Wha is a differenial equaion? A differenial equaion is any equaion conaining one or more derivaives. The simples differenial equaion, herefore, is jus a usual inegraion problem y f (). Commen: The soluion of he above is, of course, he indefinie inegral of f (), y F() + C, where F() is any aniderivaive of f () and C is an arbirary consan. Such a soluion is called a general soluion of he differenial equaion. I is a general form of a se of infiniely many funcions, each differs from ohers by one (or more) consan erm and/or consan coefficiens, which all saisfy he differenial equaion in quesion. Every differenial equaion, if i does have a soluion, always has infiniely many funcions saisfying i. All of hese soluions, differing from one anoher by one, or more, arbirary consan / coefficien(s), are given by he formula of he general soluion. Addiional auxiliary condiion(s), which migh appear as a problem demands, will be required o narrow down he soluion se o one or a few specific funcions from he formula of he general soluion. 008, 01 Zachary S Tseng A-1-1
Classificaion of Differenial Equaions Ordinary vs. parial differenial equaions An ordinary differenial equaion (ODE) is a differenial equaion wih a single independen variable, so he derivaive(s) i conains are all ordinary derivaives. A parial differenial equaion (PDE) is a differenial equaion wih wo or more independen variables, so he derivaive(s) i conains are parial derivaives. Order of a differenial equaion The order of a differenial equaion is equal o he order of he highes derivaive i conains. Examples: (1) y + y 5 e (firs order ODE) () cos() y sin() y 3 cos() (firs order ODE) (3) y 3y + y e cos(5) (second order ODE) (4) y (4) + (y ) 30 0 (fourh order ODE) (5) u xx 4u + u (second order PDE) (6) y (5) (y y ) + y 4e 7 (fifh order ODE) 008, 01 Zachary S Tseng A-1 -
Linear vs. nonlinear differenial equaions: An n-h order ordinary differenial equaion is called linear if i can be wrien in he form: y (n) a n 1 () y (n 1) + a n () y (n ) + + a 1 () y + a 0 () y + g(). Where he funcions a s and g are any funcions of he independen variable, in his insance. Noe ha he independen variable could appear in any shape or form in he equaion, bu he dependen variable, y, and is derivaives can only appear alone, in he firs power, no in a denominaor or inside anoher (ranscendenal) funcion. In oher words, he righ-hand side of he equaion above mus be a linear funcion of he dependen variable y and is derivaives. Oherwise, he equaion said o be nonlinear. In he examples above, () and (3) are linear equaions, while (1), (4) and (6) are nonlinear. (5) is a linear parial differenial equaion, as each of he parial derivaives appears alone in he firs power. The nex example looks similar o (3), bu i is a (second order) nonlinear equaion, insead. Why? (7) y 3y + y e cos(5y) 008, 01 Zachary S Tseng A-1-3
Exercises A-1.1: 1 10 Deermine he order of each equaion below. Also deermine wheher each is a linear or nonlinear equaion. 1. y + 3 y cos( ). y + 11y y + e 6 y y ln 3. y y 5 4. 5y 5 y 5. y 3 + sec() y (6) y 6. y + 5y + 4y e 4 y 7. (y ) 8 y 1 8. y cos(y) 3 sin() y (5) 9. e y (4) + 3y co(e ) y 6 + y 4 d 10. ( y ) d 4 ye 8 11. For wha value(s) of n will he following equaion be linear? y 9y n n sin(3n) Answers A-1.1: 1. 1s order, linear;. 3rd order, linear; 3. nd order, nonlinear; 4. 1s order, linear; 5. 6h order, nonlinear; 6. nd order, linear; 7. 1s order, nonlinear; 8. 5h order, nonlinear; 9. 4h order, linear; 10. 5h order, linear; 11. When n 0 or 1, he equaion is linear. 008, 01 Zachary S Tseng A-1-4
Direcion Field (a.k.a. Slope Field) Direcion field is a simple visualizaion ool ha could be used o sudy he approximaed behavior of he soluions of a firs order differenial equaion wihou having o solve i firs. y f (, y), Wha is i? Firs draw a grid on he y-plane. Then for each poin ( 0, y 0 ) on he grid compue he value f ( 0, y 0 ). Noe ha f ( 0, y 0 ) y is acually he insananeous rae of change of a soluion, y φ(), of he given equaion a he poin ( 0, y 0 ). I, herefore, represens he slope of he line angen o he soluion whose curve passes hrough ( 0, y 0 ), a he exac poin. Draw a shor arrow a each such poin ( 0, y 0 ) ha is poining in he direcion