Chaptr Ordinar Diffrntial Equations of th First Ordr and First Dgr Gnral form:., d +, d 0.a. f,.b I. Sparabl Diffrntial quations Form: d + d 0 C d d E 9 + 4 0 Solution: 9d + 4d 0 9 + 4 C E + d Solution: d tan + C tan + C E3 + 5 4 0, 0 d 4 Solution: 5 d 5 C 0 C Th solution is 5 + Southrn Taiwan Univrsit E4, 0 d Solution: d ln C + C C 0 C 0 C Th solution is O. D. E. Chaptr_4
II. Rducibl to sparabl diffrntial quations. Homognous quation Dfinition: If ft, t t r f,, thn f, is a homognous function of dgr r. Substituting t / into ft, t t r f,, w hav r r f, f, f, f, F. r In particular r 0, thn f, F If, d +, d 0, whn, and, ar of th sam dgr in and, / is a homognous function of dgr zro, and th diffrntial quation can b writtn as d F. d. Lt u, quation. bcoms du d du u F u is sparabl. d F u u E5 + 0 Solution: Lt u, th quation bcoms uu + u u + 0 uu + u u + 0 uu + u + 0 udu d ln + u ln + C + u C/ u + C + C Southrn Taiwan Univrsit E6 + cos d cos d 0 Solution: Lt u + ucosud cosuud + du 0 cosudu d sinu ln + C sin ln C O. D. E. Chaptr_5
., and, ar linar in and Form: a + b + c d + a + b + c d 0.3 If a a b, lt X +, Y +, quation.3 bcoms b [a X + b Y + a + b + c ]dx + [a X + b Y + a + b + c ]dy 0..4 a b c 0 W choos a b c 0 Thn quation.4 rducs to a X + b Y dx + a X + b Y dy 0 is a homognous quation. If a a b c k, quation.3 bcoms b c [ka + b + c ] d + a + b + c d 0..5 Lt v a + b d dv a d, quation.5 bcoms b kv + c d + v + c dv a d 0 b a v c [ kv c v c ] d dv 0 is a sparabl on. b b 3 If Southrn a b c Taiwan Univrsit k, quation.3 bcoms a b c ka + b + c d + a + b + c d 0..6 a. If a + b + c 0, w gt onl a trivial solution. b. If a + b + c 0, quation.6 rducs to kd + d 0 k + C. O. D. E. Chaptr_6
E7 4 + 3 + d + + + d 0 4 3 0 Solution:, 3 0 Lt X, Y + 3, w hav 4X + 3YdX + X + YdY 0 Lt Y ux 4X + 3uXdX + X + uxudx + Xdu 0 dx u 4 4u u dx X u du 0 du 0 X u dx du X u C X u u 0 ln ln u ln[ X u ] C ln C u E8 4 + 5 + + 3 0 u Solution : Lt u u u Th quation bcoms u 5 u 3 0 u 5 u u 6 0 4u u 5 u 0 u 5 4u d u 5 du 0 d du 0 4u / / d du 0 d du C u 4 4u u ln4u C 8 4u ln4u 8C 8 8 4 ln[4 ] 8C 4 8 ln4 8 C Southrn Taiwan Univrsit E9 d + d 0 Solution: Dividing b d + d 0 + C O. D. E. Chaptr_7
3. Tp of fa + b + c.7 du a Lt u a + b + c u a + b d, quation.7 bcoms b du a d du du du f u a bf u a bf u d is sparabl. b d d a bf u E0 + 7.8 Solution: Lt u + 7 u +, quation.8 has th rsult u u du d tan u + C + 7 tan + C u [Erciss]. + 4. sin [Answrs]. ln + + C. tan + C 4. Isobaric quations Southrn Taiwan Univrsit Dfinition: If ft, t m, t m t r f,,, w sa that f,, is an isobaric function of wight r. In particular, if t /, thr follows f,, f,, or m m r r r f,, f,, F,..9 m m m m A diffrntial quation can b put into th form.9, thn F, 0. m m O. D. E. Chaptr_8
Solving this for, and multipling b m, w hav m m m m d m du m Lt u u mu,.0 is rducibl to m d d m du m m du d mu u is sparabl. d u mu [ot] If th wights m and m ar assignd to and rspctivl, th trm a b c has th.0 wight a + bm + cm. If an prssion is to b isobaric, all its trms must b isobaric and of th sam wight. E 3 + 4. Solution: Th wights of ach trm ar 3 + m, 0, 0, + m, if m, vr trm has th sam wight. Lt u u u 3, th quation bcoms 3 uu 3 + 4u u4u+ 4u du d 4u 4u ln 4u ln C du d d 4u 4u 4 C 4u 4u d E + + 0. Solution: wight m,m, + m + m, so m m m. Lt u u u Southrn Taiwan Univrsit u + + u u u 0 u + u u + uu u 0 u du d d u u u du ln u u ln C u u u ln u C ln C O. D. E. Chaptr_9
