a . 13.3.1 C´acd.inh ngh˜ıaco ba’n . . . Chuˆo˜il˜uy th`uadˆo´iv´oibiˆe´n thu. c x l`achuˆo˜ida.ng n 2 n X anx = a0 + a1x + a2x + ···+ anx + (13.6) n>0 hay n n X an(x − a) = a0 + a1(x − a)+···+ an(x − a) + (13.7) n>0 . trong d´oc´achˆe. sˆo´ a0,a1, ,an, l`anh˜ung h˘a`ng sˆo´.B˘a`ng ph´epdˆo’i . . . . biˆe´n x bo’ i x − a t`u (13.6) thu duo. c (13.7). Do d´odˆe ’ tiˆe.n tr`ınhb`ay ta chı’ cˆa` n x´et(13.6) l`adu’ (t´u.c l`axem a = 0). Chuˆo˜i (13.6) luˆonhˆo.itu. ta.idiˆe’m x = 0, c`on(13.7) hˆo.itu. ta.i x = a. . . Do d´otˆa.pho. pdiˆe’m m`achuˆo˜il˜uy th`uahˆo.itu. luˆonluˆon =6 ∅. . . . Dˆo´iv´oichuˆo˜il˜uy th`uabˆa´t k`y(13.6) luˆonluˆontˆo`nta.isˆo´ thu. c R :06 R 6 +∞ sao cho chuˆo˜id´ohˆo.itu. tuyˆe.tdˆo´i khi |x| R.Sˆo´ R d´oduo. cgo.il`ab´ank´ınhhˆo. itu. cu’a chuˆo˜i www.matheducare.com
MATH-EDUCARE 200 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i . . (13.6) v`akhoa’ng I(R)=(−R, R)duo. cgo.il`akhoa’ng hˆo. itu. cu’achuˆo˜i l˜uy th`u.a (13.6). . B´ank´ınhhˆo.itu. R cu’a chuˆo˜il˜uy th`uac´othˆe’ t´ınhthˆongqua c´ac . . hˆe. sˆo´ cu’an´obo’ imˆo.t trong c´accˆongth´uc |a | R = lim n , (13.8) n→∞ |an+1| ho˘a.c 1 R = lim (13.9) n→∞ n p|an| . . nˆe´u gi´oiha.no’ vˆe´ pha’icu’a (13.8) v`a(13.9) tˆo`nta.i. . . . . D- .inh ngh˜ıa13.3.1. Ngu`oi ta n´oir˘a`ng h`am f(x) khai triˆe’nduo. c . n th`anhchuˆo˜il˜uy th`ua P anx trˆenkhoa’ng (−R, R)nˆe´u trˆenkhoa’ng n>0 . d´o c h u˜ ˆo id˜anˆeuhˆo.itu. v`atˆo’ng cu’a n´ob˘a`ng f(x), t´ucl`a n f(x)=X anx ,x∈ (−R, R). n>0 + . D.inh ngh˜ıa13.3.2. 1 Chuˆo˜il˜uy th`uada.ng f 0(x ) f (n)(x ) f(x )+ 0 (x − x )+···+ 0 (x − x )n + 0 1! 0 n! 0 f (n)(x ) = X 0 (x − x )n (13.10) n! 0 n>0 . . . . duo. cgo.i l`achuˆo˜i Taylor cu’a h`am f(x)v´oi tˆamta.idiˆe’m x0 (o’ dˆay (0) 0! = 1, f (x0)=f(x0)). 2+ C´achˆe. sˆo´ cu’a chuˆo˜i Taylor f 0(x ) f (n)(x ) a = f(x ),a = 0 , ,a = 0 (13.11) 0 0 1 1! n n! . . duo. cgo.il`ac´achˆe. sˆo´ Taylor cu’a h`am f(x). www.matheducare.com
MATH-EDUCARE 13.3. Chuˆo˜il˜uy th`u.a 201 + 3 Khi x0 = 0, chuˆo˜i Taylor f 0(0) f (n)(0) f (n)(0) f(0) + x + ···+ xn + ···= X xn (13.12) 1! n! n! n>0 . . duo. cgo.il`achuˆo˜i Maclaurin. . . 13.3.2 D- iˆe`ukiˆe.n khai triˆe’nv`aphuong ph´apkhai triˆe’n . . D- .inh l´y13.3.1 (Tiˆeuchuˆa’n khai triˆe’n). H`am f(x) khai triˆe’nduo. c th`anhchuˆo˜il˜uy th`u.a n X anx n>0 trˆenkhoa’ng (−R, R) khi v`achı’ khi trˆenkhoa’ng d´oh`am f(x) c´oda. o . h`ammo. icˆa´p v`atrong cˆongth´uc Taylor f 0(0) f (n)(0) f(x)=f(0) + x + ···+ xn + R (x) 1! n! n . phˆa`nduRn(x) → 0 khi n →∞∀x ∈ (−R, R). . . . . . . . Trong thu. c h`anhngu`oi ta thu`ong su’ du.ng dˆa´uhiˆe.udu’ nhu sau. . . . D- .inh l´y13.3.2. Dˆe ’ h`am f(x) khai triˆe’nduo. c th`anhchuˆo˜il˜uy th`ua n X anx ,x∈ (−R, R) n>0 diˆe`ukiˆe.ndu’ l`atrˆenkhoa’ng d´oh`am f(x) c´oda. o h`ammo. icˆa´p v`ac´ac . da. oh`amd´obi. ch˘a. n, t´ucl`a∃ M>0:∀ n =0, 1, 2, v`a ∀ x ∈ (−R, R) th`ı |f (n)(x)| 6 M. Ta nˆeu ra dˆayhai phu.o.ng ph´apkhai triˆe’n h`amth`anhchuˆo˜il˜uy th`u.a www.matheducare.com
MATH-EDUCARE 202 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i . . . . . . . 1. Phuong ph´apI (phuong ph´aptru. ctiˆe´p) gˆo`m c´acbu´oc sau: . a) T´ınhc´achˆe. sˆo´ theo cˆongth´uc (13.11) . b) Ch´ung to’ r˘a`ng lim Rn(x)=0. n→∞ . . . . Nhuo. cdiˆe’mcu’aphuong ph´apn`ayl`at´ınhto´anqu´acˆo`ng kˆe`nh v`a . . . sau n˜ua l`aviˆe.c kha’o s´atgi´oiha.n Rn(x) → 0(n →∞)la.i c`angph´uc . ta.phon. . . . . . . . 2. Phuong ph´apII (phuong ph´apgi´antiˆe´p) l`aphuong ph´apdu. a trˆenba’ng c´ackhai triˆe’n “c´os˘a˜n” (hay Khai triˆe’nba’ng)c`ung v´o.i c´ac ph´ept´ınhdˆo´iv´o.i chuˆo˜il˜uy th`u.a. x2 xn xn I. ex =1+x + + ···+ −···= P , x ∈ R. 2! n! n>0 n! x3 x5 x2n+1 II. sin x = x − + −···+(−1)n + ···= 3! 5! (2n + 1)! x2n+1 = P (−1)n , x ∈ R. n>0 (2n + 1)! x2 x4 x2n III. cos x =1− + −···+(−1)n + ···= 2! 4! (2n)! x2n = P (−1)n , x ∈ R. n>0 (2n)! IV. α(α − 1) α(α − 1) ···(α − n +1) (1 + x)α =1+αx + x2 + ···+ xn + 2! n! α! =1+X xn, −1 1 α! α! α(α − 1) ···(α − n +1) α! =1, = , = Cα nˆe´u α ∈ N. 0 n n! n n www.matheducare.com
MATH-EDUCARE 13.3. Chuˆo˜il˜uy th`u.a 203 Khi α = −1 ta c´o 1 =1− x + x2 −···+(−1)nxn + 1+x = X(−1)nxn, −1 0 1 =1+x + x2 + ···+ xn + ···= X xn, −1 0 V. x2 x3 xn ln(1 + x)=x − + −···+(−1)n−1 + ; −1 1 + . n−1 Gia’i. 1 Tas˜e´apdu.ng cˆongth´uc (13.8). V`ı an =(−1) n v`a n an+1 =(−1) (n + 1) nˆenta c´o |a | n R = lim n = lim =1. n→∞ |an+1| n→∞ n +1 . . Nhu vˆa.y chuˆo˜ihˆo.itu. b´oi −1 1 n>1 n>1 Do d´o chuˆo˜id˜acho phˆank`yta.idiˆe’m x = −1 (khˆongtho’a m˜andiˆe`u kiˆe.ncˆa` n!) www.matheducare.com
MATH-EDUCARE 204 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i V´o.i x = 1 ta c´o X(−1)n−1n ⇒ lim (−1)n−1n khˆongtˆo`ntai n→∞ . n>1 Do d´o chuˆo˜i phˆank`yta.idiˆe’m x = 1. Vˆa.ymiˆe`nhˆo.itu. cu’a chuˆo˜il`a (−1, 1). N V´ı du. 2. T`ım khoa’ng hˆo.itu. cu’a chuˆo˜i (−1)n(x − 2)n X · nn n>1 . . . . . Gia’i. Trong tru`o ng ho. p n`ayta su’ du. ng cˆongth´uc (13.9) v`athu . . duo. c 1 1 R = lim = lim = lim n =+∞. n→∞ n |a | n→∞ (−1)n n→∞ p n rn n n . . Diˆe`ud´oc´ongh˜ıa l`achuˆo˜id˜achohˆo.itu. v´oimo.i gi´atri. x,t´ucl`a I(R)=(−∞, +∞). V´ı du. 3. T`ım khoa’ng hˆo.itu. cu’a chuˆo˜i X n!xn, 0! ≡ 1. n>0 . Gia’i. Ap´ du.ng cˆongth´uc (13.8) ta c´o |a | n! 1 R = lim n = lim = lim =0. n→∞ |an+1| n→∞ n!(n +1) n→∞ n +1 Vˆa.y R =0.Diˆe`ud´oc´ongh˜ıar˘a`ng chuˆo˜id˜acho hˆo.itu. ta.idiˆe’m x =0. N 1 V´ı du 4. Khai triˆe’n h`am th`anhchuˆo˜il˜uy th`u.atai lˆancˆandiˆe’m . 4 − x . . . . x0 =2(c˜ung t´uc l`a:theo c´acl˜uyth`uacu’ahiˆe.u x − 2 hay chuˆo˜il˜uy . . th`uav´oi tˆamta.idiˆe’m x0 = 2). www.matheducare.com
MATH-EDUCARE 13.3. Chuˆo˜il˜uy th`u.a 205 Gia’i. Ta biˆe´ndˆo’i h`amd˜achodˆe ’ c´othˆe’ ´apdu.ng khai triˆe’nba’ng: 1 = X tn, −1 0 Ta c´o 1 1 1 = · 4 − x 2 x − 2 1 − 2 x − 2 Xem t = ta c´o: 2 1 1 x − 2 x − 2 2 x − 2 n = h1+ + + ···+ + i 4 − x 2 2 2 2 1 1 ⇒ = X (x − 2)n. 4 − x 2n+1 n>0 Khai triˆe’n n`aychı’ d´ungkhi − x 2 0 (−1)nπ2n = X (x − 2)2n,x∈ R. N 42n(2n)! n>0 www.matheducare.com
MATH-EDUCARE 206 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i 1+x V´ı du 6. Khai triˆe’n h`am f(x)=ln th`anhchuˆo˜i Maclaurin. . 1 − x 1+x Gia’i. Ta c´oln = ln(1 + x) − ln(1 − x). M˘at kh´ac 1 − x . x2 xn ln(1 + x)=x − + ···+(−1)n−1 + , −1 0 n! 1 2n 1 2. f(x)=ln .(DS. P xn, R = ) 1 − 2x n>1 n 2 x3 (−1)nx3n+1 √ 3. f(x)=x ln 1+ .(DS. xn, R = 3 3) P n 3 n>1 3 n √ 4 4n · 2 · 5 ···(3n − 4) 3 − − n 4. f(x)= 1 4x.(DS. 1 x + P n x ; 3 n>2 3 · n! R =1) (−1)n52n+1 5. f(x) = sin 5x.(DS. P x2n+1; R =+∞) n>0 (2n + 1)! x3 (−1)nx6n ∞ 6. f(x) = cos .(DS. P 2n ; R =+ ) 3 n>0 3 (2n)! www.matheducare.com
MATH-EDUCARE 13.3. Chuˆo˜il˜uy th`u.a 207 . . . B˘a`ng c´ach biˆe´ndˆo ’i (trong tru`ong ho. pcˆa` n thiˆe´t) sao cho c´othˆe’ ´a p . du.ng c´ackhai triˆe’nba’ng dˆe ’ khai triˆe’n h`am f(x) th`anhchuˆo˜il˜uy th`ua . v´oi tˆamta.idiˆe’m x0. H˜aychı’ ra b´ank´ınhhˆo.itu. cu’a chuˆo˜i n n − x −5 (−1) (x − 10) 7. f(x)=e 2 , x = 10. (DS. e , R =+∞) 0 P n n>0 n! 2 n x a ln 2 n 8. f(x)=2 , x0 = a.(DS. 2 P (x − a) , R =+∞) n>0 n! 2 n ln x −x 3 n 9. f(x)=2 3 , x0 = 0. (DS. P x , R =+∞) n>0 n! n n 1−2x3 (−1) 2 e 3n 10. f(x)=e , x0 = 0. (DS. P x , R =+∞) n>0 n! 1 (x +2)n+1 11. f(x)=(2+x)ex−1, x = −2. (DS. , R =+∞) 0 3 P e n>0 n! 12. f(x) = sin(a + x), x0 =0. (−1)nx2n (−1)nx2n+1 (DS. sin a X + cos a X ,R=+∞) 2n! (2n + 1)! n>0 n>0 13. f(x) = sin x cos 3x, x0 =0. 1 (−1)n(4x)2n+1 1 (−1)n(2x)2n+1 (DS. X − X ,R=+∞) 2 (2n + 1)! 2 (2n + 1)! n>0 n>0 cos x − 1  ,x=06 14. f(x)= x ; x0 =0. 0,x=0  (−1)n(x)2n−1 (DS. P , R =+∞) n>0 (2n)! π 2n−1 x − π n 2 15. f(x) = cos x, x0 = .(DS. P (−1) , R =+∞) 2 n>1 (2n − 1)! 16. f(x) = sin2 x cos2 x, x =0. www.matheducare.com
MATH-EDUCARE 208 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i (−1)n+1 · 24n−3 (DS. P x2n, R =+∞) n>1 (2n)! 2 17. f(x) = ln(x +3x + 2), x0 =0. xn (DS. ln 2 + X(−1)n−1[1 + 2−n] ,R=+∞) n n>1 2 18. f(x) = ln(4 + 3x − x ), x0 =2. 1 DS. ln 6 + X (−1)n−13−n − 2−n(x − 2)n) n n>1 1 19. f(x)= , x =0;x =4. x2 − 2x − 3 0 0 (−1)n+1 1 1 n (DS. 1) X h − ixn, |x| 0 (−1)n 1 (−1)n 2) X h − i(x − 4)n, |x − 4| 0 1 1 1 Chı’ dˆa˜n. Biˆe’udiˆe˜n f(x)= h − i. 4 x − 3 x +1 x2 + x +1 20. f(x)= , x =0. (x − 1)(x +2) 0 2n+1 − (−1)n − n (DS. 1 P n+1 x , R =1) n>0 2 1 21. f(x)= , x = −2. x2 +4x +7 0 (x +2)2n √ (DS. (−1)n , R = 3) P n+1 n>0 3 . . . . Su’ du.ng phuong ph´apda.o h`amho˘a.c t´ıch phˆancu’achuˆo˜il˜uy th`ua . . dˆe ’ khai triˆe’n h`am f(x) th`anhchuˆo˜il˜uy th`uav´oi tˆam x0 = 0 v`achı’ ra b´ank´ınhhˆo.itu π x3 (−1)nx2n−1 22. f(x) = arcctgx.(DS. x + + ···+ + , |x| 6 1) 2 3 2n − 1 www.matheducare.com
MATH-EDUCARE 13.3. Chuˆo˜il˜uy th`u.a 209 π Chı’ dˆa˜n. Khai triˆe’n arctgx v`asu’. dung hˆe th´u.c arcctgx = − . . 2 arctgx. 1 1 23. f(x)= .(DS. (n + 1)(n +2)xn, R =1) 3 P (1 − x) 2 n>0 (2n − 1)!!x2n+1 24. f(x) = arc sin x.(DS. x + P , R =1) n>1 (2n)!!(2n +1) x 2 n 4n+2 Z sin t (−1) x 25. f(x)= dt.(DS. P , R =+∞) t n>1 (2n + 1)!(4n +2) 0 x n−1 n Z ln(1 + t) (−1) x 26. f(x)= dt.(DS. P 2 , R =1) t n>1 n 0 x 1 − cos 2t (−1)n4nx2n−1 27. f(x)=Z dt.(DS. , R =+∞) 2 P t n>1 (2n − 1)(2n)! 0 x ln(1 + x)  ,x=06 , 28. f(x)=Z g(t)dt, g(x)= x 1,x=0. 0  xn (DS. X(−1)n−1 ,R=1) n2 n>1 √ 29. f(x) = ln(x + x2 + 1). (2n − 1)!! (DS. x + P (−1)n x2n+1, R =1) n>1 (2n)!! x dt Chı’ dˆa˜n. f(x)=Z √ . 1+t2 0 . . . Ap´ du.ng c´acphuong ph´apth´ıch ho. pdˆe ’ khai triˆe’n h`am f(x) th`anh . . chuˆo˜il˜uy th`uav´oi tˆamta.i x0 v`achı’ ra b´ank´ınhhˆo.itu. cu’a chuˆo˜i n x−1 3 (x − 4) 30. f(x)=e , x0 = 4. (DS. e P , R =+∞) n>0 n! www.matheducare.com