given by he slope of he angen line. Afer an arrow is drawn for every poin of he grid, we can do connecing-he-dos and race curves by connecing one arrow o he nex arrow in he grid where he firs is poining a. Those curves raced his way are called inegral curves (so called because, in effec, hey each approximaes an aniderivaive of he funcion f (, y)). Each inegral curve approximaes he behavior of a paricular funcion ha saisfies he given differenial equaion. The collecion of all inegral curves approximaes he behaviors of he general soluion of he equaion. Example: y Wha we are doing is consrucing he graphs of some funcions ha saisfy he given differenial equaion by firs approximaing each soluion funcion s local behavior a a poin ( 0, y 0 ) using is linearizaion (i.e. he angen line approximaion). Then we obain he longer-erm behavior by connecing hose local approximaions, poin-by-poin moving among he grid, ino curves ha are fairly accuraely resemble he acual graphs of hose funcions. We will look a his ool in more deails in a laer secion, when we sudy Auonomous Equaions. 008, 01 Zachary S Tseng A-1-5
Figure 1. The direcion field of y 008, 01 Zachary S Tseng A-1-6
Figure. The direcion field of y (wih a few inegral curves raced approximaing curves of he form y + C). 008, 01 Zachary S Tseng A-1-7
Commen: Each inegral curve, represening a specific funcion ha saisfies he given differenial equaion colloquially, such funcion is called a paricular soluion of he equaion is analogously a cerain aniderivaive. (In his presen example, i is acually an aniderivaive, ha of f (, y).) The enire direcion field shows he differen behaviors of a collecion of hose paricular soluions. In oher words, i gives us a rough idea abou he general soluion of he differenial equaion. The direcion field, in is enirey, is hus analogous o he indefinie inegral of f (, y). Nex, le us examine a slighly more ineresing direcion field of anoher simple firs order differenial equaion, y y. Even wihou knowing wha is general soluion is (ye), we can neverheless readily deduce from is direcion field he long-erm behavior of is soluions, which all seem o behave like he line y 1. 008, 01 Zachary S Tseng A-1-8
Figure 3. Anoher example: he direcion field of y y Commen: We will learn shorly how o solve his equaion. The exac soluions are funcions of he form y 1 + Ce. When C 0, he soluion is jus he line y 1, which appears as he slan asympoe of all oher soluions in he above graph. 008, 01 Zachary S Tseng A-1-9
Firs Order Linear Differenial Equaions A firs order ordinary differenial equaion is linear if i can be wrien in he form y + p() y g() where p and g are arbirary funcions of. This is called he sandard form or canonical form of he firs order linear equaion. We ll sar by aemping o solve a couple of very simple equaions of such ype. 008, 01 Zachary S Tseng A-1-10
Example: Find he general soluion of he equaion y y 0. Firs le s rewrie he equaion as dy d y. Then, assuming y 0, divide boh sides by y: Muliply boh sides by d: 1 dy y d dy y d Now wha we have here are wo derivaives which are equal. I implies (as a consequence of he Mean Value Theorem) ha he aniderivaives of he wo sides mus differ only by a consan of inegraion. Inegrae boh sides: ln y + C or, Where C 1 e C y e ( + C ) e C e C 1 e is an arbirary, bu always posiive consan. To simplify one sep farher, we can drop he absolue value sign and relax he resricion on C 1. C 1 can now be any posiive or negaive (bu no zero) consan. Hence y() C 1 e, C 1 0. (1) 008, 01 Zachary S Tseng A-1-11
Lasly, wha happens if our eariler assumpion ha y 0 is false? Well, if y 0 (ha is, when y is he consan funcion zero), hen y 0 and he equaion is reduced o 0 0 0 which is an expression ha is always rue. Therefore, he consan zero funcion is also a soluion of he given equaion. No exacly by a coinciden, i corresponds o he missing case of C 1 0 in (1). As a resul, he general soluion is in he form y() C e, for any consan C. Tha is, any funcion of his form, regardless of he value of C, will saisfy he equaion y y 0. While here are infiniely many such funcions, no oher ype of funcions could saisfy he equaion. The similar echnique could also be used o solve his nex example. 008, 01 Zachary S Tseng A-1-1