E3 3 + 0. Solution: wight 3 + m, + m, m, so m + m m, Lt u, u + u 3 u + u u u u + u u u du d d u 0 du 0 ln C u u u u u / C C / [Erciss]. + + 0. 3 + 4 + 0 3. + + + 3 0 4. + 4 4 [Answrs]. + C. + 4 C 3. + 3 C 4. C C III. Eact diffrntial quations. Eact diffrntial quations A first ordr diffrntial quations, d +, d 0 is calld act if its lft sid is u u du d d, its solution is u C [Condition] Sinc u u u u,, b th assumption of continuit, w hav Southrn Taiwan Univrsit. Th solution can b obtaind from th following : u u and from d k, w can dtrmin k. O. D. E. Chaptr_0
E4 3 + 3 d + 3 + 3 d 0. 3 3 3 3 Solution: Sinc 6, it is act, and u 3 + 3 d + k 4 /4 + 3 / + k u 3 + k 3 + 3, k 4 /4 th solution is 4 /4 + 3 / + 4 /4 C E5 sin cosh d cos sinh d 0, 0 0. Solution: Sinc sin cosh cos sinh sinsinh, it is act, and u sin cosh d + k cos cosh + k u cos sinh + k cos sinh k C u cos cosh C 0 0 cos0 cosh0 C C, Th solution is cos cosh d [Erciss ]. d d. 0 d 3. 0 3 d d 0 [Answrs]. + C C. log tan C 3. log C Southrn Taiwan Univrsit. Intgrating factors If, d +, d 0 is not act, w multipl b,,, d +, d 0 is act. Th function, is calld an intgrating factor. O. D. E. Chaptr_
O. D. E. Chaptr_ [Cas ]If,, from th condition of actnss, w gt d d d hav w, on dpnds onl If [Cas ] If,, from th condition of actnss, w gt d d d hav w, on onl dpnds If E6 d d 0. C C f f f u f f d f d u d ln ln Th solution is 0 0,, Solution : Southrn Taiwan Univrsit
E7 d + 4 + 3 d 0, 0..5 / / 6 Solution: Sinc d dpnds onl on, intgrating factor is and u 3 d + k 3 + k u 3 + k 4 + 3 k, and th gnral solution is 3 + C Substituting 0..5, w hav 0..5 3 +.5 4 C C 4.975 Th solution is 3 + 4.975 [Cas 3] Form: r s md + nd0. Sinc d a b a b ad + bd,. has an intgrating factor, whr +r m, + s n. or gnrall, suppos is th intgrating factor of th quation r s md + nd + p q ud + vd 0..3 ultipling.3 b and rarranging th trms, w hav m +r +s+ + u +p +q+ d + n +r+ +s +v +p+ +q d 0. This is act, providd, i.., m + s + +r +s + u + q + +p +q n + r + +r +s + v + p + +p +q for all valus of and, w hav Southrn m + s + n + r +, Taiwan u + q + v + p +. Univrsit Th two quations can b solvd for and. O. D. E. Chaptr_3
E8 4 3d + d + 4d + 3d 0 Solution: 3 4 + 4 d + 5 + 3 3 d 0 ultipling 3 +4 + + 4 + + d + +5 + + 3 +3 d 0 3 +4 + + 4 + +, +5 + + 3 +3 To b act 3 + +4 + + 4 + + + 5 +4 + + 3 + 3 + 3 + + 5, 4 + 3 + 3, u d + f 3 5 4 + 4 3 3 d + f 6 4 + 4 3 + f u 6 3 + 3 4 + f 6 3 + 3 4 f 0 f 0 Th solution is 6 4 + 4 3 C E9 d + d + 3d + d 0 Solution: ultipling and rarranging th trms, w gt + + 3 + + d + + + + + d 0 For actnss + + 3 + + + + + + + + + +, 3 + +, 0 u + 3 d + k + 3 + k u + 3 + k + 3 k C Th gnral solution is + 3 C Southrn Taiwan Univrsit O. D. E. Chaptr_4
IV. Linar diffrntial quations of first ordr. Form: + p r.4 If r 0,.4 is said to b homognous, othrwis it is said to b nonhomognous..4 is quivalnt to d [F] Fr.5 d F + F Fr + F F r Comparing with.4, w hav F df p pd F F F Intgrating.5, w gt th solution: pd F Frd + C F Frd + CF pd pd rd C or h [ h rd + C], whr h pd E0 Solution: h d [ d + C] + C E + 3sin + cos Solution: h d [ 3sin + cosd + C] sin + C Southrn Taiwan Univrsit E + tan sin, 0 Solution: h tand lnsc lnsc [ lnsc sind + C] cossind + Ccos cos + Ccos 0 cos 0 + Ccos0 C 3 cos + 3cos O. D. E. Chaptr_5
. Rducibl to linar form: Brnoulli quations Form: + p g a a is an ral numbr.6.6 a + p a g.7 Lt u a u a a,.7 bcoms u + pu g u + apu ag a This is linar in u. E3 A B Solution: A B A, B ar constants. Lt u u, thr follows u Au B h B / A C A Ad A u A A Bd C B A C A E4 + Solution: Th quation can b writtn in th form d which is linar in. d h d, [ d + C] + C Southrn Taiwan Univrsit O. D. E. Chaptr_6