MATH-EDUCARE 210 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i n n 3x −x 3 − 2(−1) n 31. f(x)=e − 2e , x0 = 0. (DS. P x , R =+∞) n>0 n! 1 (ln 8)n 32. f(x)= , x = −1. (DS. 32 (x +1)n, R =+∞) 3x−2 0 P 2 n>0 n! n x x−1 (ln 2 + 1) n 33. f(x)=2 e , x0 = 1. (DS. 2 P (x − 1) , R =+∞) n>0 n! π 34. f(x) = sin 3x, x = . 0 4 √ 2 (−1)n32n π 2n (DS. X x − 2 (2n)! 4 n>0 √ 2 (−1)n32n+1 π 2n+1 − X x − ,R=+∞) 2 (2n + 1)! 4 n>0 x π 35. f(x) = cos , x = − . 2 0 3 π 2n √ (−1)nx + (DS. 3 X 3 (2n)!22n+1 n>0 (−1)n π π 2n+1 + X x + x + ,R=+∞) (2n + 1)!(22n+1 3 3 n>0 π 36. f(x) = cos2 x, x = . 0 4 π 2n−1 n n−1 1 (−1) 4 x − (DS. + X 4 ,R=+∞) 2 (2n − 1)! n>1 2 37. f(x) = sin x cos x, x0 =0. (−1)n (DS. X (1+32n+1)x2n+1,R=+∞) 4(2n)! n>0 38. f(x) = cos x cos 2x, x0 =0. 1 1 32n (DS. X(−1)nh + ix2n,R=+∞) 2 (2n)! (2n)! n>0 www.matheducare.com
MATH-EDUCARE 13.4. Chuˆo˜i Fourier 211 39. f(x)=x ln x, x0 =1. (−1)n−1(x − 1)n+1 (−1)n−1(x − 1)n (DS. X + X ,R=1) n n n>1 n>1 40. f(x) = ln(2x + 3), x0 =4. 2 n+1 1 11 (DS. ln 11 + X(−1)n (x − 4)n,R= ) 11 n +1 2 n>1 41. f(x) = ln(3 − 4x), x0 = −2. 4 n (x +2)n −19 3 (DS. ln 11 − X ,x∈ h , i) 11 n 4 4 n>1 x +3 42. f(x) = arctg , x =0. x − 3 0 π (−1)n+1 x2n+1 (DS. − + X ,R=3) 4 32n+1 2n +1 n>0 3 x2n Chı’ dˆa˜n. f 0(x)=− = (−1)n+1 .T`u. d´o 2 P 2n+1 x +9 n>0 3 x x (−1)n+1 Z f(t)dt = f(x) − f(0) = X Z t2ndt. 32n+1 0 n>0 0 13.4 Chuˆo˜i Fourier . 13.4.1 C´acd.inh ngh˜ıaco ba’n Hˆe. h`am 1 πx πx nπx nπx , cos , sin , ,cos , sin , (13.13) 2 ` ` ` ` . . . . . . . duo. cgo.il`ahˆe. luo. ng gi´acco so’ .D´o l `a h ˆe. tru. c giao trˆendoa.n[−`, `] . (t´uc l`at´ıch phˆantheo doa.nd´ocu’a t´ıch hai h`amkh´acnhau bˆa´tk`ycu’a hˆe. b˘a`ng 0). www.matheducare.com
MATH-EDUCARE 212 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i D- .inh ngh˜ıa13.4.1. 1) Chuˆo˜i h`amda.ng a nπx nπx 0 + X a cos + b sin , (13.14) 2 n ` n ` n>1 trong d´o `>0; an, bn l`ac´ach˘a`ng sˆo´ (go.il`ahˆe. sˆo´ cu’a chuˆo˜i (13.14)) . . . . duo. cgo.il`achuˆo˜iluo. ng gi´ac. . . . . 2) Chuˆo˜iluo. ng gi´ac(13.14) duo. cgo.i l`achuˆo˜i Fourier cu’a h`am . . . . f(x) theo hˆe. luo. ng gi´acco so’ (13.13) (go.it˘a´t l`achuˆo˜i Fourier) nˆe´u . . . c´achˆe. sˆo´ cu’an´oduo. c t´ınhtheo cˆongth´uc ` 1 nπx a = Z f(x) cos dx, n =0, 1, n ` ` −` (13.15) ` 1 nπx b = Z f(x) sin dx, n =1, 2, n ` ` −` . . . . . C´achˆe. sˆo´ an, bn duo. c t´ınhtheo cˆongth´uc (13.15) duo. cgo.il`ahˆe. sˆo´ . . . . Fourier cu’a h`am f theo hˆe. luo. ng gi´acco so’ . ` . Z Gia’ su’ h`am f(x) tho’a m˜andiˆe`ukiˆe.n |f(x)|dx 1 . . . . . . . . o’ dˆaydˆa´u“∼”duo. cd`ung khi d˘a’ng th´ucchuaduo. cch´ung minh (v`a . . . . . duo. cgo.il`adˆa´utuong ´ung). . 13.4.2 Dˆa´uhiˆe.udu’ vˆe` su. hˆo. itu. cu’ achuˆo˜i Fourier . . . . Dˆa´utuong ´ung “∼” trong hˆe. th´uc (13.16) c´othˆe’ thay b˘a`ng dˆa´ud˘a’ng . th´ucnˆe´u h`am f(x) tho’a m˜andˆa´uhiˆe.udu’ sau dˆayvˆe` khai triˆe’n h`am th`anhchuˆo˜i Fourier. www.matheducare.com
MATH-EDUCARE 13.4. Chuˆo˜i Fourier 213 . D- .inh l´yDirichlet. Gia’ su’ f(x) l`ah`amtuˆa`n ho`anchu k`y T (go. il`a 0 . h`am T -tuˆa`n ho`an), f(x) v`a f (x) ho˘a. cliˆen tu. c kh˘a´pnoi ho˘a. cliˆen tu. c . . . t`ung kh´uc(t´uc l`achı’ c´omˆo. tsˆo´ h˜uuha. ndiˆe’m gi´andoa. n loa. i I trong mˆo˜i chu k`y). . Khi d´ochuˆo˜i Fourier cu’a h`am f(x) hˆo. itu. v´oimo. i x dˆe´ntˆo’ng S(x) v`a 1) Ta. imo. idiˆe’mliˆen tu. ccu’a h`am f(x), chuˆo˜ihˆo. itu. dˆe´nch´ınh h`am f(x) t´u.cl`aS(x)=f(x). . . 2) Ta. imo. idiˆe’m gi´andoa. n, chuˆo˜ihˆo. itu. dˆe´nnu’ atˆo’ng c´acgi´oi . ha. nmˆo. tph´ıa bˆentr´aiv`abˆenpha’icu’a h`am,t´ucl`a f(x +0)+f(x − 0) S(x )= 0 0 ,xl`adiˆe’m gi´andoan. 0 2 0 . . 3) Nˆe´u f(x) liˆentu. c kh˘a´pnoi th`ıchuˆo˜i Fourier cu’an´ohˆo. itu. tuyˆe. t dˆo´iv`ad`ˆe u. . . . C´ohai tru`ong ho. pd˘a.cbiˆe.t sau dˆay: . 1) Nˆe´u f(x) l`ah`amch˘a˜n, t´ucl`af(x)=f(−x) ∀ x th`ı bn =0 ∀ n > 1 v`achuˆo˜i Fourier cu’a n´ochı’ ch´u.a c´ach`amcosin: ` a nπx 2 nπx f(x)= 0 + X a cos ,a= Z f(x) cos dx n =0, 1, 2, 2 n ` n ` ` n>1 0 . 2) Nˆe´u f(x) l`ah`amle’,t´ucl`af(x)=−f(−x) ∀ x th`ı an =0 ∀ n =0, 1, v`achuˆo˜i Fourier cu’a n´ochı’ ch´u.a c´ach`amsin: ` nπx 2 nπx f(x)=X b sin ,b= Z f(x) sin dx n =1, 2, n ` n ` ` n>1 0 Nˆe´u h`am f(x)chı’ x´acd.inh trˆendoa. n[a, b] ⊂ [−`, `] v`akhˆongx´ac . d.inh trˆen[−`, `] \ [a, b] th`ıc´othˆe’ du. ng h`amphu. F (x) sao cho f(x),x∈ [a, b] F (x)= g(x),x∈ [−`, `] \ [a, b],  www.matheducare.com
MATH-EDUCARE 214 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i trong d´o g(x) l`ah`amkha’ t´ıch t`uy ´y. . . Sau d´ogia’ thiˆe´t F (x +2`)=F (x) ∀ x ∈ R v`athu duo. c h`amtuˆa` n . . . . ho`an F (x)c´ochuk`y2`. Ph´epdu. ng h`am F (x)nhuvˆa.yduo. cgo.il`a ph´epth´actrıˆe’n tuˆa` n ho`an f(x). . . H`am f(x)duo. c cho trˆen[0,`]c´othˆe’ th´actriˆe’nt`uy ´ysang khoa’ng . . . kˆ`e [−`, 0] v`ado vˆa.yn´oduo. cbiˆe’udiˆ˜enbo’ i c´acchuˆo˜i Fourier kh´ac . . . nhau (m`athu`ong g˘a.p l`achuˆo˜i Fourier chı’ ch´ua ho˘a.c c´ach`amcosin ho˘a.c c´ach`amsin). . . . Ta x´ethai tru`ong ho. pd˘a.cbiˆe.t sau dˆay: . . 1) Chuˆo˜i Fourier theo c´ach`amcosin thu duo. c khi th´actriˆe’nch˘a˜n . . . h`amd˜acho trˆendoa.n[0,`] sang khoa’ng kˆe` [−`, 0]. Trong tru`ong ho. p . n`ayd`ˆo thi. cu’a h`am F (x)dˆo´ix´ung qua tru.c tung. . . 2) Chuˆo˜i Fourier theo c´ach`amsin thu duo. c khi th´actriˆe’nle’ h`am . . . d˜acho sang khoa’ng kˆe` [−`, 0). Trong tru`o ng ho. p n`ayd`ˆo thi. cu’a h`am . F (x)l`adˆo´ix´ung qua gˆo´cto.adˆo . Gia’ su’ h`am f(x)liˆen tu.c trˆendoa.n[−`, `]v`af(−`)=f(`)v`aa0, an, bn, n ∈ N l`ac´achˆe. sˆo´ Fourier cu’a h`am f(x). Khi d´ota c´od˘a’ng th´u.c ` 1 a2 Z f 2(x)dx = 0 + X(a2 + b2 ). (13.17) ` 2 n n n>1 −` . . . . D˘a’ng th´uc (13.17) duo. cgo.il`ad˘a’ng th´uc Parseval. ´ ´ CAC VIDU. V´ı du. 1. Khai triˆe’n h`an f(x) = signx, −π 1 www.matheducare.com
MATH-EDUCARE 13.4. Chuˆo˜i Fourier 215 Gia’i. V`ı h`am f(x)le’ nˆen an =0,n =0, 1, v`a π 2 2 cos nx π Z bn = signx sin nxdx = − π π n 0 0 4  ´ 2  nˆeu n =2m − 1 = ([1 − cos nπ]) = π(2m − 1) ,m∈ N. nπ 0nˆe´u n =2m  Do d´ov´o.i −π 1 π Nˆe´ud˘at x = th`ıt`u. chuˆo˜i Fourier thu du.o.c, ta c´o . 2 . 4 (−1)m+1 (−1)m π 1= X ⇒ X = · N π 2m − 1 2m +1 4 m>1 m>0 3 2 V´ı du. 2. T`ım khai triˆe’n Fourier cu’a h`am f(x)=x + x − x +1, x ∈ [−1, 1]. . . . Gia’i. Nhu vˆa.y ta pha’i khai triˆe’n h`amd˜acho theo hˆe. co so’ {1, cos πx,sin πx,cos 2πx, sin 2πx, }. Ta c´o 1 4 a = Z (x3 + x2 − x − 1)dx = − , 0 3 1 www.matheducare.com
MATH-EDUCARE 216 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i 1 Z 3 2 an = (x + x − x − 1) cos nπxdx −1 1 1 2 1 Z = h(x − 1) sin nπx − 2 x sin nπxdxi nπ −1 −1 1 2 1 Z 4 n = hx cos nπx − cos nπxdxi = (−1) ,n=1, 2, π2n2 −1 π2n2 −1 1 12 b = Z (x3 + x2 − x − 1) sin nπxdx = (−1)n,n=1, 2, n π3n3 −1 V`ı h`am f(x) kha’ vi trˆenkhoa’ng (−1, 1) nˆenta c´o 2 4 (−1)n 3(−1)n x3 + x2 − x − 1=− + X h cos nπx + sin nπxi, 3 π2 n2 πn3 n>1 ∀x ∈ (−1, 1). V´ı du. 3. Khai triˆe’n h`am f(x)=x, x ∈ [0,`]: 1) theo c´ach`amcosin; 2) theo c´ach`amsin. Gia’i. 1) Dˆe ’ khai triˆe’n h`am f(x) th`anhchuˆo˜i Fourier theo c´ach`am . cosin, ta thu. chiˆe.n ph´epth´actriˆe’nch˘a˜n h`am f(x)=x, x ∈ [0, 1] sang . . ∗ . doa. nkˆe` [−`, 0] v`athu duo. c h`am f (x)=|x|, x ∈ [−`, `]. T`u d´oth´ac ∗ ∗ triˆe’n2`-tuˆa` n ho`anh`am f (x) ra to`antru.csˆo´.Hiˆe’n nhiˆenh`am f (x) liˆentu. c ∀ x ∈ R. Ta c´o: ` 2 a = Z xdx = `, 0 ` 0 ` 0nˆe´u n ch˘a˜n, 2 Z nπx  an = x cos dx = 4` ` ` − nˆe´u n le’. 0  n2π2 www.matheducare.com
MATH-EDUCARE 13.4. Chuˆo˜i Fourier 217 . . . T`u d´ota thu duo. c ` 4` cos(2n − 1)x x = − X , 0 6 x 6 `. 2 π2 (2n − 1)2 n>1 . 2) Dˆe ’ khai triˆe’n h`am f(x) theo c´ach`amsin, ta thu. chiˆe.nph´ep . . th´actriˆe’nle’ h`am f(x)=x, x ∈ [0,`] sang doa.nkˆe` [−`, 0] v`athu duo. c h`am f ∗(x)=x, x ∈ [−`, `]. T`u. d´oth´actriˆe’n2`-tuˆa` n ho`anh`am f ∗(x) . . ra to`antru.csˆo´. H`amd˜aduo. c th´actriˆe’n tho’am˜and.inh l´yDirichlet. Do vˆa.y ta c´o: a0 =0,an =0,∀ n =1, 2, ` ` 2 nπx 2 x` nπx ` ` nπx Z h Z i bn = x sin dx = − cos + cos dx ` ` ` nπ ` 0 nπ ` 0 0 2` = (−1)n+1. nπ Do d´o 2` (−1)n+1 nπx x = X sin , 0 6 x 6 `, π n ` n>1 V´ı du. 4. Khai triˆe’n h`am π 0nˆe´u x 6 2 f(x)= π π |x|− nˆe´u < |x| <π  2 2 v`a f(x +2π)=f(x), x ∈ R th`anhchuˆo˜i Fourier. . . Gia’i. H`am f(x) l`ah`amch˘a˜n nˆenn´okhai triˆe’nduo. c th`anhchuˆo˜i Fourier theo c´ach`amcosin. Ta c´o π π 2 2 π π a = Z f(x)dx = Z x − dx = , 0 π π 2 4 0 π/2 π π 2 2 π a = Z f(x) cos nxdx = Z x − cos nxdx n π π 2 0 π/2 nπ 2 cos nπ cos = h − 2 i,n∈ N. π n2 n2 www.matheducare.com
MATH-EDUCARE 218 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i T`u. d´osuy ra r˘a`ng v´o.i x ∈ (−π,π) nπ π 2 cos nπ cos f(x)= + X h − 2 i cos nx 8 π n2 n2 n>1 π 2 (−1)n 1 (−1)n = + X cos nx − X cos 2nx, 8 π n2 2π n2 n>1 n>1 . . trong d´otad˜asu’ du.ng c´achˆe. th´uc nπ cos 2 cos nx =0 v´o.i n =2m − 1 n2 nπ cos (−1)m 2 cos nx = cos 2mx, n =2m. N n2 4m2 www.matheducare.com