Example: For arbirary consans r and k, r 0, solve he equaion y r y k. We will proceed as before o rewrie he equaion ino equaliy of wo derivaives. Then inegrae boh sides. Assuming r y + k 0: dy d ry+ k dy ry+ k d dy ry+ k d Therefore, 1 ln r ry+ k + C Simplifying: ln ry + k r + C 1 ry + k r e + C 1 ry k C + e r e 1 C1, where e is an arbirary posiive consan. Dropping he absolue value sign: r C1 + k C e, C ±e is any nonzero consan. ry 1 r Tha is, y ( C e k) r C r e r k r. 008, 01 Zachary S Tseng A-1-13
Lasly, i can be easily checked ha if r y + k 0, implying ha y is k he consan funcion, he given differenial equaion is again r saisfied. This consan soluion corresponds o he above general soluion for he case C 0. Hence, he general soluion now includes all possible values of he unknown arbirary consan: C r k y e, C is any consan. r r 008, 01 Zachary S Tseng A-1-14
The Inegraing Facor Mehod In he previous examples of simple firs order ODEs, we found he soluions by algebraically separae he dependen variable- and he independen variable- erms, and wrie he wo sides of a given equaion as derivaives, each wih respec o one of he wo variables. Then jus inegrae boh sides and simplify o find he soluion y. However, his process was feasible only because he equaions in quesion were a special ype, namely ha hey were boh separable, in addiion o being firs order linear equaions. They do, however, illusraed he main goal of solving a firs order ODE, namely o use inegraion o removed he y -erm. Mos firs order linear ordinary differenial equaions are, however, no separable. So he previous mehod will no work because we will be unable o rewrie he equaion o equae wo derivaives. In such insances, a more elaborae echnique mus be applied. How do we, hen, inegrae boh sides? Le s look again a he firs order linear differenial equaion we are aemping o solve, in is sandard form: y + p() y g(). Wha we will do is o muliply he equaion hrough by a suiably chosen funcion µ(), such ha he resuling equaion µ() y + µ()p() y µ()g() (*) would have inegrae-able expressions on boh sides. Such a funcion µ() is called an inegraing facor. 008, 01 Zachary S Tseng A-1-15
Commen: The idea of inegraing facor is no really new. Recall how you have inegraed sec(x) in Mah 141. The inegral as given could no be inegraed. However, afer he inegrand has been muliplied by a suiable from of 1, in his case (an(x) + sec(x))/ (an(x) + sec(x)), he inegraion could hen proceed quie easily. an x+ sec x an x+ sec x sec x an x+ sec sec x+ an x sec x dx sec x dx dx x du u ln u + C ln sec x+ an x + C Now back o he equaion µ() y + µ()p() y µ()g() (*) On he righ side here is explicily a funcion of. So i could always, in heory a leas, be inegraed wih respec o. The lef hand side is he more ineresing par. Take anoher look of he lef side of (*) and compare i wih his following expression lised side-by-side: µ() y + µ()p() y µ() y + µ () y The second expression is, by he produc rule of differeniaion, nohing more han (µ() y). Noice he similariy beween he wo expressions. Suppose he simple differenial equaion µ()p() µ () could be saisfied, we would hen have µ() y + µ()p() y µ() y + µ () y (µ() y) Trivially, hen, he lef side of (*) could be inegraed wih respec o. (µ() y + µ()p() y) d (µ() y) d µ() y 008, 01 Zachary S Tseng A-1-16
Hence, o solve (*) we inegrae boh sides: (µ() y + µ()p() y) d µ()g() d µ() y µ()g() d (**) Therefore, he general soluion is found afer we divide he las equaion hrough by he inegraing facor µ(). Bu before we can solve for he general soluion, we mus ake a sep back and find his (almos magical!) inegraing facor µ(). We have seen on he las page ha i mus saisfies he equaion µ()p() µ (). This is a simpler equaion ha can be solved by our firs mehod of separae he variables hen inegrae: p( ) µ ( ) µ ( ) p() d ln µ() + C e p ( ) d ln µ ( ) e e C p( ) d e Cµ ( ) 1 008, 01 Zachary S Tseng A-1-17