MATH-EDUCARE 13.4. Chuˆo˜i Fourier 219 ` ˆ BAI TA. P Khai triˆe’n h`amth`anhchuˆo˜i Fourier trˆendoa.n (khoa’ng) d˜a c h o v `a . . . . . . trong mˆo.tsˆo´ tru`ong ho. p h˜aysu’ du.ng khai triˆe’nthuduo. cdˆe ’ t´ınh tˆo’ng cu’a chuˆo˜isˆo´: x π sin nx 1. f(x)= , x ∈ (0, 2π). (DS. − P ) 2 2 n>1 n 6nˆe´u0 1 (−1)n−1 2 ∈ − ’ ˜ ´ 4. f(x)=x , x [ π,π). T´ınhtˆong c´acchuˆoisˆo P 2 v`a n>1 n 1 P 2 . n>1 n π2 cos nx (DS. x2 = − 4 X(−1)n−1 . Thay x = 0, ta thu du.o.c 3 n2 0 . n>1 (−1)n−1 π2 1 π2 X = . Thay x = π thu du.o.c X = ) n2 12 . n2 6 n>1 n>1 5. f(x)=x2, x ∈ (0,π), f(x)=f(x + pi). π2 1 π (DS. + X cos 2nx − sin 2nx) 3 n2 n n>1 www.matheducare.com
MATH-EDUCARE 220 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i x 6. f(x) = cos , x ∈ (0, 2π], f(x)=f(x +2π). 2 x 8 n sin nx (DS. cos = X ) 2 π (2n − 1)(2n +1) n>1 0nˆe´u − 3 1 (2n − 1)  3 6 1 (2n − 1)nx 3 (−1)n nπx (DS. − X cos − X sin , 4 π2 (2n − 1)2 3 π n 3 n>1 n>1 π2 tˆo’ng cu’a chuˆo˜ib˘a`ng v´o.i x =0) 8 0 8. f(x)=x sin x, x ∈ [−π,π]. cos x (−1)n cos nx (DS. 1 − +2X ) 2 n2 − 1 n>2 H˜aykhai triˆe’n c´ach`amsau dˆayth`anhchuˆo˜i Fourier theo c´ach`am sin ho˘a.c h`amcosin 4n2 +1 ∈ ’ ˜ 9. f(x)=x cos x, x (0,π). T´ınhtˆong chuˆoi P 2 2 . n>0 (4n − 1) 2 π 4 4n2 +1 (DS. 1) x cos x = − + cos x − X cos 2nx. π 2 π (4n2 − 1)2 n>1 sin x n 2) x cos x = − +2X(−1)n sin nx, 0 6 x 2 4n2 +1 π2 1 3) X = + (thˆe´ x = 0 v`ao1)) (4n2 − 1) 8 2 n>0 10. f(x)=x(π − x), 0 6 x 1 www.matheducare.com
MATH-EDUCARE 13.4. Chuˆo˜i Fourier 221 π (−1)n 11. f(x) = cos x, x ∈ h0, i.T´ınh S = P . 2 n>1 (2n − 1)(2n +1) 4 1 cos 2nx 1 π (DS. h − X(−1)n i; S = − khi x =0) π 2 4n2 − 1 2 4 0 n>1 1nˆe´u06 x 6 1 12. f(x)= T´ınh 0nˆe´u1 1 n>1 2 1 sin n π − 1 (DS. h + X cos nxi,σ = ; π 2 n 1 2 n>1 1 khi x =0; σ = − v´o.i x = π) 0 2 2 13. f(x)=| cos x|.T´ınh 1 1 S = X(−1)n v`a S = X . 1 4n2 − 1 2 4n2 − 1 n>1 n>1 2 4 (−1)n+1 (DS. + X cos 2nx; π π 4n2 − 1 n>1 π − 2 1 π S = khi x =0,S = khi x = ) 1 4 0 2 2 0 2 | | ∈ − 1 14. f(x)= sin x , x [ π,π]. T´ınh S = P (2n−1)(2n+1). n>1 2 4 cos 2nx 1 (DS. − X ,S= khi x =0) π π (2n − 1)(2n +1) 2 0 n>1 (−1)n−1 15. f(x) = sign(sin x). T´ınh S = P . n>1 2n − 1 4 sin(2n − 1)x (−1)n+1 π π (DS. X , X = v´o.i x = ) π 2n − 1 2n − 1 4 0 2 n>1 n>1 www.matheducare.com
MATH-EDUCARE 222 Chu.o.ng 13. L´ythuyˆe´tchuˆo˜i 16. f(x)=x − [x]={x} - phˆa` n thˆa.p phˆancu’asˆo´ x. 1 1 sin 2nπx (DS. {x} = − X ) 2 π n n>1 −x nˆe´u − π 6 x 6 0  17. f(x)= x2 nˆe´u0 1 0 khi − 2 6 x 6 0 18. f(x)=1 x khi 0 1 2 19. f(x)=x, x ∈ [3, 5]. (DS. 4 + P (−1)n+1 sin nπx) n>1 nπ Chı’ dˆa˜n. Ta hiˆe’u khai triˆe’n Fourier cu’a h`am f(x)=x trˆenkhoa’ng (3, 5) l`akhai triˆe’n Fourier cu’a h`amtuˆa` n ho`anv´o.ichuk`y2` =5−3=2 tr`ungv´o.i h`am f(x) trˆenkhoa’ng (3, 5). π π 20. f(x) = sin 2x + cos 5x, x ∈ h − , i. 2 2 2 20 (−1)n (DS. − + sin 2x + X cos 2nx) 5π π 4n2 − 25 n>1 π π 21. f(x)=x sin 2x, x ∈ h − , i. 4 4 1 8 (−1)n+1n (DS. + X cos 4nx) π π 4n2 − 1 n>1 www.matheducare.com
MATH-EDUCARE 13.4. Chuˆo˜i Fourier 223 0, −π 6 x 6 0 22. f(x)= trˆendoa.n[−π,π]. sin x, 0 1 π 0, 6 |x| 6 π  2 23. f(x)= π trˆendoa.n[−π,π]. cos x, |x| 1 π cos x, x ∈ h0, i  2 24. f(x)= π trˆendoa. n[0,π]. − cos x, x ∈ ,πi  2 4 1 cos 2nx (DS. h + X(−1)n+1 i) π 2 4n2 − 1 n>1 n . a0 . 25. Gia’ su’ Sn(f,x)= + P(ak cos kx+ bk sin kx) l`atˆo’ng riˆength´u 2 k=1 n cu’a chuˆo˜i Fourier cu’ah`am2π-tuˆa` n ho`an f(x). Ch´u.ng minh r˘a`ng π 1 1 sin n + (t − x) S (f,x)= Z f(t) 2 dt. n t − x π 2 sin −π 2 . . Chı’ dˆa˜n. Su’ du.ng c´accˆongth´uc t´ınhhˆe. sˆo´ Fourier cu’a h`am f(x) . v`ahˆe. th´uc 1 sin n + α 1 2 + cos α + cos 2α + ···+ cos nα = α · 2 2 sin 2 www.matheducare.com
MATH-EDUCARE Chu.o.ng 14 Phu.o.ng tr`ınhvi phˆan 14.1 Phu.o.ng tr`ınhvi phˆancˆa´p1 225 14.1.1 Phu.o.ng tr`ınht´ach biˆe´n 226 14.1.2 Phu.o.ng tr`ınhd ˘a’ng cˆa´p 231 14.1.3 Phu.o.ng tr`ınhtuyˆe´nt´ınh 237 14.1.4 Phu.o.ng tr`ınhBernoulli . . . . . . . . . . . 244 14.1.5 Phu.o.ng tr`ınhvi phˆanto`anphˆa` n 247 14.1.6 Phu.o.ng tr`ınh Lagrange v`aphu.o.ng tr`ınh Clairaut 255 14.2 Phu.o.ng tr`ınhvi phˆancˆa´pcao 259 . . 14.2.1 C´acphuong tr`ınhcho ph´epha. thˆa´pcˆa´p . . 260 14.2.2 Phu.o.ng tr`ınhvi phˆantuyˆe´n t´ınhcˆa´p2v´o.i hˆe. sˆo´ h˘a`ng 264 14.2.3 Phu.o.ng tr`ınh vi phˆantuyˆe´nt´ınh thuˆa` n . nhˆa´tcˆa´p n (ptvptn cˆa´p n)v´oihˆe. sˆo´ h˘a`ng . 273 . . 14.3 Hˆe. phuong tr`ınhvi phˆantuyˆe´n t´ınhcˆa´p . 1v´oihˆe. sˆo´ h˘a`ng 290 www.matheducare.com
MATH-EDUCARE 14.1. Phu.o.ng tr`ınhvi phˆancˆa´p1 225 14.1 Phu.o.ng tr`ınhvi phˆancˆa´p1 . . . . . Trong mu.c n`ayta x´etc´acphuong tr`ınhvi phˆanthu`ong cˆa´p1,t´ucl`a . . 0 . . . . . 0 phuong tr`ınhda.ng F (x, y, y ) = 0 ho˘a.cdu´oida.ng gia’iduo. cv´oi y l`a y0 = f(x, y). . . . . H`amkha’ vi y = ϕ(x)duo. cgo.i l`anghiˆe.mcu’aphuong tr`ınhvi . . . . phˆannˆe´u khi thay n´ocho ˆa’n h`amcu’aphuong tr`ınhta s˜ethu duo. c d`ˆo ng nhˆa´tth´u.c. . . 0 Nghiˆe. mtˆo’ng qu´at cu’aphuong tr`ınhvi phˆancˆa´p1: y = f(x, y) trong miˆe`n D ⊂ R2 l`ah`am y = ϕ(x, C) tho’a m˜anc´act´ınhchˆa´t sau: + . . . 1 N´ol`anghiˆe.mcu’aphuong tr`ınhd˜acho∀ C thuˆo.ctˆa.pho. p n`ao d´o ; + . 2 V´oimo.idiˆe`ukiˆe.nband`ˆau y(x0)=y0 sao cho (x0,y0) ∈ D chı’ c´omˆo.t gi´atri. duy nhˆa´t C = C0 l`amcho nghiˆe.m y = ϕ(x, C0) tho’a m˜andiˆe`ukiˆe.n ban d`ˆa ud˜acho. . . . Mo.i nghiˆe.m y = ϕ(x, C0) nhˆa.nduo. ct`u nghiˆe.mtˆo’ng qu´at y = . . . . ϕ(x, C)´ung v´oi gi´atri. cu. thˆe’ C = C0 duo. cgo.il`anghiˆe. mriˆeng. . . 0 B`aito´ant`ımnghiˆe.m riˆengcu’aphuong tr`ınh y = f(x, y) tho’a m˜an . . diˆe`ukiˆe.nband`ˆau y(x0)=y0 duo. cgo.i l`ab`aito´anCauchy. . . . Tuy nhiˆen,ta c`ong˘a.pnh˜ung phuong tr`ınhvi phˆanc´oc´acnghiˆe.m . . . . . khˆongthˆe’ thu duo. ct`u nghiˆe.mtˆo’ng qu´atv´oibˆa´tc´u gi´atri. C n`ao. . . . . . Nghiˆe.mnhuvˆa.ygo.il`anghiˆe. mk`ydi. (bˆa´t thu`ong!) Ch˘a’ng ha.nphuong 0 tr`ınh y = p1 − y2 c´onghiˆe.mtˆo’ng qu´at y = sin(x + C) v`ah`am y =1 . . . . . c˜ung l`anghiˆe.mnhung khˆongthˆe’ thu duo. ct`u nghiˆe.mtˆo’ng qu´atv´oi . bˆa´tc´u gi´atri. C n`ao.D´ol`anghiˆe.mk`ydi www.matheducare.com
MATH-EDUCARE 226 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan 14.1.1 Phu.o.ng tr`ınht´ach biˆe´n . . Phuong tr`ınhvi phˆancˆa´p1da.ng P (x, y)dx + Q(x, y)dy =0 . . . . duo. cgo.il`aphuong tr`ınht´achbiˆe´n nˆe´u c´ach`am P (x, y)v`aQ(x, y) . . . . phˆant´ıch duo. c th`anht´ıch c´acth`uasˆo´ m`amˆo˜ith`uasˆo´ chı’ phu. thuˆo.c v`aomˆo.tbiˆe´n: f1(x)f2(y)dx + ϕ1(x)ϕ2(y)dy =0. (14.1) Dˆe ’ t´ıch phˆanphu.o.ng tr`ınhn`ayta cˆa` nchia2vˆe´ cu’aphu.o.ng tr`ınh . . (14.1) cho f2(y)ϕ1(x) =6 0 v`athu duo. c f (x) ϕ (y) 1 dx + 2 dy = 0 (14.2) ϕ1(x) f2(y) . . . v`at`u d´othuduo. c t´ıch phˆantˆo’ng qu´at f (x) ϕ (y) Z 1 dx + Z 2 dy = C. (14.3) ϕ1(x) f1(y) Trong ph´epchia dˆe ’ c´o(14.2), c´othˆe’ l`ammˆa´t nghiˆe.mcu’a c´ac . . . . phuong tr`ınh f1(y)=0v`aϕ1(x) = 0. Do vˆa.ydˆe ’ thu duo. c to`anbˆo. . nghiˆe.mcu’a (14.1) ta cˆa` nho. p nhˆa´t v`ao(14.3) c´ackhˆongdiˆe’mcu’a h`am f1(y)v`aϕ1(x). . . Nhˆa. nx´et. Phuong tr`ınhda.ng dy = f(ax + by), dx . . . trong d´o f(ax + by) h`amliˆentu.c, c´othˆe’ duavˆe` phuong tr`ınht´ach biˆe´n [a + bf(t)]dx − dt =0 bo’.i ph´epdˆo’ibiˆe´n t = ax + by. www.matheducare.com
MATH-EDUCARE 14.1. Phu.o.ng tr`ınhvi phˆancˆa´p1 227 ´ ´ CAC VIDU. . . 2 V´ı du. 1. Gia’iphuong tr`ınh py +1dx = xydy. . . Gia’i. D´ol`aphuong tr`ınh da.ng (14.1). Chia hai vˆe´ cho t´ıch xpy2 +1tac´o dx ydy = ,x=06 x py2 +1 v`at`u. d´o dx ydy Z = Z + C ⇒ ln |x|−py2 +1=C. x py2 +1 . . . Nhu vˆa.y, mo.i nghiˆe.mcu’aphuong tr`ınhd˜acho l`a ln |x|−py2 +1=C, v`a x =0. N . . 2 0 2 V´ı du. 2. Gia’iphuong tr`ınh(x − 1)y +2xy =0,y(0) = 1. . . Gia’i. D`ˆau tiˆenta t`ımnghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınh.Ta c´o dy 2xdx (x2 − 1)dy +2xy2dx =0⇒ + =0. y2 x2 − 1 . . . T`u d´othuduo. c 1 − +ln|x2 − 1| = C. (14.4) y . . Do vˆa.ymo.i nghiˆe.mcu’aphuong tr`ınhd˜acho l`a 1 − +ln|x2 − 1| = C, v`a y =0. y . . . . . . . . T`u tˆa.pho. pmo.idu`ong cong t´ıch phˆanthu duo. ctat`ımdu`o ng cong qua diˆe’m(0, 1). Thay x =0v`ay = 1 v`ao(14.4) ta c´o C = −1. . Nhu vˆa.y h`am 1 y = 1+ln|x2 − 1| www.matheducare.com
MATH-EDUCARE 228 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan l`anghiˆe.mcu’a b`aito´anCauchy d˜acho. N . . x 0 x V´ı du. 3. Gia’iphuong tr`ınh(1 + e )yy = e , y(0) = 1. Gia’i. Ta c´o dy exdx y2 (1 + ex)y = ex ⇒ ydy = ⇒ = ln(1 + ex)+C. (14.5) dx 1+ex 2 . . Thay x =0v`ay = 1 v`ao(14.5) ta thu duo. c nghiˆe.m riˆeng 1+ex 2 r 1+ex 2 y2 =1+ln ⇒ y = ± 1+ln . 2 2 . ` ` ` . . T`u diˆeukiˆe.n ban dˆa u suy r˘ang y>0(yx=0 =1> 0) do vˆa.y tru´oc . dˆa´u c˘anta lˆa´ydˆa´u +. Nhu vˆa.y nghiˆe.m riˆengcˆa` nt`ıml`a r 1+ex 2 y = 1+ln . N 2 . . V´ı du. 4. Gia’iphuong tr`ınh dy = cos(x + y). dx Gia’i. Ap´ du.ng nhˆa.nx´etd˜anˆeu,ta d˘a.t z = x + y. Khi d´o t a c ´o dz dy =1+ dx dx . . . v`at`u d´ophuong tr`ınhd˜acho c´oda.ng dz dz dz = 1 + cos z ⇒ = dx ⇒ Z = Z dx + C dx 1 + cos z 1 + cos z Z dz z ⇒ z = x + C ⇒ tg = x + C ⇒ z = 2[arctg(x + C)+nπ]. 2 cos2 2 2 . . V`ı khi gia’iphuong tr`ınhta d˜achia hai vˆe´ cho 1 + cos z. Do vˆa.y, cˆa` nkiˆe’m tra xem c´omˆa´t nghiˆe.m hay khˆong.V`ı1 + cos z =0⇔ zn = . . (2n +1)π, n ∈ Z.Thˆe´ zn v`aophuong tr`ınhta thˆa´y zn l`anghiˆe.mcu’a . . . phuong tr`ınhd˜acho. Tro’ vˆ`e biˆe´nc˜u ta c´onghiˆe.m: y = −x + 2arctg(x + C)+2nπ, www.matheducare.com
MATH-EDUCARE 14.1. Phu.o.ng tr`ınhvi phˆancˆa´p1 229 v`a y = −x +(2n +1)π. N BAI` TAˆ. P Gia’i c´acphu.o.ng tr`ınhsau 1. (1 + y2)dx +(1+x2)dy = 0. (DS. arctgx + arctgy = C) 2. (1 + y2)dx + xydy = 0. (DS. x2(1 + y2)=C) 3. (1 + y2)dx = xdy.(DS. y = tg ln Cx) √ √ 4. xp1+y2 + yy0 1+x2 = 0. (DS. 1+x2 + p1+y2 = C) 5. e−y(1 + y0)=1. (DS. ex = C(1 − e−y)) 6. y0 = ax+y (0 <a=6 1). (DS. ax + a−y = C) 7. ey(1 + x2)dy − 2x(1 + ey)dx = 0. (DS. 1 + ey = C(1 + x2)) 8. y2 sin xdx + cos2 x ln ydy = 0. (DS. (1 + Cy +lny) cos x) 1 9. ex sin3 y +(1+e2x) cos y · y0.(DS. arctgex = + C) 2 sin2 y y − x π 10. y0 = sin(x − y). (DS. x + C = cotg + ) 2 4 y − x 11. y0 = cos(y − x). (DS. x − cotg = C, y − x =2kπ, k ∈ Z) 2 12. y − y0 =2x − 3. (DS. y =1− 2x + Cex) Chı’ dˆa˜n. Dˆo’ibiˆe´n z = y +2x − 3. 1 1 13. (x +1)3dy − (y − 2)2dx = 0. (DS. − + = C) y − 2 2(x +1)2 √ √ √ √ 14. ( xy + x)y0 − y = 0. (DS. 2 y +ln|y|−2 x = C) 2 x 18−y 15. 2x+y +3x−2yy0 = 0. (DS. 3 − = C) 2 ln 18 ln 3 www.matheducare.com
MATH-EDUCARE 230 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan 16. (y + xy)dx +(x − xy)dy = 0. (DS. x − y +ln|xy| = C) 17. yy0 + x = 1. (DS. (x − 1)2 + y2 = C2) 18. 1+(1+y0)ey = 0. (DS. (ey +1)ex = C) 1 19. y0 = cos(x − y) − cos(x + y). 2 − cos x (DS. y =2arctg(C1e + nπ) dy 20. 3extgydx +(2− ex) = 0. (DS. tgy − C(2 − ex)3 =0) cos2 y . . T`ım nghiˆe.mriˆeng cu’a c´acphuong tr`ınhtho’a m˜anc´acdiˆe`ukiˆe.n ban d`ˆaud˜achı’ ra. π 21. ydx + cotgxdy =0,y = −1. (DS. y = −2 cos x) 3 22. y2 + x2y0 =0,y(−1) = 1. (DS. x + y =0) 23. 2(1 + ex)yy0 = ex, y(0) = 0. (DS. 2ey2 = ex +1) 1 24. xy0 + y = y2, y(1) = .(DS. 4xy(1 − y) − 1=0) 2 π π 25. y0 sin x = y ln y,a)y = e,b)y =1. 2 2 tg x (DS. a) y = e 2 , b) y ≡ 1) π 26. y0 sin x − y cos x =0,y = 1. (DS. y = sin x) 2 27. y ln ydx + xdy =0,y(1) = 1. (DS. y =1) √ 28. xp1 − y2dx + y 1 − x2dy =0,y(0) = 1. √ (DS. 1 − x2 + p1 − y2 =1,y=1) 2 1 1 29. (2x +1)dy + y2dx =0,y(4) = 1. (DS. ln x + =2 − 1) 9 9 y 30. (1 + y2)dx − xydy =0,y(2) = 1. (DS. x2 =2+2y2) π 1 1 31. y0 =(2y + 1)cotgx, y = .(DS. y = 2 sin2 x − ) 4 2 2 www.matheducare.com
MATH-EDUCARE 14.1. Phu.o.ng tr`ınhvi phˆancˆa´p1 231 π 32. y0tgx − y =1,y = 1. (DS. y = 2 sin x − 1) 2 π √ 33. sin y cos xdy = cos y sin ydx, y(0) = .(DS. cos x = 2 cos y) 4 π 34. y0 sin x = y ln y, y = 1. (DS. y =1) 2 √ √ 35. xydx +(1+y2) 1+x2dy =0,y( 8) = 1. √ (DS. 2 1+x2 +lny2 + y2 =7) 14.1.2 Phu.o.ng tr`ınhd˘a’ng cˆa´p . . . . . 1. Tru´ochˆe´tluu´yr˘a`ng h`am f(x, y)duo. cgo.il`ah`amd˘a’ng cˆa´p cˆa´p m dˆo´iv´o.i c´acbiˆe´ncu’an´onˆe´u n´otho’a m˜and`ˆong nhˆa´tth´u.c f(tx, ty)= tmf(x, y). dy Phu.o.ng tr`ınhvi phˆan = f(x, y)du.o.cgoi l`aphu.o.ng tr`ınhd˘a’ng dx . . cˆa´pdˆo´iv´o.i c´acbiˆe´n x v`a y nˆe´u h`am f(x, y) l`ah`amd˘a’ng cˆa´pcˆa´p0 dˆo´iv´o.i c´acbiˆe´ncu’a n´o. dy Phu.o.ng tr`ınhd˘a’ng cˆa´p luˆonluˆonc´othˆe’ biˆe’udiˆe˜ndu.´o.idang = . dx y ϕ . x Nh`o. ph´epdˆo’ibiˆe´n y u = x . . . . . . . ta duaduo. cphuong tr`ınhd˘a’ng cˆa´pvˆe` phuong tr`ınht´ach biˆe´nd˜abiˆe´t c´ach gia’i: du x = ϕ(u) − u. dx . . . . Nˆe´u u = u0 l`anghiˆe.mcu’aphuong tr`ınh ϕ(u) − u = 0 th`ıphuong tr`ınhd˘a’ng cˆa´p c`onc´onghiˆe.ml`ay = u0x. . . . . . . . 2. C´acphuong tr`ınhduaduo. cvˆ`e phuong tr`ınhd˘a’ng cˆa´p www.matheducare.com
MATH-EDUCARE 232 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan . . i) Phuong tr`ınhvi phˆanda.ng a1x + b1y + c1 y = f ,ai = const,bi = const,i=1, 2. (14.6) a2x + b2y + c2 . . . . . c´othˆe’ duaduo. cvˆ`e phuong tr`ınhd˘a’ng cˆa´pnˆe´u a1 b1 = a1b2 − a2b1 =06 . a2 b2 Dˆe ’ l`amviˆe.cd´o,ta d˘a.t x = u + α, y = v + β v`acho.n α v`a β sao cho vˆe´ a u + b v pha’icu’aphu.o.ng tr`ınh(14.6) c´odang f 1 1 . N´oic´ach kh´ac . a u + b v . . 2 2 α v`a β l`anghiˆe.mcu’ahˆe. phuong tr`ınh a1α + b1β + c1 =0, (14.7) a2α + b2β + c2 =0. . . T`ım nghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınh dv a u + b v = f 1 1 du a2u + b2v . . . . rˆo`i thay u bo’ i x − α, thay v bo’ i y − β ta thu duo. c nghiˆe.mtˆo’ng qu´at cu’a (14.6). a b ´ 1 1 ii) Nˆeu = a1b2 − a2b1 =0th`ıa1x + b1y = λ(a2x + b2y), a2 b2 . . . . . . . . λ = const. Trong tru`o ng ho. p n`ayphuong tr`ınh(14.6) duaduo. cvˆ`e . . phuong tr`ınht´ach biˆe´nb˘a`ng c´ach d˘a.t z = a2x + b2y. ´ ´ CAC VIDU. . . 2 2 2 V´ı du. 1. Gia’iphuong tr`ınh2x dy =(x + y )dx. . . 2 . . Gia’i. Chia hai vˆe´ cu’aphuong tr`ınhcho x dx ta thu duo. c dy y 2 2 =1+ . dx x www.matheducare.com
MATH-EDUCARE 14.1. Phu.o.ng tr`ınhvi phˆancˆa´p1 233 0 0 . . D˘a.t y = ux ⇒ y = xu + u v`athu duo. c dy 2du dx 2xu0 +2u =1+u2 ⇒ 2x = u2 − 2u +1⇒ = · dx (u − 1)2 x T´ıch phˆanphu.o.ng tr`ınhn`ayta c´o 2 2 − 2x − =ln|x| +lnC ⇒−y =lnCx ⇒ Cx = e y−x u − 1 − 1 x . Khi thu. chiˆe.nviˆe.c chia cho x v`a u − 1tacˆa` n xem x =0v`a6 u =1.6 . . Kiˆe’m tra tru. ctiˆe´p ta thˆa´y x =0v`au =1(t´ucl`ay = x)c˜ung l`a . . nghiˆe.mcu’aphuong tr`ınhd˜acho. Vˆa.y − 2x Cx = e y−x ,y= x, x =0. N . . V´ı du. 2. Gia’iphuong tr`ınh 0 y − 1 xy = y1+ln ,y(1) = e 2 . x y y y Gia’i. Trong phu.o.ng tr`ınhd˘a’ng cˆa´p y0 = 1+ln ta d˘at u = , x x . x 0 0 . . . . y = u + xu .Tathuduo. cphuong tr`ınht´ach biˆe´n du du u + x = u(1 + ln u) ⇒ x = u ln u dx dx du dx ⇒ Z = Z +lnC ⇒ ln | ln u| =ln|x| +lnC u ln u x ⇒ ln u = Cx. y Thay u bo’.i ta c´o x y ln = Cx ⇒ y = xeCx. x − 1 D´ol`anghiˆe.mtˆo’ng qu´at. Thay diˆe`ukiˆe.n ban d`ˆau y(1) = e 2 ta c´o 1 − x C = − v`ado d´onghiˆem riˆengcˆa` n t`ıml`a y = xe 2 . N 2 . www.matheducare.com
MATH-EDUCARE 234 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan . . V´ı du. 3. Gia’iphuong tr`ınh(x + y − 2)dx +(x − y +4)dy =0. . . Gia’i. Phuong tr`ınhd˜acho tho’a m˜andiˆe`ukiˆe.n a1b2 − a2b1 = −2 =6 0. Ta t`ımc´acsˆo´ α v`a β b˘a`ng c´ach gia’ihˆe. x + y − 2=0) ⇔ α = x0 = −1,β = y0 =3. x − y +4=0 . . . Thu. chiˆe.nph´ep dˆo’ibiˆe´n x = u − 1, y = v + 3. Khi d´o p h u ong tr`ınhd˜acho tro’. th`anh (u + v)du +(u − v)dv =0. (14.8) . . . . Phuong tr`ınh(14.8) l`aphuong tr`ınhd˘a’ng cˆa´p. D˘a.t v = zu ta thu . . duo. c (u + uz)du +(u − uz)(udz + zdu)=0, (1 + 2z − z2)du + u(1 − z)dz =0, du 1 − z + dz =0, u 1+2z − z2 1 ln |u| + ln |1+2z − z2| =lnC 2 hay l`a u2(1 + 2z − z2)=C. Tro’. vˆ`e biˆe´nc˜u x v`a y ta c´o y − 3 (y − 3)2 (x +1)2h1+2 − i = C x +1 (x +1)2 1 hay l`a 2 2 x +2xy − y − 4x +8y = C. (C = C1 + 14). N www.matheducare.com
MATH-EDUCARE 14.1. Phu.o.ng tr`ınhvi phˆancˆa´p1 235 . . V´ı du. 4. Gia’iphuong tr`ınh (x + y +1)dx +(2x +2y − 1)dy =0. . . . 11 . Gia’i. R˜or`angl`adˆo´iv´oiphuong tr`ınhd˜acho ta c´o =0,t´uc 22 l`ahˆe. x + y +1 =0) 2x +2y − 1=0 . . . vˆonghiˆe.m. Trong tru`ong ho. p n`ayta d˘a.t z = x + y, dy = dz − dx v`a 2z − 1 (2 − z)dx +(2z − 1)dz =0⇒ dx − dz =0 z − 2 ⇒ x − 2z − 3ln|z − 2| = C. Tro’. vˆ`e biˆe´nc˜u ta c´o x +2y +3ln|x + y − 2| = C. N ` ˆ BAI TA. P Gia’i c´acphu.o.ng tr`ınhsau 1. (x − y)dx + xdy = 0. (DS. y = x(C − ln x)) 2. xy0 = y(ln y − ln x). (DS. y = xe1+Cx) 3. (x2 + y2)dx − xydy = 0. (DS. y2 = x2(ln x − C)) y y y 4. xy0 cos = y cos − x.(DS. sin +lnx = C) x x x 0 y y − y 5. y = e x + .(DS. ln Cx = −e x ) x x 6. xy0 = y ln .(DS. y = xeCx+1) y www.matheducare.com
MATH-EDUCARE 236 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan 7. x2dy =(y2 − xy + x2)dx.(DS. (x − y)lnCx = x) 8. xy0 = y + py2 − x2.(DS. y + py2 − x2 = Cx2, y = x) 9. (4x − 3y)dx +(2y − 3x)dy = 0. (DS. y2 − 3xy +2x2 = C) 10. (y − x)dx +(y + x)dy = 0. (DS. x2 +2xy − x2 = C) 11. xy0 = y(1 + ln y − ln x). (DS. y = xeCx) x 12. y − xy0 = y ln .(DS. y = xeCx) y y 13. y − xy0 = x + yy0.(DS. arctg +lnCpx2 + y2 =0) x 14. ydy +(x − 2y)dx.(DS. x =(y − x)lnC(y − x)) √ √ √ 15. ydx +(2 xy − x)dy = 0. (DS. x + y ln Cy =0) y y y 16. xy0 cos = y cos − x.(DS. sin +lnx = C) x x x 1 17. (y + px2 + y2)dx − xdy = 0. (DS. y = (x2 − C2)) 2C 18. (x + y)dx +(x − y)dy = 0. (DS. x2 +2xy − y2 = C) . . . . . . . Gia’i c´acphuo. ng tr`ınhvi phˆanduaduo. cvˆ`e phuong tr`ınhd˘a’ng cˆa´p sau x − 2y +5 y − x − 3 19. y0 = − .(DS. = C) 2x − y +4 (y + x +1)3 20. (2x − y +1)dx +(2y − x − 1)dy =0. (DS. x2 − xy + y2 − x − y = C) 2x + y − 1 21. y0 = . 4x +2y +5 (DS. 10y − 5x + 7 ln(10x +5y +9)=C) 22. (x + y +2)dx +(2x +2y − 1)dy =0. (DS. x +2y +5ln|x + y − 3| = C) 23. (x − 2y +3)dy +(2x + y − 1)dx =0. (DS. x2 + xy − y2 − x +3y = C) 24. (x − y +4)dy +(x + y − 2)dx =0. www.matheducare.com