This is he general soluion, of course. We jus need one insance of i. Since any nonzero funcion of he above form can be used as he inegraing facor, we will jus choose he simples one, ha of C 1 1. As a resul e p( ) d µ ( ). Once i is found, we can immediaely divide boh sides of he equaion (**) by µ() o find y(), using he formula y( ) µ ( ) g( ) d + µ ( ) ( C) Noe: In order o use his inegraing facor mehod, he equaion mus be pu ino he sandard form firs (i.e. y -erm mus have coefficien 1). Else our formulas won work. Commen: As i urns ou, wha we have jus discovered is a very powerful ool. As long as we are able o inegrae he wo required inegrals, his inegraing facor mehod can be used o solve any firs order linear ordinary differenial equaion. 008, 01 Zachary S Tseng A-1-18
Example: We will use our new found general purpose mehod o again solve he equaion y r y k, r 0. The equaion is already in is sandard form, wih p() r and g() k. The inegraing facor is r d r µ ( ) e e. The general soluion is r r k r k r ( e k d) e e + C Ce 1 y + r e r r Tha is i! (I looks slighly differen, bu his is indeed he same soluion we found a lile earlier using a differen mehod.) 008, 01 Zachary S Tseng A-1-19
Example: We have previously seen he direcion field showing he approximaed graph of he soluions of y y. Now le us apply he inegraing facor mehod o solve i. The equaion has as is sandard form, y + y. Where p() 1 and g(). The inegraing facor is d µ ( ) e e. The general soluion is, herefore, y 1 e ( e d) e ( e e d) e ( e e C) + 1 + Ce. 008, 01 Zachary S Tseng A-1-0
Summary: Solving a firs order linear differenial equaion y + p() y g() 0. Make sure he equaion is in he sandard form above. If he leading coefficien is no 1, divide he equaion hrough by he coefficien of y -erm firs. (Remember o divide he righ-hand side as well!) 1. Find he inegraing facor: µ ( ) e p( ) d. Find he soluion: y( ) µ ( ) g( ) d + µ ( ) ( C) This is he general soluion of he given equaion. Always remember o include he consan of inegraion, which is included in he formula above as (+ C) a he end. Like an indefinie inegral (which gives us he soluion in he firs place), he general soluion of a differenial equaion is a se of infiniely many funcions conaining one or more arbirary consan(s). 008, 01 Zachary S Tseng A-1-1
Iniial Value Problems (I.V.P.) Every ime we solve a differenial equaion, we ge a general soluion ha is really a se of infiniely many funcions ha are all soluions of he given equaion. In pracice, however, we are usually more ineresed in finding some specific funcion ha saisfies a given equaion and also saisfies some addiional behavioral requiremen(s), raher han jus finding an arbirary funcion ha is a soluion. The behavioral requiremens are usually given in he form of iniial condiions ha say he specific soluion (and is derivaives) mus ake on cerain given values (he iniial values) a some prescribed iniial ime 0. For a firs order equaion, he iniial condiion comes simply as an addiional saemen in he form y( 0 ) y 0. Tha is o say, once we have found he general soluion, we will hen proceed o subsiue 0 ino y() and find he consan C in he general soluion such ha y( 0 ) y 0. The resul, if i could be found, is a specific funcion (or funcions) ha saisfies boh he given differenial equaion, and he condiion ha he poin ( 0, y 0 ) is conained on is graph. Such a problem where boh an equaion and one or more iniial values are given is called an iniial value problem (abbreviaed as I.V.P. in he exbook). The specific soluion husly found is ofen called a paricular soluion of he differenial equaion. Graphically, he general soluion of a firs order ordinary differenial equaion is represened by he collecion of all inegral curves in a direcion field, while each paricular soluion is represened individually by one of he inegral curves. To summarize, an iniial value problem consiss of wo pars: 1. A differenial equaion, and. A se of iniial condiion(s). We firs solve he equaion o find he general soluion (which conains one or more arbirary consans or coefficiens). Then we use he iniial condiion(s) o deermine he exac value(s) of hose consan(s). The resul is a paricular soluion of he equaion. 008, 01 Zachary S Tseng A-1 -
008, 01 Zachary S Tseng A-1-3 Example: Solve he iniial value problem y y 3 e 4, y(1). Firs divide boh sides by. e y y 4 p ) (, and e g 4 ) (. The inegraing facor is ln ln ) ( e e e d µ. The general soluion is ( ) ( ) C e d e d e y + + 3 4 4 1 C e + + Apply he iniial condiion y(1) 1 e 1 + + C 1 e + + C 0 e + C C e Therefore, e e y +.