MATH-EDUCARE 14.1. Phu.o.ng tr`ınhvi phˆancˆa´p1 237 (DS. x2 +2xy − y2 − 4x +8y = C) . . . . . T`ım nghiˆe.m riˆengcu’a c´acphuong tr`ınhd˘a’ng cˆa´p ho˘a.cduaduo. c vˆ`e d˘a’ng cˆa´p sau 25. xdy − ydx = ydy, y(−1) = 1. (DS. x = −y(1 + ln |y|)) 26. xydx +(y2 − x2)dy =0,y(1) = 1. (DS. x2 + y2(ln y2 − 1) = 0) y π 27. xy0 − y = xtg , y(1) = .(DS. y = xarc sin x) x 2 28. x2 − y2 +2xyy0 =0,y(1) = 1. (DS. x2 +2x + y2 =0) 14.1.3 Phu.o.ng tr`ınhtuyˆe´n t´ınh . . Phuong tr`ınhda.ng dy + P (x)y = Q(x) (14.9) dx . . . . . trong d´o P (x)v`aQ(x) l`anh˜ung h`amliˆentu.c, duo. cgo.i l`aphuong tr`ınhvi phˆantuyˆe´n t´ınhcˆa´p1.T´ınh chˆa´t tuyˆe´n t´ınho’. dˆayc´ongh˜ıa 0 . . l `a ˆa ’n h`am y v`ada.o h`am y cu’a n´otham gia trong phuong tr`ınhl`a . tuyˆe´n t´ınh,t´uc l`ac´obˆa.cb˘a`ng 1. . . . . Nˆe´u Q(x) ≡ 0 th`ı(14.9) duo. cgo.il`aphuong tr`ınhtuyˆe´n t´ınh thuˆa`n . . . . nhˆa´t cˆa´p1.Nˆe´u Q(x) 6≡ 0 th`ı(14.9) duo. cgo.i l`aphuong tr`ınhtuyˆe´n t´ınhkhˆongthuˆa` n nhˆa´t. . . . . . . . . . Phuong ph´apgia’i. Hai phuong ph´apthu`ong duo. csu’ du.ng l`a 1+ Phu.o.ng ph´apbiˆe´n thiˆenh˘a`ng sˆo´. . . D`ˆautiˆent`ım nghiˆe.mcu’aphuong tr`ınhthuˆa` n nhˆa´t dy + P (x)y =0. (14.10) dx . Sau d´otrong cˆongth´uc nghiˆe.mtˆo’ng qu´atcu’a (14.10) ta xem h˘a`ng . . sˆo´ C l`ah`amkha’ vi cu’a x: C = C(x). Ta thu duo. c h`am C = C(x) . . . t`u phuong tr`ınh vi phˆant´ach biˆe´n sau khi thˆe´ nghiˆe.mtˆo’ng qu´at www.matheducare.com
MATH-EDUCARE 238 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan . . . . . cu’a (14.10) v`ao(14.9). Phuong ph´apv`ua nˆeugo.i l`aphuong ph´ap Lagrange. + . . . . 2 Phuong ph´apdˆo’ibiˆe´n c`ongo.il`aphuong ph´apBernoulli. . . . Dˆe ’ gia’i (14.9) ta t`ımh`am y du´oida.ng t´ıch cu’a hai h`amchuabiˆe´t cu’a x: y = u(x)v(x). Thˆe´ y v`ao(14.9) ta c´o v[u0 + P (x)u]+v0u = Q(x). (14.11) V`ı y l`at´ıch cu’a hai h`amnˆenmˆo.t trong hai c´othˆe’ cho.nt`uy ´y,c`on . . . . . h`amkia duo. c x´acd.inh bo’ i (14.11). Thˆongthu`o ng ta cho.n u(x) sao . . 0 cho biˆe’uth´uc trong dˆa´u ngo˘a.c vuˆongb˘a`ng 0, t´ucl`au + P (x)u =0. . . Dˆe ’ c´odiˆe`ud´ota chı’ cˆa` nlˆa´y u(x) l`anghiˆe.m riˆengcu’aphuong tr`ınh 0 . . . . u + P (x)u = 0. Gia’iphuong tr`ınhn`ayta thu duo. c u(x). Thˆe´ u(x) v`ao(14.11) ta c´o v0u = Q(x) . . . v`athu duo. c nghiˆe.mtˆo’ng qu´at v = v(x, C). Nhu vˆa.y y = u(x)v(x, C) l`anghiˆe.mtˆo’ng qu´atcu’a (14.9). . . . . . Trong nhiˆe`u tru`ong ho. pphuong tr`ınhvi phˆancˆa´p 1 khˆongtuyˆe´n t´ınhdˆo´iv´o.i y m`al`atuyˆe´nt´ınhdˆo´iv´o.i x,t´u.c l`aphu.o.ng tr`ınhc´othˆe’ . duavˆe` da.ng dx + F (y)x = R(y). (14.12) dy . . . . . . . . Viˆe.c gia’iphuong tr`ınh(14.12) tuong tu. nhu gia’iphuong tr`ınh(14.9) v´o.ich´u´yl`a:y l`adˆo´isˆo´, x = x(y) l `a ˆa ’n h`am. www.matheducare.com
MATH-EDUCARE 14.1. Phu.o.ng tr`ınhvi phˆancˆa´p1 239 ´ ´ CAC VIDU. . . 0 2x V´ı du. 1. Gia’iphuong tr`ınh y +3y = e . Gia’i. Ta s˜egia’iphu.o.ng tr`ınhd˜acho b˘a`ng phu.o.ng ph´apbiˆe´n thiˆen h˘a`ng sˆo´. D`ˆau tiˆengia’iphu.o.ng tr`ınhthuˆa` n nhˆa´t dy y0 +3y =0⇒ = −3dx. y . . . T`u d´othuduo. c: −3x −3x ln |y| = −3x +ln|C1|⇒y = ±C1e = Ce . . . Nghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınhkhˆongthuˆa` n nhˆa´td˜a cho s˜e . . . . −3x 0 duo. c t`ımdu´oida.ng y = C(x)e .Lˆa´yda.o h`am y rˆo`ithˆe´ c´acbiˆe’u th´u.ccu’a y v`a y0 v`aophu.o.ng tr`ınhd˜acho ta c´o 1 C0(x)e−3x = e2x ⇒ C0(x)=e5x ⇒ C(x)= e5x + C 5 2 . . . trong d´o C2 l`ah˘a`ng sˆo´ t`uy ´y.T`u d´othuduo. c nghiˆe.mtˆo’ng qu´atcu’a phu.o.ng tr`ınhd˜acho 1 1 y = C(x)e−3x = e5x + C = e5x + C e−3x. N 5 2 5 2 . . . . V´ı du. 2. Gia’iphuong tr`ınhtrong v´ıdu. 1b˘a`ng phuong ph´apdˆo’i biˆe´n. 0 0 0 . . Gia’i. D˘a.t y = uv. Khi d´o y = u v + v u. Thay v`aophuong tr`ınh . . ta thu duo. c u0v + uv0 +3uv = e2x ⇒ u[v0 +3v]+u0v = e2x. (14.13) T`u. d´ota c´ohai phu.o.ng tr`ınhdˆe ’ t`ım u v`a v: v0 +3v =0, (14.14) vu0 = e2x. (14.15) www.matheducare.com
MATH-EDUCARE 240 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan T`u. (14.14) suy ra dv = −dx ⇒ v = e−3x. 3v −3x . . Thˆe´ v = e v`ao(14.15) ta duo. c 1 e−3xu0 = e2x → u0 = e5x ⇒ u = e5x + C 5 v`ado d´o 1 1 y = e−3x e5x + C = e2x + Ce−3x. 5 5 R˜or`angl`aca’ hai c´ach gia’id`ˆe uchomˆo.tkˆe´t qua’. N . . V´ı du. 3. Gia’iphuong tr`ınh dy 1 = · (14.16) dx x cos y + a sin 2y Gia’i. Phu.o.ng tr`ınhd˜acho khˆongpha’i l`aphu.o.ng tr`ınhtuyˆe´n t´ınh dˆo´iv´o.i y. Tuy nhiˆen,b˘a`ng ph´epbiˆe´ndˆo’ido.n gia’n ta biˆe´ndˆo’in´ovˆ`e phu.o.ng tr`ınhtuyˆe´n t´ınhdˆo´iv´o.i x v`a x0: dx − x cos y = a sin 2y. dy dx dv du D˘at x = u(y)v(y) ⇒ = u + v .Thˆe´ x v`aophu.o.ng tr`ınhv`u.a . dy dy dy . . . . thu duo. c ta c´ohai phuong tr`ınhdˆe ’ x´acd.inh u v`a v du − u cos y =0, (14.17) dy dv u = a sin 2y. (14.18) dy . . . . sin y Gia’iphuong tr`ınh(14.17) ta thu duo. c u = e .Thˆe´ kˆe´t qua’ n`ayv`ao (14.18) dˆe ’ t`ım v.Tac´o dv esiny = a sin 2y dy v(y)=2a Z sin y cos ye− sin ydy + C = −2a(sin y +1)e− sin y + C. www.matheducare.com
MATH-EDUCARE 14.1. Phu.o.ng tr`ınhvi phˆancˆa´p1 241 . . . T`u d´othuduo. c nghiˆe.mtˆo’ng qu´at x(y)=u · v = esin y − 2a(sin y +1)e− sin y + C = −2a(sin y +1)+Cesiny. N V´ı du. 4. Gia’i b`aito´anCauchy x(x − 1)y + y = x2(2x − 1),y(2) = 4. . . Gia’i. T`ım nghiˆe.mtˆo’ng qu´atdu´oida.ng y = u(x)v(x) ⇒ y0 = u0v + v0u. Thˆe´ y v`a y0 v`aophu.o.ng tr`ınhd˜acho ta s˜ec´ohai phu.o.ng tr`ınhdˆe ’ x´ac d.inh u(x)v`av(x): x(x − 1)v0 + v =0, (14.19) x(x − 1)vu0 = x2(2x − 1). (14.20) x Gia’i (14.19) ta thu du.o.c v = .Thˆe´ v`ao(14.20) ta c´o . x − 1 u0 =2x − 1 ⇒ u(x)=x2 − x + C. Do d´onghiˆe.mtˆo’ng qu´atc´oda.ng x Cx y = uv =(x2 − x + C) ⇒ y = + x2. x − 1 x − 1 . . . v`asu’ du. ng diˆe`ukiˆe.n ban d`ˆa u ta thu duo. c 2 4=C · +22 ⇒ C =0⇒ y = x2. 2 − 1 . 2 Nhu vˆa.y nghiˆe.mcu’a b`aito´anCauchy l`a y = x . N BAI` TAˆ. P www.matheducare.com
MATH-EDUCARE 242 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan Gia’i c´acphu.o.ng tr`ınhsau 1. y0 − y = ex.(DS. y =(x + C)ex) 2. y0 +2y = e−x.(DS. y = Ce−2x + e−x) 3. y0 = x + y.(DS. y = Cex − x − 1) 3 0 2 2 − x 4. y + x y = x .(DS. y =1+Ce 3 ) C 5. xy0 + y = 3. (DS. y =3+ ) x ex + C 6. xy0 + y = ex.(DS. y = ) x 7. y0 +2xy =2xe−x2 .(DS. y =(x2 + C)e−x2 ) 8. y0 − 2xy =2xex2 .(DS. y =(x + C)ex2) 9. y0 +2xy = e−x2 .(DS. y =(x + C)e−x2 ) 1 10. y0 + y = cos x.(DS. y = Ce−x + (cos x + sin x)) 2 C − cos 2x 11. y0 cos x − y sin x = sin 2x.(DS. y = ) 2 cos x 12. xy0 − 2y = x3 cos x.(DS. y = Cx2 + x2 sin x) C 13. xy0 + y =lnx + 1. (DS. y =lnx + ) x 2y e−x2 C − e−x2 14. y0 + = .(DS. y = ) x x 2x2 x ln Ctg 15. y0 − ytgx = cotgx.(DS. y =1+ 2 ) cos x 16. y0x ln x − y =3x3 ln2 x.(DS. y =(C + x3)lnx) 17. y0 + y cos x = sin 2x.(DS. y = 2(sin x − 1) + Ce− sin x) 2 ex(x − 2) 18. y0 − y = .(DS. y = Cx2 + ex) x x 3 2 2 2x 19. y0 +2xy =2x2e−x .(DS. y = e−x + C)) 3 www.matheducare.com
MATH-EDUCARE 14.1. Phu.o.ng tr`ınhvi phˆancˆa´p1 243 Gia’i c´acphu.o.ng tr`ınhtuyˆe´nt´ınhdˆo´iv´o.i x sau dˆay y C 20. y0 = .(DS. x = + y ln y) 2y ln y + y − x y 2 2 − y − y 21. (e 2 − xy)dy − dx = 0. (DS. x =(C + y)e 2 ) 22. (sin2 y + xcotgy)y0 = 1. (DS. x =(− cos y + C) sin y) 23. (x + y2)dy = ydx.(DS. x = Cy + y2, y =0) 24. (2ey − x)y0 = 1. (DS. x = Ce−y + ey) 1 25. (y2 − 6x)y0 +2y = 0. (DS. x = y2 + Cy3) 2 26. y = xy0 + y0 ln y.(DS. x = Cy − 1 − ln y) 1 27. (x2 ln y − x)y0 = y.(DS. x = ) ln y +1− Cy 1 28. (2xy +3)dy − y2dx = 0. (DS. x = Cy2 − ) y y4 29. (y4 +2x)y0 = y.(DS. x = Cy2 + ) 2 1 30. ydx +(x + x2y2)dy = 0. (DS. x = ) y(y + C) dy 1 31. e−x − e−x = ey.(DS. e−y = Ce−x − ex) dx 2 − Chı’ dˆa˜n. D˘a.t z(x)=e y. 1 x 32. 3dy +(1+ex+3y)dx = 0. (DS. y = − ln(C + x) − ) 3 3 − Chı’ dˆa˜n. D˘a.t z(x)=e 3y. Gia’i c´acb`aito´anCauchy sau 1 33. ydx − (3x +1+lny)dy =0. y − =1. 3 y3 − 4 1 (DS. x = − ln y) 9 3 Chı’ dˆa˜n. Xem x l `a ˆa ’n h`am. 34. x2 + xy0 = y, y(1) = 0. (DS. y = x − x2) www.matheducare.com
MATH-EDUCARE 244 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan x2 35. y0 cos x − y sin x =2x, y(0) = 0. (DS. y = ) cos x 1 sin x 36. y0 − ytgx = , y(0) = 0. (DS. y = ) cos3 x cos2 x 37. y0 + y cos x = cos x, y(0) = 1. (DS. y =1) 38. (1 − x)(y0 + y)=e−x, y(2) = 0. (DS. −e−x ln |1 − x|) 1 2 39. y0 +3ytg3x = sin 6x, y(0) = .(DS. y = cos 3x[1 − cos 3x]) 3 3 π 40. y0 sin x − y cos x =1;y = 0. (DS. y = − cos x) 2 1 x 41. y0 − ytgx = , y(0) = 1. (DS. y = +1) cos x cos x 42. y0 + x2y = x2, y(2) = 1. (DS. y =1) √ 1 1 π 2 43. y0 − y = − − sin x, y =1+ . sin x cos x sin x 4 2 (DS. tgx + cos x) 44. y0 + y cos x = sin x cos x, y(0) = 1. (DS. y =2e− sin x + sin x − 1) 14.1.4 Phu.o.ng tr`ınhBernoulli . . Phuong tr`ınhda.ng y0 + P (x)y = Q(x)yα,α= const,α=06 ,α=16 , (14.21) . . . . . trong d´o P (x)v`aQ(x) l`anh˜ung h`amliˆentu.c, duo. cgo.i l`aphuong tr`ınhBernoolli. C˜ung giˆo´ng nhu. phu.o.ng tr`ınhtuyˆe´n t´ınh,phu.o.ng tr`ınhBernoulli . . . . . duo. c gia’i nh`o phuong ph´ap a) dˆo’ibiˆe´n y = u(x)v(x), b) biˆe´n thiˆenh˘a`ng sˆo´ t`uy ´y. Phu.o.ng tr`ınh(14.21) c´othˆe’ du.avˆe` phu.o.ng tr`ınhtuyˆe´nt´ınhbo’.i ph´epdˆo’ibiˆe´n z = y1−α www.matheducare.com
MATH-EDUCARE 14.1. Phu.o.ng tr`ınhvi phˆancˆa´p1 245 ´ ´ CAC VIDU. . . 2 2 0 3 V´ı du. 1. Gia’iphuong tr`ınh x y y + xy =1. Gia’i. Chia hai vˆe´ cho x2y2: y 1 y0 + = y−2 · · (14.22) x x2 D´ol`aphu.o.ng tr`ınhBernolli. Thay y = uv v`ao(14.22) ta c´o: uv 1 u0v + v0u + = , x x2u2v2 v 1 ⇒u0v + uv0 + = · x x2u2v2 T`u. d´odˆe ’ t`ım u v`a v ta c´ohai phu.o.ng tr`ınh v 1 1) v0 + = 0; 2) vu0 = · x x2u2v2 1 Phu.o.ng tr`ınh1) cho ta nghiˆem v = v`at`u. d´o . x u0 1 = ⇒ u2u0 = x ⇒ u2du = xdx x u2 3 2 2 u x C r3 3x ⇒ = + ⇒ u = + C. 3 2 3 2 . . Do vˆa.y nghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınhd˜acho c´oda.ng r 3 C y = uv = 3 + . N 2x x3 . . 0 3 2 V´ı du. 2. Gia’iphuong tr`ınh y − 2xy =3x y . Gia’i. D´ol`aphu.o.ng tr`ınhBernolli. Chia hai vˆe´ cu’aphu.o.ng tr`ınh cho y2: y−2y0 − 2xy−1 =2x3. − − 0 0 D˘a.t z = y 1 →−y 2y = z .Dod´o z0 +2xz = −2x3. www.matheducare.com
MATH-EDUCARE 246 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan . . . . Gia’iphuong tr`ınhn`ayta thu duo. c z = Ce−x2 +1− x2 . . v`ado d´onghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınhd˜acho l`a 1 y = · N Ce−x2 +1− x2 . . 0 2 V´ı du. 3. Gia’iphuong tr`ınh xy + y = y ln x. Gia’i. Phu.o.ng tr`ınhd˜acho l`aphu.o.ng tr`ınhBertnoulli. Ta s˜eb˘a`ng phu.o.ng ph´apbiˆe´n thiˆenh˘a`ng sˆo´. + . . . . . 1 Nghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınhthuˆa` n nhˆa´ttuong ´ung l`a C y = . x + . . . . 2 Nghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınhkhˆongthuˆa` n nhˆa´ts˜eduo. c C(x) t`ımdu.´o .idang y = , trong d´o C(x) l`ah`amm´o.i chu.abiˆe´t. Thay . x C(x) y = v`aophu.o.ng tr`ınhd˜acho ta thu du.o.c x . ln x dC ln x C0(x)=C2(x) ⇒ = dx x2 C2 x2 1 ln x 1 x ⇒ = + + C ⇒ C(x)= C(x) x x 1+Cx+lnx v`ado d´onghiˆe.mtˆo’ng qu´atcu’a n´ol`a 1 y = · N 1+Cx +lnx ` ˆ BAI TA. P Gia’i c´acphu.o.ng tr`ınhBernoulli sau 1 1. y0 +2xy =2xy2.(DS. y = ) 1+Cex2 2. 3xy2y0 − 2y3 = x3.(DS. y3 = x3 + Cx2) www.matheducare.com
MATH-EDUCARE 14.1. Phu.o.ng tr`ınhvi phˆancˆa´p1 247 3. (x3 + ey)y0 =3x2.(DS. x3e−y = C + y) −x2 2 e 4. y0 +2xy = y2ex .(DS. y = ) C − x 1 5. y0 − y cos x = y2 cos x.(DS. y = ) Ce− sin x − 1 6. 2y0 sin x + y cos x = y3 sin2 x.(DS. y2(C − x) sin x =1) . . . . Su’ du. ng ph´epdˆo’ibiˆe´ndua c´acphuong tr`ınhphi tuyˆe´n sau dˆay . . . . vˆ`e phuong tr`ınhtuyˆe´n t´ınhho˘a.cphuong tr`ınhBernoulli. Gia’i c´ac phu.o.ng tr`ınhd´o. 1 7. y0 − tgy = ex .(DS. sin y =(x + C)ex. cos y Chı’ dˆa˜n. D˘a.t z = sin y. 8. y0 = y(ex +lny). (DS. ln y =(x + C)ex) Chı’ dˆa˜n. D˘a.t z =lny. 9. y0 cos y + sin y = x + 1. (DS. sin y = x + Ce−x) Chı’ dˆa˜n. D˘a.t z = sin y. 2 2 0 − y −x y 10. yy +1=(x − 1)e 2 .(DS. x − 2+Ce = e 2 ) y2 Chı’ dˆa˜n. D˘a.t z = e 2 11. y0 + x sin 2y =2xe−x2 cos2 y.(DS. tgy =(C + x2)e−x2 ) Chı’ dˆa˜n. D˘a.t z =tgy. 14.1.5 Phu.o.ng tr`ınhvi phˆanto`anphˆa`n I. Nˆe´u trong phu.o.ng tr`ınhvi phˆancˆa´p1 P (x, y)dx + Q(x, y)dy = 0 (14.23) c´achˆe. sˆo´ P v`aQ tho’am˜andiˆe`ukiˆe.n ∂Q ∂P = (14.24) ∂x ∂y www.matheducare.com
MATH-EDUCARE 248 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan th`ıvˆe´ tr´aicu’a n´ol`avi phˆanto`anphˆa` ncu’a h`am V (x, y) n`aod´o v `a . . . . . . . . . trong tru`ong ho. p n`ayphuong tr`ınh(14.23) duo. cgo.il`aphuong tr`ınh vi phˆanto`anphˆa`n (ptvptp) v`a P (x, y)dx + Q(x, y)dy = dV (x, y) = 0 (14.25) ∂Q Phu.o.ng tr`ınh(14.23) l`aptvptp khi v`achı’ khi c´ach`am P , Q, , ∂x ∂P liˆentuc trong miˆe`ndo.nliˆen D ⊂ R2 v`atho’a m˜andiˆe`ukiˆen ∂y . . (14.24). Nˆe´u dV (x, y) = 0 th`ınghiˆe.mtˆo’ng qu´atcu’a (14.25) c´oda.ng V (x, y)=C, . . . . trong d´o C l`ah˘a`ng sˆo´ t`uy ´yv`aduo. c t`ımtheo c´acphuong ph´apsau. 1+ T´ıch phˆanbiˆe’uth´u.c dV (x, y) theo du.`o.ng L(A, M) ⊂ D bˆa´tk`y . . . gi˜ua hai diˆe’m A(x0,y0)v`aM(x, y) v`athu duo. c nghiˆe.mcu’a ptvptp: (x,y) V (x, y)= Z Pdx+ Qdy = C. (x0,y0) . . . . . Thˆongthu`ong lˆa´y L(A, M)l`adu`ong gˆa´pkh´uc v´oi c´acca.nh song song . v´oi tru.cto.adˆo. A(x0,y0), B(x, y0), M(x, y)): x y Z Z V (x, y)= P (t, y0)dt + Q(x, t)dt = C. x0 y0 + 2 V`ı dV (x, y)=dxV (x, y)+dy V (x, y), trong d´o dxV v`a dyV l`a c´acvi phˆanriˆengcu’a V (x, y)nˆen b˘a`ng c´ach t´ıch phˆanriˆengbiˆe.tmˆo˜i . . biˆe’uth´ucd´o,ta c´o V (x, y)biˆe’udiˆe˜nbo’ i hai da.ng sau 1) V (x, y)=R Pdx+ ϕ(y), xem y l`akhˆongdˆo’i, 2) V (x, y)=R Qdy + ψ(x), xem x khˆongdˆo’i. www.matheducare.com
MATH-EDUCARE 14.1. Phu.o.ng tr`ınhvi phˆancˆa´p1 249 . . Dˆe´ndˆay, ho˘a.c xuˆa´t ph´att`u 1) ta t`ım ϕ(y)du. a v`aodiˆe`ukiˆe.nl`a ∂V 0 = Q(x, y) ⇒ Z Pdx + ϕ0(y)=Q(x, y) ∂y y . . . v`at`u d´othuduo. c ϕ(y). . Ho˘a.ct`ımψ(x)du. a v`aodiˆe`ukiˆe.nl`a ∂V 0 = P (x, y) ⇒ Z Qdy + ψ0(x)=P (x, y) ∂x x . . . v`at`u d´othuduo. c ψ(x). II. Th`u.asˆo´ t´ıchphˆan . . . Nˆe´uphuong tr`ınh (14.23) khˆongl`aptvptp nhung tˆo`nta.i h`am µ = µ(x, y) 6≡ 0 sao cho sau khi nhˆanhai vˆe´ cu’a (14.23) v´o.i µ m`a . . . . . . phuong tr`ınhthu duo. c trong kˆe´t qua’ l`aptvptp th`ıh`am µ(x, y)duo. c . . . . go.il`ath`uasˆo´ t´ıchphˆan.Tachı’ ha. nchˆe´ x´et hai tru`ong ho. p sau. ∂P ∂Q 1+ Nˆe´u − /Q l`ah`amchı’ cu’abiˆe´n x th`ıphu.o.ng tr`ınh ∂y ∂x . . . (14.23) c´oth`uasˆo´ t´ıch phˆan µ = µ(x)chı’ phu. thuˆo.c x v`aduo. c x´ac . . . d.inh bo’ iphuong tr`ınh ∂P ∂Q − d ln µ ∂y ∂x = · (14.26) dx Q ∂Q ∂P 2+ Tu.o.ng tu. nˆe´u − /P l`ah`amchı’ cu’abiˆe´n y th`ı(14.23) . ∂x ∂y . . . . . . c´oth`uasˆo´ t´ıch phˆan µ = µ(y)chı’ phu. thuˆo.c y v`aduo. ct`ımt`u phuong tr`ınh ∂Q ∂P − d ln µ ∂x ∂y = · (14.27) dy P CAC´ V´IDU. www.matheducare.com
MATH-EDUCARE 250 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan . . V´ı du. 1. Gia’iphuong tr`ınh (x + y +1)dx +(x − y2 +3)dy =0. . Gia’i. O’ dˆa y P = x + y +1,Q = x − y2 +3.V`ı ∂P ∂Q =1= ∂y ∂x nˆen vˆe´ tr´aicu’aphu.o.ng tr`ınhd˜acho l`avi phˆanto`anphˆa` ncu’a h`am V (x, y) n`aod´o v `a ∂V 1) = x + y +1, ∂x ∂V 2) = x − y2 +3. ∂y . . . T`u 1) thu duo. c V = Z P (x, y)dx + ϕ(y)=Z (x + y +1)dx + ϕ(y) x2 = + yx + x + ϕ(y). (*) 2 . . . . Dˆe ’ t`ım ϕ(y) ta cˆa` nsu’ du.ng 2) v`akˆe´t qua’ v`uathuduo. c ∂ x2 + yx + x + ϕ(y) = x − y2 +3⇒ ϕ0(y)=−y2 +3 ∂y 2 y3 ⇒ ϕ(y)=Z (−y2 +3)dy ⇒ ϕ(y)=− +3y + C . 3 1 . . . Thˆe´ biˆe’uth´uc ϕ(y) v`ao(*) ta thu duo. c x2 y3 V (x, y)= + xy + x − +3y + C . 2 3 1 . . Phuong tr`ınhd˜acho c´oda.ng dV (x., y) = 0 v`anghiˆe.mtˆo’ng qu´atcu’a . . . . . n´oduo. c x´acd.inh bo’ iphuong tr`ınh x2 y3 V (x, y)=C hay + xy + x − +3y + C = C . 2 2 3 1 2 www.matheducare.com
MATH-EDUCARE 14.1. Phu.o.ng tr`ınhvi phˆancˆa´p1 251 . . D˘a.t6(C2 − C1)=C -h˘a`ng sˆo´ t`uy ´yv`athu duo. c nghiˆe.mtˆo’ng qu´at cu’aphu.o.ng tr`ınhd˜acho l`a 3x2 +6xy +6x − 2y3 +18y = C. N . . V´ı du. 2. Gia’iphuong tr`ınh (1 + xpx2 + y2)dx +(−1+px2 + y2)ydy =0. . . Gia’i. Dˆ˜e kiˆe’m tra r˘a`ng phuong tr`ınhd˜acho l`aptvptp. Thˆa.tvˆa.y ∂P ∂Q xy . v`ı ≡ = liˆentu.c kh˘a´pnoi trong m˘a.t ph˘a’ng nˆen ∂y ∂x px2 + y2 . . phuong tr`ınhd˜acho l`aptvptp. Nghiˆe.mtˆo’ng qu´atcu’a n´oc´othˆe’ viˆe´t . . . . du´oida.ng t´ıch phˆandu`o ng (x,y) Z [Pdx+ Qdy]=C. (x0,y0) . . . . . Cho.ndu`o ng t´ıch phˆanl`adu`ong gˆa´pkh´uc c´oc´acca.nh song song v´oi . . tru.cto.adˆo. MNK, trong d´o M(x0,y0), K(x, y0), N(x, y) v`athu duo. c x y Z Z P (t, y0)dt + Q(x, t)dt = C. x0 y0 . . . Trong tru`ong ho. p n`ayta cho.n M(0; 1) v`ac´o √ P (x, 1) = 1 + x x2 +1; Q(x, y)=(−1+px2 + y2)y. Do d´ot´ıch phˆantˆo’ng qu´atc´oda.ng x y √ √ Z [1 + t t2 +1]dt + Z [−1+ x2 + t2]tdt = C 0 1 hay l`a 1 y2 x + (x2 + y2)3/2 − = C. 3 2 www.matheducare.com
MATH-EDUCARE 252 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan . . Nˆe´u ta cho.n M l`adiˆe’m kh´accu’am˘a.t ph˘a’ng th`ıthu duo. ckˆe´t qua’ kh´ac . kˆe´t qua’ trˆenbo’ ida.ng cu’ah˘a`ng sˆo´ t`uy ´y. N . . V´ı du. 3. Gia’iphuong tr`ınh (x + y2)dx − 2xydy =0. . Gia’i. O’ dˆay P = x + y2, Q = −2xy. Ta thˆa´y ngay phu.o.ng tr`ınh d˜acho khˆongl`aphu.o.ng tr`ınhvi phˆanto`anphˆa` n. Ta c´o ∂P ∂Q − ∂y ∂x 2y +2y 2 = = − Q −2xy x v`ado vˆa.y d ln µ 2 1 = − ⇒ ln µ = −2ln|x|⇒µ = · dx x x2 Phu.o.ng tr`ınh x + y2 xy dx − 2 dy =0 x2 x2 l`aphu.o.ng tr`ınhvi phˆanto`anphˆa` n. Vˆe´ tr´aicu’a n´oc´othˆe’ viˆe´tdu.´o.i da.ng dx 2xydy − y2dx y2 − =0⇒ d ln |x|− =0. x x2 x . . . T`u d´othuduo. c t´ıch phˆantˆo’ng qu´at x = Cey2/x. N . . V´ı du. 4. Gia’iphuong tr`ınh 2xy ln ydx +(x2 + y2py2 +1)dy =0. . Gia’i. O’ dˆa y P =2xy ln y, Q = x2 + y2py2 + 1. Ta c´o ∂Q ∂P − ∂x ∂y 2x − 2x(ln y +1) 1 d ln µ 1 1 = = − ⇒ = − ⇒ µ = · P 2xy ln y y dy y y www.matheducare.com
MATH-EDUCARE 14.1. Phu.o.ng tr`ınhvi phˆancˆa´p1 253 1 Nhˆan µ = v´o.i hai vˆe´ cu’aphu.o.ng tr`ınhd˜acho ta thu du.o.cphu.o.ng y . tr`ınhvi phˆanto`anphˆa` n 2xy ln ydx x2 + y2py2 +1 + dy =0 y y hay l`a 1 d(x2 ln y)+ypy2 +1dy =0⇒ x2 ln y + (y2 +1)3/2 = C. N 3 BAI` TAˆ. P Gia’i c´acphu.o.ng tr`ınhsau 1. (3x2 +6xy2)dx +(6x2y +4y3)dy = 0. (DS. x3 +3x2y2 + y4 = C) 2. 3xeydx +(x3ey − 1)dy = 0. (DS. x3ey − y = C) 3. e−ydx +(1− xe−y)dy = 0. (DS. y + xe−y = C) 4. 2x cos2 ydx +(2y − x2 sin 2y)dy = 0. (DS. x2 cos2 y + y2 = C) 5. (3x2 +2y)dx +(2x − 3)dy = 0. (DS. x3 +2xy − 3y = C) 6. (3x2y − 4xy2)dx +(x3 − 4x2y +12y3)dy =0. (DS. x3y − 2x2y2 +3y4 = C) x2 cos 2y 7. (x cos 2y +1)dx − x2 sin 2ydy = 0. (DS. + x = C) 2 8. (3x2ey)dx +(x3ey − 1)dy = 0. (DS. x3ey − y = C) 9. (2y − 3)dx +(2x +3y2)dy = 0. (DS. 2xy − 3x + y3 = C) x 10. (x +ln|y|)dx + 1+ + sin y)dy =0. y x2 (DS. + x ln |y| + y − cos y = C) 2 11. (3x2y2 +7)dx +2x3ydy = 0. (DS. x3y2 +7x = C) 12. (ey + yex +3)dx =(2− xey − ex)dy. www.matheducare.com
MATH-EDUCARE 254 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan (DS. xey + yex +3x − 2y = C) 13. sin(x + y)dx + x cos(x + y)(dx + dy)=0. (DS. x sin(x + y)=C) 14. e−ydx +(1− xe−y)dy = 0. (DS. xe−y + y = C) 15. (12x +5y − 9)dx +(5x +2y − 4)dy =0. (DS. 6x2 +5xy + y2 − 9x − 4y = C) 16. (3xy2 − x2)dx +(3x2y − 6y2 − 1)dy =0. (DS. 6y +12y3 − 9x2y2 +2x3 = C) x 17. (ln y − 2x)dx + − 2ydy =0. y (DS. x ln y − x2 − y2 = C) sin 2x sin2 x 18. + xdx + y − dy =0. y y2 sin2 x x2 + y2 (DS. + = C) y 2 19. (3x2 − 2x − y)dx +(2y − x +3y2)dy =0. (DS. x3 + y3 − x2 − xy + y2 = C) 1 1 20. sin y + y sin x + dx + x cos y − cos x + dy =0. x y (DS. x sin y − y cos x +ln|xy| = C) T`ım th`u.asˆo´ t´ıch phˆandˆe ’ gia’i c´acphu.o.ng tr`ınh 21. (1 − x2y)dx + x2(y − x)dy =0. 1 (DS. xy2 − 2x2y − 2=Cx, µ = ) x2 22. (x2 + y)dx − xdy =0,µ = µ(x). y 1 (DS. x − = C, µ = − ) x x2 23. (x + y2) − 2xydy =0,µ = µ(x). 1 (DS. x ln |x|−y2 = Cx, µ = ) x2 24. (2x2y +2y +5)dx +(2x3 +2x)dy =0. www.matheducare.com