Example: Solve he iniial value problem cos() y sin() y 3 cos(), y(π) 0. Divide hrough by cos(): y an() y 3 p() an() and g() 3 The inegraing facor is ( ) e an( ) d µ. (Wha is his funcion?) Use he u-subsiuion, le u cos() hen du sin()d: sin( ) d du an( ) d ln u + C ln cos( + C cos( ) u ) Near 0 π, cos() is posiive, so we could drop he absolue value. an( ) d ln(cos( )) Hence, µ ( ) e e cos( ). ( sin( ) sin( ) ) 1 3 y( ) 3 cos( ) d d cos( ) cos( ) 3 cos( ) ( sin( ) + cos( ) + C) 3 an( ) + 3+ C sec( ) y(π) 0 6π an(π) + 3 + C sec(π) 0 + 3 + C 3 + C C 3 Therefore, y() 3 an() + 3 3sec(). 008, 01 Zachary S Tseng A-1-4
The Exisence and Uniqueness Theorem (of he soluion a firs order linear equaion iniial value problem) Does an iniial value problem always a soluion? How many soluions are here? The following heorem saes a precise condiion under which exacly one soluion would always exis for a given iniial value problem. Theorem: If he funcions p and g are coninuous on he inerval I: α < < β conaining he poin 0, hen here exiss a unique funcion y φ() ha saisfies he differenial equaion y + p() y g() for each in I, and ha also saisfies he iniial condiion y( 0 ) y 0, where y 0 is an arbirary prescribed iniial value. Tha is, he heorem guaranees ha he given iniial value problem will always have (exisence of) exacly one (uniqueness) soluion, on any inerval conaining 0 as long as boh p() and g() are coninuous on he same inerval. The larges of such inervals is called he inerval of validiy of he given iniial value problem. In oher words, he inerval of validiy is he larges inerval such ha (1) i conains 0, and () i does no conain any disconinuiy of p() nor g(). Conversely, neiher exisence nor uniqueness of a soluion is guaraneed a a disconinuiy of eiher p() or g(). Noe ha, unless 0 is acually a disconinuiy of eiher p() or g(), here always exiss a non-empy inerval of validiy. If, however, 0 is indeed a disconinuiy of eiher p() or g(), hen he inerval of validiy will be empy. Clearly, in such a case he condiions ha he inerval mus conain 0 and ha i mus no conain a disconinuiy of p() or g() will be conradicing. 008, 01 Zachary S Tseng A-1-5
If so, such an iniial value problem is no guaraneed o have a unique soluion a all. Example: Consider he iniial value problem solved earlier cos() y sin() y 3 cos(), y(π) 0. The sandard form of he equaion is y an() y 3 wih p() an() and g() 3. While g() is always coninuous, p() has disconinuiies a ±π/, ±3π/, ±5π/, ±7π/, According o he Exisence and Uniqueness Theorem, herefore, a coninuous and differeniable soluion of his iniial value problem is guaraneed o exis uniquely on any inerval conaining 0 π bu no conaining any of he disconinuiies. The larges such inervals is (3π/, 5π/). I is he inerval of validiy of his problem. Indeed, he acual soluion y() 3 an() + 3 3sec() is defined everywhere wihin his inerval, bu no a eiher of is endpoins. 008, 01 Zachary S Tseng A-1-6
How o find he inerval of validiy For an iniial value problem of a firs order linear equaion, he inerval of validiy, if exiss, can be found using his following simple procedure. Given: y + p() y g(), y( 0 ) y 0. 1. Draw he number line (which is he -axis).. Find all he disconinuiies of p(), and he disconinuiies of g(). Mark hem off on he number line. 3. Locae on he number line he iniial ime 0. Look for he longes inerval ha conains 0, bu conains no disconinuiies. Sep 1: Draw he -axis. Sep : Mark off he disconinuiies. Sep 3: Locae 0 and deermine he inerval of validiy. 008, 01 Zachary S Tseng A-1-7