MATH-EDUCARE 14.1. Phu.o.ng tr`ınhvi phˆancˆa´p1 255 1 (DS. 5arctgx +2xy = C, x =0,µ = ) 1+x2 25. (x4 ln x − 2xy3)dx +3x2y2dy =0. 1 (DS. y3 + x3(ln x − 1) = C, x =0,µ = ) x4 26. (x + sin x + siny)dx + cos ydy =0. (DS. 2ex sin y +2ex(x − 1) + ex(sin x − cos x)=C, µ = ex) 27. (2xy2 − 3y3)dx +(7− 3xy2)dy =0. 7 1 (DS. x2 − − 3xy = C, µ = ) y y2 Gia’i c´acb`aito´anCauchy sau 28. (2x + yexy)dx +(1+xexy)dy =0,y(0) = 1. (DS. x2 + y + exy =2) 29. 2x cos2 ydx +(2y − x2 sin 2y)dy =0,y(0) = 0. (DS. 2y2 + x2 cos 2y + x2 =0) 30. 3x2ey +(x3ey − 1)y0 =0,y(0) = 1. (DS. x3ey − y = −1) 2xdx y2 − 3x2 31. + dy =0,y(1) = 1. (DS. y = x) y3 y4 14.1.6 Phu.o.ng tr`ınhLagrange v`aphu.o.ng tr`ınh Clairaut . . Phuong tr`ınhvi phˆanda.ng y = xϕ(y0)+ψ(y0) (14.28) . . . . 0 0 duo. cgo.i l`aphuong tr`ınhLagrange, trong d´o ϕ(y )v`aψ(y ) l`ac´ach`am d˜a b i ˆe´tcu’a y0. . . . 0 0 Trong tru`ong ho. p khi ϕ(y )=y th`ı(14.28) c´oda.ng y = xy0 + ψ(y0) (14.29) www.matheducare.com
MATH-EDUCARE 256 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan . . . . v`aduo. cgo.i l`aphuong tr`ınhClairaut. 1+ Phu.o.ng ph´apgia’i phu.o.ng tr`ınhLagrange a) Gia’iphu.o.ng tr`ınh(14.28) dˆo´iv´o.i y0 (dˆe ’ t`ımbiˆe’uth´u.c y0 qua x . . . . . v`a y)nˆe´u ph´epgia’id´othu. chiˆe.nduo. c. Tiˆe´pdˆe´n l`agia’iphuong tr`ınh vi phˆan y0 = f(x, y). b) Phu.o.ng ph´apdu.a tham sˆo´.Du.a v`aotham sˆo´ dy = p dx . . . . v`athu duo. chˆe. th´ucth´u nhˆa´tliˆen hˆe. x, y v`atham sˆo´ p: y = xϕ(p)+ψ(p). (14.30) . . Dˆe ’ c´ohˆe. th´ucth´u hai cˆa` n thiˆe´tdˆe ’ x´acd.inh x v`a y ta lˆa´yda.o h`amhai dy vˆe´ cu’a (14.30) rˆo`i thay = p v`ao: dx dp dp p = ϕ(p)+xϕ0(p) + ψ0(p) , dx dx dp p − ϕ(p)=[xϕ0(p)+ψ0(p)] , (14.31) dx dx [p − ϕ(p)] = xϕ0(p)+ψ0(p). (14.32) dp Dˆayl`aphu.o.ng tr`ınhtuyˆe´n t´ınhdˆo´iv´o.iˆa’n h`am x. Gia’i (14.32) ta c´o . . . x = W (p, C). Nhu vˆa.y nghiˆe.mcu’aphuong tr`ınhLagrange l`a x = W (p, C), y = xϕ(p)+ψ(p). . Nhˆa. nx´et. Ch´u´yr˘a`ng khi chuyˆe’nt`u (14.31) dˆe´n (14.32) ta cˆa` n dp dp chia cho v`akhi d´o b i mˆa´t c´acnghiˆemm`ap l`ah˘a`ng sˆo´ =0. dx . . dx Do d´onˆe´u xem p khˆongdˆo’i th`ı(14.31) chı’ tho’a m˜ankhi p l`anghiˆe.m cu’aphu.o.ng tr`ınh p − ϕ(p)=0. (14.33) www.matheducare.com
MATH-EDUCARE 14.1. Phu.o.ng tr`ınhvi phˆancˆa´p1 257 . . . . . Do vˆa.yphuong tr`ınhLagrange c`onch´ua nghiˆe.mbˆa´tthu`o ng da.ng y = xϕ(α)+ψ(α) . . . . trong d´o α l`anghiˆe.mcu’a (14.33). D´ol`aphuong tr`ınhdu`o ng th˘a’ng. 2+ Phu.o.ng ph´apgia’i phu.o.ng tr`ınhClairaut 0 D˘a.t y = p ta c´o y = xp + ψ(p). (14.34) . Lˆa´yda.o h`amd˘a’ng th´uc n`aytheo x ta c´o dp dp p = p + x + ψ0(p) → dx dx dp ⇒ [x + ψ0(p)] =0⇔ dx dp =0⇔ p = const = C, (14.35) " dx x + ψ0(p)=0. (14.36) . . . . Trong tru`ong ho. p (14.35) t`u (14.34) ta c´o y = Cx+ ψC . . l`anghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınh(14.29). . . . . . . . Trong tru`ong ho. p (14.36) nghiˆe.mcu’aphuong tr`ınhClairaut duo. c . x´acd.inh bo’ i y = xp + ψ(p),) (14.37) x = −ψ0(p). . . . . . . . Phuong tr`ınhClairaut c´othˆe’ c´onghiˆe.mbˆa´tthu`ong thu duo. cbo’ i . . viˆe.ckhu’ p t`u (14.37). www.matheducare.com
MATH-EDUCARE 258 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan ´ ´ CAC VIDU. . . 0 0 V´ı du. 1. Gia’iphuong tr`ınh y =2xy +lny . 0 Gia’i. D˘a.t y = p ⇒ dy = pdx v`a y =2xp +lnp.Lˆa´yda.o h`amta c´o dp dx 1 pdx =2pdx +2xdp + ⇒ p = −2x − p dp p hay l`a dx 2 1 C 1 = − x − ⇒ x = − · dp p p2 p2 p . . . . . Thay gi´atri. x v`uathuduo. c v`aobiˆe’uth´ucdˆo´iv´oi y ta c´o C 1 2C x = − ,y=lnp + − 2. N p2 p2 p a V´ı du 2. Gia’iphu.o.ng tr`ınh y = xy0 + (a = const). . 2y0 0 Gia’i. D˘a.t y = p ta c´o a y = xp + · 2p . . Lˆa´yda.o h`amd˘a’ng th´uc n`ayrˆo`i thay dy bo’ i pdx ta c´o a a pdx = pdx + xdp − dp ⇒ x − dp =0 2p2 2p2 . . a) dp =0⇒ p = const = C ⇒ nghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınhl`a y = Cx+ a . a 2C a a b) x − =0⇒ x = .Khu’. p t`u. hai phu.o.ng tr`ınh x = 2p2 2p2 2p2 a v`a y = xp + ta c´o y2 =2ax.D´ol`anghiˆembˆa´tthu.`o.ng. N 2p . www.matheducare.com
MATH-EDUCARE 14.2. Phu.o.ng tr`ınhvi phˆancˆa´p cao 259 ` ˆ BAI TA. P Gia’i c´acphu.o.ng tr`ınhsau 1. y =2xy0 + sin y0. C cos p sin p 2C 2 cos p (DS. x = − − , y = − − sin p, y =0) p2 p2 p p p 2. y = x(1 + y0)+y02. (DS. x = 2(1 − p)+Ce−p, y = [2(1 − p)+Ce−p](1 + p)+p2) 1 3. y = xy02 − . y0 Cp2 +2p − 1 Cp2 +2p − 1 1 (DS. x = , y = − .) 2p2(p − 1)2 2(p − 1)2 p a a 4. y = xy0 + .(DS. y = Cx+ ,4y3 =27ax2) y02 C2 x2 5. y = xy0 + y02.(DS. y = Cx+ C2, y = − ) 4 C − 1 6. xy02 − yy0 − y0 + 1 = 0. (DS. y = Cx − ,(y +1)2 =4x) C 7. py02 +1+xy0 − y =0. p √ x = − , (DS. y = Cx+ 1+C2 ,  p1 − p2 ) 2 y = px + p1+p . p − ln p + C 8. y0 + y = xy02.(DS. x = , y = xp − p2) (1 − p)2 14.2 Phu.o.ng tr`ınhvi phˆancˆa´p cao . . . . Phuong tr`ınhvi phˆancˆa´p n l`aphuong tr`ınhda.ng F (x, y, y0, ,y(n)) = 0 (14.38) www.matheducare.com
MATH-EDUCARE 260 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan . . . (n) hay (nˆe´u n´ogia’iduo. cdˆo´iv´oida.o h`am y ): y(n) = f(x, y, y0, ,y(n−1)). (14.39) . . B`aito´ant`ımnghiˆe.m y = ϕ(x)cu’aphuong tr`ınh(14.39) tho’a m˜an diˆe`ukiˆe.n ban d`ˆau 0 0 (n−1) (n−1) y(x0)=y0,y(x0)=y0, ,y (x0)=y0 (14.40) . . . . . duo. cgo.il`ab`aito´anCauchy dˆo´iv´oiphuong tr`ınh(14.39). . . Nghiˆe.mtˆo’ng qu´at cu’aphuong tr`ınhvi phˆancˆa´p n (14.39) l`atˆa.p . . . ho. pmo.i nghiˆe.m x´acd.inh bo’ i cˆongth´uc y = ϕ(x, C1,C2, ,Cn) . . . ch´ua n h˘a`ng sˆo´ t`uy ´y C1,C2, ,Cn sao cho nˆe´u cho tru´o c c´acdiˆe`u . . ˜ ˜ ˜ kiˆe.n ban d`ˆau (14.40) th`ıt`ımduo. c c´ach˘a`ng sˆo´ C1, C2, ,Cn dˆe ’ y = ˜ ˜ . . ϕ(x, C1, ,Cn) l`anghiˆe.mcu’aphuong tr`ınh(14.39) tho’a m˜anc´ac diˆe`ukiˆe.n (14.40). . . . . Mo.i nghiˆe.mthuduo. ct`u nghiˆe.mtˆo’ng qu´atv´oi c´acgi´atri. cu. thˆe’ . . cu’a c´ach˘a`ng sˆo´ t`uy ´y C1,C2, ,Cn duo. cgo.il`anghiˆe. mriˆeng cu’a phu.o.ng tr`ınhvi phˆan(14.39). . . Phuong tr`ınhda.ng φ(x, y, C1,C2, ,Cn)=0x´acd.inh nghiˆe.m . . . . . . tˆo’ng qu´atcu’aphuong tr`ınhvi phˆandu´oida.ng ˆa’nduo. cgo.il`at´ıch phˆantˆo’ng qu´atcu’a phu.o.ng tr`ınh. . . 14.2.1 C´acphuong tr`ınhcho ph´epha. thˆa´pcˆa´p . . (n) Da.ng I. Phuong tr`ınhda.ng y = f(x). . . Sau n lˆa` n t´ıch phˆanta thu duo. c nghiˆe.mtˆo’ng qu´at n−1 n−1 Z Z x x y = f(x) dx ···dx +C + C + ···+ C − x + C . 1 (n − 1)! 2 (n − 2)! n 1 n | {zn } | {zn } . . V´ı du. 1. T`ım nghiˆe.m riˆengcu’aphuong tr`ınh ln x y(3) = ,y(1) = 0,y0(1) = 1,y00(1) = 2. x2 www.matheducare.com
MATH-EDUCARE 14.2. Phu.o.ng tr`ınhvi phˆancˆa´p cao 261 Gia’i. T´ıch phˆan3 lˆa` nphu.o.ng tr`ınhd˜acho ta c´o ln x ln x 1 y00 = Z dx = − − + C , x2 x x 1 1 y0 = − ln2 x − ln x + C x + C , 2 1 2 x x2 y = − ln2 x + C + C x + C . 2 1 2 2 3 . Su’ du.ng c´acdiˆe`ukiˆe.n ban d`ˆa u ta c´o C1  + C2 + C3 =0 2  1  ⇒ C1 =3,C2 = −2,C3 = · C1 + C2 =1 2  −1+C1 =2 . T`u d´osuy ra nghiˆe.m riˆengcˆa` nt`ıml`a x 3 1 y = − ln2 x + x2 − 2x + · N 2 2 2 . . . Da.ng II. Phuong tr`ınh(14.38) khˆongch´uaˆa’n h`amv`ada.oh`amdˆe´n cˆa´p k − 1: F (x, y(k),y(k+1), ,y(n))=0. Khi d´ob˘a`ng c´ach d˘a.t y(k) = p(x) . . . cˆa´pcu’aphuong tr`ınhs˜eha. xuˆo´ng k donvi . . (5) (4) V´ı du. 2. Gia’iphuong tr`ınh xy − y =0. . . . Gia’i. Phuong tr`ınhd˜acho khˆongch´uaˆa’n h`amv`ac´acda.o h`am dˆe´ncˆa´p 3. Do d´otad˘a.t y(4) = p www.matheducare.com
MATH-EDUCARE 262 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan . . v`athu duo. c dp x − p =0⇒ p = C x ⇒ y(4) = C x. dx 1 1 T´ıch phˆanliˆentiˆe´p ta c´o x2 y(3) = C + C , 1 2 2 x3 y(2) = C + C x + C , 1 6 2 3 x4 x2 y0 = C + C + C x + C , 1 24 2 2 3 4 x5 x3 x2 y = C + C + C + C x + C 1 120 2 6 3 2 4 5 5 3 2 = C1x + C2x + C3x + C4x + C5; C C C C = 1 , C = 2 , C = 3 · N 1 120 2 6 3 2 . . . Da.ng III. Phuong tr`ınh(14.38) khˆongch´uabiˆe´ndˆo.clˆa.p F (y,y0,y00, ,y(n))=0. (14.41) Khi d´ob˘a`ng c´ach d˘a.t y0 = p . . . . (trong d´o p duo. c xem l`ah`amcu’a y : p = p(y)) cˆa´pcu’aphuong tr`ınh . ha. xuˆo´ng1donvi 0 00 Dˆe ’ gia’i ta cˆa` nbiˆe’udiˆe˜nc´acda.o h`am y ,y , ,y(n) qua c´acda.o h`amcu’a h`am p = p(y). Ta c´o dy y0 = = p, dx dp dp dy dp y00 = = · = p , dx dy dx dy d dp d dp dy d2p dp 2 y000 = p = hp i = p2 + p , dx dy dy dy dx dy2 dy www.matheducare.com
MATH-EDUCARE 14.2. Phu.o.ng tr`ınhvi phˆancˆa´p cao 263 . . . . Thˆe´ c´ackˆe´t qua’ n`ayv`aovˆe´ tr´aicu’a (14.41) ta thu duo. cphuong tr`ınh cˆa´p n − 1. . . 00 02 −y V´ı du. 3. Gia’iphuong tr`ınh y + y =2e . . . . 0 Gia’i. Dˆayl`aphuong tr`ınhkhˆongch´uabiˆe´ndˆo.clˆa.p x.D˘a.t y = dp p ⇒ y00 = p v`athu du.o.c dy . dp p + p2 =2e−y. dy D´ol`aphu.o.ng tr`ınhBernoulli. B˘a`ng ph´epdˆo’iˆa’n h`am p2 = z ta thu . . . . duo. cphuong tr`ınhtuyˆe´n t´ınh dz +2z =4e−y,z= y02. dy −y −2y . 02 N´oc´onghiˆe.mtˆo’ng qu´atl`a z =4e + C1e . Thay z bo’ i y ta thu . . duo. c dy = ±p4e−y + C e−2y. dx 1 T´ach biˆe´n v`at´ıch phˆanta c´o 1 x + C = ± p4ey + C 2 2 1 C ⇒ ey + C =(x + C )2, C = 1 · e1 2 e1 4 D´ol`at´ıch phˆantˆo’ng qu´atcu’aphu.o.ng tr`ınhd˜acho. N ` ˆ BAI TA. P Gia’i c´acphu.o.ng tr`ınhsau x2 1. y(3) =60x2.(DS. y = x5 + C + C x + C ) 1 2 2 3 00 0 2. (x − 3)y − y = 0. (DS. y = C1 ln |x − 3| + C2) x4 3. y(3) = x + cos x.(DS. y = − sin x + C x2 + C x + C ) 24 1 2 3 www.matheducare.com