Example: Consider he iniial value problems (a) ( 81) y + 5e 3 y sin(), y(1) 10π (b) ( 81) y + 5e 3 y sin(), y(10π) 1 The equaion is firs order linear, so he heorem applies. The sandard form of he equaion is 3 5e sin( ) y + y 81 81 3 5e sin( ) wih p( ) and g ( ). Boh have disconinuiies a ± 9. 81 81 Hence, any inerval such ha a soluion is guaraneed o exis uniquely mus conain he iniial ime 0 bu no conain eiher of he poins 9 and 9. In (a), 0 1, so he inerval conains 1 bu no ± 9. The larges such inerval is ( 9, 9). In (b), 0 10π, so he inerval conains 10π bu neiher of ± 9. The larges such inerval is (9, ). Remember ha he value of y 0 does no maer a all, 0 alone deermines he inerval. Suppose he iniial condiion is y( 100) 5 insead. Then he larges inerval on which he iniial value problem s soluion is guaraneed o exis uniquely will be (, 9). Lasly, suppose he iniial condiion is y( 9) 88. Then we would no be assured of a unique soluion a all. Since 9 is boh 0 and a disconinuiy of p() and g(). The inerval of validiy would be, herefore, empy. 008, 01 Zachary S Tseng A-1-8
Depending on he problem, he inerval of validiy, if exiss, could be as large as he enire real line, or arbirarily small in lengh. The following example is an iniial value problem ha has a very shor inerval of validiy for is unique soluion. Example: Consider he iniial value problems ( 10 000000 ) y + y 0, y(0) α. Wih he sandard form y ' y 0 000000, 10 he disconinuiies (of p()) are ±10 1000000. The iniial ime is 0 0. Therefore, he inerval of validiy for is soluion is he inerval ( 10 1000000, 10 1000000 ), an inerval of lengh 10 1000000 unis! However, he imporan hing is ha somewhere on he -axis a unique soluion o his iniial value problem exiss. Differen iniial value α will give differen paricular soluion. Bu he soluion will each uniquely exis, a a minimum, on he inerval ( 10 1000000, 10 1000000 ). Again, according o he heorem, he only ime ha a unique soluion is no guaraneed o exis anywhere is whenever he iniial ime 0 jus happens o be a disconinuiy of eiher p() or g(). Now suppose he iniial condiion is y(0) 0. I should be fairly easy o see ha he consan zero funcion y() 0 is a soluion of he iniial value problem. I is of course he unique soluion of his iniial value problem. Noice ha his soluion exiss for all values of, no jus inside he inerval ( 10 1000000, 10 1000000 ). I exiss even a disconinuiies of p(). This illusraes ha, while ouside of he inerval of validiy here is no guaranee ha a soluion would exis or be unique, he heorem neverheless does no preven a soluion o exis, even uniquely, where he condiion required by he heorem is no me. 008, 01 Zachary S Tseng A-1-9
Nonlinear Equaions: Exisence and Uniqueness of Soluions A heorem analogous o he previous exiss for general firs order ODEs. Theorem: Le he funcion f and f y be coninuous in some recangle α < < β, γ < y < δ, conaining he poin ( 0, y 0 ). Then, in some inerval 0 h < < 0 + h conained in α < < β, here is a unique soluion y φ() of he iniial value problem y f (,y), y( 0 ) y 0. This is a more general heorem han he previous ha applies o all firs order ODEs. I is also less precise. I does no specify a precise region ha a given iniial value problem would have a soluion or ha a soluion, when i exiss, is unique. Raher, i saes a region ha somewhere wihin here has o be par of i in which a unique soluion of he iniial value problem will exis. (I does no preclude ha a second soluion exiss ouside of i.) The boom line is ha a nonlinear equaion migh have muliple soluions corresponding o he same iniial condiion. On he oher hand i is also possible ha i migh no have a soluion defined on pars of he region where f and f y are boh coninuous. Example: Consider he (nonlinear) iniial value problem y y 1/, y(0) 0. When 0, f y is no coninuous. Therefore, i would no necessarily 6 have a unique soluion. Indeed, boh y and y 0 are funcions ha 36 saisfy he problem. (Verify his fac!) 008, 01 Zachary S Tseng A-1-30
Exercises A-1.: 1 4 Find he general soluion of each equaion below. 1. y y 4. y + 10 y 1 3 3. y' e y 0 4. y y e 5 16 Solve each iniial value problem. Wha is he larges inerval in which a unique soluion is guaraneed o exis? 5. y + y e, y(0) 6. y 11y 4e 6, y(0) 9 7. y y +, y(1) 5 8. ( + 1) y y 3 +, y(0) 4 9. y + ( 6 ) y 0, y(0) 8 10. y + 4y, y( ) 0 11. ( 49) y + 4 y 4, y(0) 1 / 7 1. y y +, y(0) 3 13. y + y e, y(0) 1 14. y + 4y 4, y( ) 6 15. an() y sec() an () y 0, y(0) π 16. ( + 1) y + y 0, y(3) 1 008, 01 Zachary S Tseng A-1-31
17 0 Wihou solving he iniial value problem, wha is he larges inerval in which a unique soluion is guaraneed o exis for each iniial condiion? (a) y(π) 7, (b) y(1) 9, (c) y( 4) e. 17. ( + 5) y + 18. y + ( 8)( 1) 3 y y sec( / 3) + 3 ( 6)( + 1) 19. ( + 4 5) y + an() y 16 0. (4 ) y + ln(6 ) y e 1. Find he general soluion of y + y. Then show ha boh he iniial condiions y(1) 1 and y( 1) 3 resul in an idenical paricular soluion. Does his fac violae he Exisence and Uniqueness Theorem? 008, 01 Zachary S Tseng A-1-3
Answers A-1.: 1. y 4+ Ce 3 / 3 1 10. y + + Ce 10 50 500 1 3 3. y C exp e 3 4. y e + Ce 5. y e e + 3 e, (, ) 49 11 4 6 6. y e e, (, ) 5 5 7. y + ln + 4, (0, ) 8. y ( + 1)(ln( + 1) 4), (, ) 9. y 8 exp( 3 ), (, ) 1 4 10. y 4, (, 0) 4 98 + 343 11. y, ( 7, 7) ( 49) 1. y 6e 3 3, (, ) 1 1 13. y e + e cosh( ), (, ) 14. y 1 + 80 4, (, 0). π sec( ) sec( ) 1 15. y e πe, ( π/, π/) e 10 16. y, (, ) + 1 17. (a) (3, 6); (b) ( 1, 3); (c) ( 5, 1). 18. (a) (0, 3π/); (b) (0, 3π/); (c) ( 3π/, 3). 19. (a) (3π/4, 5π/4); (b) no such inerval exiss; (c) ( 5, 5π/4). 0. (a) (, 6); (b) (, ); (c) (, ). + C 1 1. y ; hey boh have y as he soluion; no, differen iniial condiions could neverheless give he same unique soluion. 008, 01 Zachary S Tseng A-1-33
Separable Differenial Equaions A firs order differenial equaion is separable if i can be wrien in he form M(x) + N(y) y 0, where M(x) is a funcion of he independen variable x only, and N(y) is a funcion of he dependen variable y only. I is called separable because he independen and dependen variables could be moved o separae sides of he equaion: dy N ( y) M ( x). dx Muliplying hrough by dx, N ( y) dy M ( x) dx. A general soluion of he equaion can hen be found by simply inegraing boh sides wih respec o each respecive variable: ) N y) dy M ( x dx+ C (. This is he implici general soluion of he equaion, where y is defined implicily as a funcion of x by he above equaion relaing he aniderivaives, wih respec o heir individual variables, of M(x) and N(y). An explici general soluion, in he form of y f (x), where y is explicily defined by a funcion f (x) which iself saisfies he original differenial equaion, could be found (in heory, alhough no always in pracice) by simplifying he implici soluion and solve for y. 008, 01 Zachary S Tseng A-1-34