MATH-EDUCARE 264 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan 4. y00 = xex; y(0) = y0(0) = 0. (DS. y =(x − 2)ex + x +2) 00 0 2 5. xy = y .(DS. y = C1x + C2) 00 0 6. xy + y = 0. (DS. y = C1 ln |x| + C2) 00 2 0 x2 7. xy =(1+2x )y .(DS. y = C1e + C2) x3 8. xy00 = y0 + x2.(DS. y = + C x2 + C ) 3 1 2 (3) 00 3 9. xy − y = 0. (DS. y = C1x + C2x + C3) 00 02 10. y = y .(DS. y = C2 − ln |C1 − x|) 00 02 11. y =1+y .(DS. y = C2 − ln | cos(x + C1)|) 12. y00 + y0 +2=0;y(0) = 0, y0(0) = −2. (DS. y = −2x) 14.2.2 Phu.o.ng tr`ınhvi phˆantuyˆe´n t´ınhcˆa´p2v´o.i hˆe. sˆo´ h˘a`ng I. Phu.o.ng tr`ınhtuyˆe´nt´ınh thuˆa` nnhˆa´t . . Phuong tr`ınhvi phˆanda.ng 00 0 y + a1y + a2y = 0 (14.42) . . . . trong d´o a1, a2 l`ac´ach˘a`ng sˆo´,duo. cgo.i l`aphuong tr`ınhtuyˆe´n t´ınh . thuˆa` n nhˆa´tcˆa´p2v´oihˆe. sˆo´ h˘a`ng. y1(x) 1. Nˆe´u y1 v`a y2 l`ac´acnghiˆe.m riˆengcu’a (14.42) sao cho =6 y2(x) const th`ı y = C1y1 + C2y2 l`anghiˆe.mtˆo’ng qu´atcu’a (14.42). . 2. Nˆe´u y = u(x)+iv(x) l`anghiˆe.mcu’a (14.42) th`ıphˆa` n thu. c u(x) v`aphˆa` na’o v(x)c˜ung l`anghiˆe.m. Dˆe ’ x´acd.inh c´acnghiˆe.m riˆeng y1(x)v`ay2(x)d`ˆau tiˆencˆa` n gia’i . . . . . λx phuong tr`ınhd˘a.c trung (ptdt) thu duo. cb˘a`ng c´ach d˘a.t y = e 2 λ + a1λ + a2 =0. (14.43) www.matheducare.com
MATH-EDUCARE 14.2. Phu.o.ng tr`ınhvi phˆancˆa´p cao 265 . Hiˆe’n nhiˆenta luˆonxem bˆo.icu’a nghiˆe.mcu’a (14.43) l`a r =1(don) . . ho˘a.c 2 (k´ep).Ta c˜ungquy u´o cbˆo.i r =0nˆe´u λ khˆongl`anghiˆe.mcu’a (14.43). Ta c´oba’ng t´omt˘a´t sau Nghiˆe.m riˆeng Nghiˆe.mtˆo’ng qu´at Nghiˆe.mcu’a ptdt cu’a (14.42) cu’a (14.42) y = eλ1x, 1 λ1x λ2x I. λ1,λ2 ∈ R, λ1 =6 λ2 y = C1e + C2e λ2x y2 = e y = eλx, y = C eλx + C xeλx II. λ = λ = λ ∈ R 1 1 2 1 2 λx λx y2 = xe = e (C1 + C2x) αx αx III. λ1,λ2 ∈ C y1 = e cos βx y = e (C1 cos βx αx λ = α ± iβ y2 = e sin βx +C2 sin βx) . . V´ı du. 1. T`ım nghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınh y00 + y0 − 2y =0. 0 00 Gia’i. T`ım nghiˆe.m trong da.ng y = eλx, ta c´o y = λeλx, y = λ2eλx. . . . . . . . Thay v`aophuong tr`ınhta thu duo. cphuong tr`ınhd˘a.c trung λ1 =1, λ2 + λ − 2=0⇔ h λ2 = −2. . . . . Ca’ hai nghiˆe.m λ1,λ2 ∈ R v`akh´acnhau nˆentheo tru`o ng ho. pIo’ ba’ng x −2x ta c´o y1 = e , y2 = e v`ado d´o x −2x y = C1e + C2e . N . . . . V´ı du. 2. C˜ung ho’inhutrˆendˆo´iv´oiphuong tr`ınh y00 +2y0 + y =0. www.matheducare.com
MATH-EDUCARE 266 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan . . . . . . Gia’i. Phuong tr`ınhd˘a.c tru. ng tuong ´ung c´oda.ng 2 λ − 2λ +1=0⇔ λ1 = λ2 =1. . . . . C´acnghiˆe.mcu’aphuong tr`ınhd˘a.c trung thu. cv`ab˘a`ng nhau nˆen y1 = x x e , y2 = xe .Dod´o x x x y = C1e + C2xe = e (C1 + C2x). N . . V´ı du. 3. T`ım nghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınh y00 − 4y0 +13y =0. . . . Gia’i. Ptdttuong ´ung c´oda.ng λ1 =2+3i, λ2 − 4λ +13=0⇔ h . λ2 =2− 3i. . . . . . C´acnghiˆe.mph´uc n`aytuong ´ung v´oi c´acnghiˆe.m riˆengdˆo.clˆa.p tuyˆe´n 2x 2x t´ınhl`a y1 = e cos 3x, y2 = e sin 3x.Dod´onghiˆe.mtˆo’ng qu´avtc´o da.ng 2x 2x y = C1e cos 3x + C2e sin 3x 2x = e (C1 cos 3x + C2 sin 3x). N II. Phu.o.ng tr`ınhtuyˆe´nt´ınh khˆongthuˆa` nnhˆa´t . . Phuong tr`ınhda.ng 00 0 y + a1y + a2y = f(x) (14.44) . . . . trong d´o a1, a2 l`anh˜ung h˘a`ng sˆo´ thu. c, f(x) l`ah`amliˆentu.cduo. cgo.i . l`a ptvp tuyˆe´n t´ınhkhˆongthuˆa`n nhˆa´tv´oihˆe. sˆo´ h˘a`ng. D- .inh l´y. Nghiˆe. mtˆo’ng qu´atcu’a (14.44) l`atˆo’ng cu’a nghiˆe. mtˆo’ng qu´at . . . . . cu’a phuong tr`ınhthuˆa`n nhˆa´ttuong ´ung v`amˆo. t nghiˆe. mriˆeng n`aod´o cu’a phu.o.ng tr`ınhkhˆongthuˆa`n nhˆa´t (14.44). www.matheducare.com
MATH-EDUCARE 14.2. Phu.o.ng tr`ınhvi phˆancˆa´p cao 267 . . T`u Iv`ad.inh l´yv`ua ph´atbiˆe’u suy r˘a`ng b`aito´ant`ımnghiˆe.mtˆo’ng . . . qu´atcu’a (14.44) duo. cduavˆ`e b`aito´ant`ımnghiˆe.m riˆeng˜y cu’a n´o.N´oi . . . . chung ph´ept´ıch phˆan(14.44) luˆonluˆonc´othˆe’ thu. chiˆe.n nh`o phuong ph´apbiˆe´n thiˆenh˘a`ng sˆo´ Lagrange. Nˆe´uvˆe´pha’ic´oda. ng d˘a. cbiˆe.t ta c´othˆe’ t`ımnghiˆe.m riˆengcu’a (14.44) . . . . . . nh`o Phuong ph´apcho. n (phuong ph´aphˆe. sˆo´ bˆa´td.inh khˆongch´ua qu´a tr`ınht´ıch phˆan).Ta c´oba’ng t´omt˘a´t sau trong d´o Pn(x),Qm(x), . . . . l`ac´acdath´ucda.isˆo´ bˆa.ctuong ´ung. Vˆe´ pha’icu’a (14.44) Nghiˆe.mcu’a ptdtDa.ng nghiˆe.m riˆeng r I. Sˆo´ 0 l`anghiˆe.m˜y = Qn(x)x . . f(x)=Pn(x)bˆo.i r cu’a ptdt Qn cˆa` nduo. c x´acd.inh r αx II. Sˆo´ α l`anghiˆe.m˜yx Qn(x)e αx . . f(x)=e Pn(x)bˆo.i r cu’a ptdt Qn(x)cˆa` nduo. c x´acd.inh III. f(x)= Sˆo´ ph´u.c iβ l`ay ˜ = xr(A cos βx a cos βx nghiˆe.mbˆo.i r +B sin βx); A, +b sin βx cu’a ptdt B -cˆa` nx´acd.inh IV. f(x)=eαx× Sˆo´ α + iβ l`ay ˜ = xreαx× [Pn(x) cos βx+ nghiˆe.mbˆo.i r [Q1(x) cos βx+ Qm(x) sin βx]cu’a ptdt Q2(x) sin βx], Q1 v`a Q2 . l`adath´ucbˆa.c s = max(m, n) V. Vˆe´ pha’icu’aphu.o.ng tr`ınhl`atˆo’ng cu’a hai h`am 00 0 y + a1y + a2y = f1(x)+f2(x). www.matheducare.com
MATH-EDUCARE 268 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan . . Khi d´onghiˆe.m riˆengc´othˆe’ t`ımdu´oida.ng y˜ =˜y1 +˜y2 . . 00 0 trong d´oy ˜ 1 l`anghiˆe.m riˆengcu’aphuong tr`ınh y + a1y + a2y = f1(x), . . 00 0 c`on˜y2 l`anghiˆe.m riˆengcu’aphuong tr`ınh y + a1y + a2y = f2(x). . . 00 0 V´ı du. 4. Gia’iphuong tr`ınh y − 2y + y = x + 1 (da.ng I) . . . . . Gia’i. Nghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınhthuˆa` n nhˆa´ttuong ´ung x c´oda.ng Y = e (C1 + C2x). V`ı λ1 = λ2 =1nˆen sˆo´ 0 khˆongl`anghiˆe.m . . cu’a ptdtv`adod´o r = 0 v`anghiˆe.m riˆengcu’aphuong tr`ınhd˜a c h o . . . . . (tru`ong ho. p I) cˆa` n t`ımdu´oida.ng y˜ = x0(Ax + B) . 0 00 trong d´o A, B l`anh˜ung h˘a`ng sˆo´ cˆa` nx´acd.inh. Thˆe´ y˜ v`a˜y ,˜y v`ao . . . phuong tr`ınhv`aso s´anhc´achˆe. sˆo´ cu’a c´acl˜uyth`uac`ung bˆa.ccu’a x ta . . thu duo. c A =1,−2A + B =1⇒ A =1,B = 3. Do vˆa.y˜y = x +3 v`a x y =˜y + Y =3+x + e (C1 + C2x). N . . 00 0 x V´ı du. 5. Gia’iphuong tr`ınh y − 4y +3y = xe (da.ng II). . . . . . Gia’i. Nghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınhthuˆa` n nhˆa´ttuong ´ung x 3x x c´oda.ng Y = C1e + C2e .V`ıvˆe´ pha’i f(x)=xe nˆen(xem II) ta . . . . c´o Pn(x)=x, α =1v`anhuvˆa.ysˆo´ α = 1 l`anghiˆe.mdoncu’aphuong . . . tr`ınhd˘a.c trung; r = 1. Do vˆa.y nghiˆe.m riˆengcˆa` nt`ımdu´oida.ng y˜ =(Ax + B)xex 1 T´ınhy ˜0,˜y00 rˆo`ithˆe´ v`aophu.o.ng tr`ınhd˜acho ta thu du.o.c A = − , . 4 1 B = − v`at`u. d´o 4 1 y =˜y + Y = − (x +1)xex + C ex + C e3x 4 1 2 1 = C ex + C e3x − x(x +1)ex. N 1 2 4 www.matheducare.com
MATH-EDUCARE 14.2. Phu.o.ng tr`ınhvi phˆancˆa´p cao 269 . . 00 V´ı du. 6. Gia’iphuong tr`ınh y + y = sin x (da.ng III). . . . 2 Gia’i. Phuong tr`ınh d˘a.c trung λ + 1 = 0 c´onghiˆe.m λ1 = i, . . λ2 = −i.Dod´onghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınhthuˆa` n nhˆa´tl`a Y = C1 cos x+C2 sin x.V`ı f(x)=sinx =0·cos x+1·sin x nˆen a =0, . . . . b =1,β =1.V`ı iβ = i l`anghiˆe.mdoncu’aphuong tr`ınhd˘a.c trung . . nˆen r = 1 v`anghiˆe.m riˆengcˆa` nt`ımdu´oida.ng y˜ =(A cos x + B sin x)x. 1 Thˆe´ y˜ v`aophu.o.ng tr`ınhta thu du.o.c A = − , B =0v`a . 2 1 y = − cos x + C cos x + C sin x. N 2 1 2 . . 00 V´ı du. 7. X´et phuong tr`ınh y + y = sin 2x (da.ng III). . . . . Gia’i. Tu ong tu. nhu trong v´ıdu. 6 ta c´o Y = C1 cos x + C2 sin x. . . Phuong tr`ınhd˜acho c´o β =2.V`ıiβ =2i khˆongl`anghiˆe.mcu’a . . . phuong tr`ınhd˘a.c trung nˆen r =0v`a y˜ = A cos 2x + B sin 2x 1 Thˆe´ y˜ v`aophu.o.ng tr`ınhd˜achoc`ung v´o.i˜y0,˜y00 ta c´o A =0,B = − . 3 Do d´o 1 y =˜y + Y = − sin 2x + C cos x + C sin x. N 3 1 2 . . 00 0 −x V´ı du. 8. Gia’iphuong tr`ınh y − 2y + y = sin x + e (da.ng V). . . . Gia’i. Phuong tr`ınhd˘a.c trung c´onghiˆe.m λ1 = λ2 = 1. Do vˆa.y . . . . . nghiˆe.mtˆo’ng qu´atcu’aphuong tr`ınhthuˆa` n nhˆa´ttuong ´ung c´oda.ng x Y = e (C1 + C2x). www.matheducare.com
MATH-EDUCARE 270 Chu.o.ng 14. Phu.o.ng tr`ınhvi phˆan V`ıvˆe´ pha’icu’aphu.o.ng tr`ınhd˜acho l`atˆo’ng cu’a hai h`amsin x v`a −x . . e nˆennghiˆe.m riˆengcˆa` n t`ımdu´oida.ngy ˜ =˜y1 +˜y2, trong d´oy ˜ 1 l`a nghiˆe.m riˆengcu’a y00 − 2y0 + y = sin x (14.45) . . c`on˜y2 l`anghiˆe.m riˆengcu’aphuong tr`ınh y00 − 2y0 + y = e−x (14.46) T`ımy ˜1.V`ı f(x) = sin x ⇒ β = 1. Tiˆe´pd´ov`ıiβ = i khˆongl`a . . . nghiˆe.mcu’aphuong tr`ınhd˘a.c trung nˆen r = 0 v`anghiˆe.m riˆeng˜y cˆa` n . . t`ımdu´oida.ngy ˜1 = A cos x + B sin x. 1 Thayy ˜ ,˜y0 ,˜y00 v`ao(14.45) ta thu du.o.c A = , B =0: 1 1 1 . 2 1 y˜ = cos x. 1 2 −x T`ımy ˜2.Vˆe´ pha’i f(x)=e (xem II). Sˆo´ α = −1 khˆongl`anghiˆe.m . . . . . cu’aphuong tr`ınhd˘a.c trung nˆen r =0nˆen nghiˆe.m riˆengcˆa` nt`ımdu´oi da.ng −x y˜2 = Ae . 1 Thayy ˜ ,˜y0 ,˜y00 v`ao(14.46) ta thu du.o.c A = v`ado vˆay 2 2 2 . 4 . 1 y˜ = e−x. 2 4 1 1 Nhu. vˆay˜y =˜y +˜y = cos x + e−x v`a . 1 2 2 4 1 1 y =˜y + Y = cos x + e−x + ex(C + C x). N 2 4 1 2 BAI` TAˆ. P www.matheducare.com