dy Example: Solve e y x x 3 0 dx Firs, separae he x- and y-erms. dy e y x 3 + x dx Then muliply boh sides by dx and inegrae e y dy ( x 3 + x) dx or, 4 x x e y + + C 4 4 x x y ln + + C 4 (implici soluion) (explici soluion) Suppose here is, in addiion, an iniial condiion of y(1). We can solve for he consan C by applying his iniial condiion: 4 1 1 e + + C 4 C e 3 4 Finally, or, e y 4 x x + + e 4 3 4 (implici soluion) 4 x x 3 y ln + + e (explici soluion) 4 4 008, 01 Zachary S Tseng A-1-35
Example: Solve he iniial value problem y' 1 y cos( ), y(0) 0. Separae he variables and inegrae, we have dy 1 y dy 1 y cos( ) d cos( ) d arcsin(y) sin() + C. Apply he iniial condiion o solve for C, he (implici) paricular soluion is arcsin(0) sin(0) + C 0 0 + C C 0 arcsin(y) sin(). The explici paricular soluion can be found easily: y sin(sin()). Quesion: How would he soluion differ if he iniial condiion is y(0) 1? (Wha happens when y 0 1?) 008, 01 Zachary S Tseng A-1-36
Example: Solve he iniial value problem 3 + 4 5 y, y 10 y(1). (y 10) y 3 + 4 5 (y 10) dy (3 + 4 5) d (y 10) dy (3 + 4 5) d (Implici) general soluion is y 10y 3 + 5 + C. The iniial condiion says ha when 1, y, so subsiue hose wo values ino he general soluion: ( ) 10( ) 1 3 + (1) 5 + C 4 + C C 6 The (implici) paricular soluion is y 10y 3 + 5 + 6. Wha is he explici soluion? We will solve explicily for y by firs using compleing-he-square o simplify he lef side: y 10y + 5 3 + 5 + 6 + 5 (y 5) 3 + 5 + 51 y 5 ± 3 + 5 + 51 y() 5 ± 3 + 5 + 51 (Which one?) 008, 01 Zachary S Tseng A-1-37
I is necessary o deermine which one of he wo expressions above is he acual soluion of his problem. Boh expressions are derived from he same implici soluion of he given equaion. Therefore, hey would boh saisfy he equaion. However, here is a unique soluion o his iniial value problem, as we know. We do have a clue regarding he rue ideniy of he soluion. The clue is in he form of he iniial condiion, y(1). Le us check. Apply he iniial condiion o boh expressions: 3 y(1) 5 ± 1 + (1) 5+ 51 5 ± 49. Since 5 49, he correc explici soluion mus be he expression wih he minus sign. y() 5 3 + 5 + 51. 008, 01 Zachary S Tseng A-1-38
Summary: Solving a separable differenial equaion M(x) + N(y) y 0 1. Rearrange he equaion ino he form below, separaing i ino a dependen variable par and an independen variable par: N(y) y M(x). Then conver boh sides ino derivaives by muliplying hrough wih dx. N(y) dy M(x) dx.. Inegraing boh sides o find he implici general soluion: ) N ( y) dy M ( x dx+ C The consans of inegraion should be combined and pu ino only one side (by convenion, he independen variable side) of he equaion. 3. If necessary / feasible, an explici general soluion, y f (x) can be found by simplifying he implici soluion and solve for y. 008, 01 Zachary S Tseng A-1-39
Exercises A-1.3: 1 6 Find he general soluion of each sysem below. 1. y y 1/. y y sin 3. y cos y sec y 4. y x + 4 5. y y 3 + 4y 6. y co y 7 17 Solve he following iniial value problem. 7. y 1 y cos( ), y(0) 1 8. y y, x 1 y() 1 9. 1 y xy, y(1) 3 10. y y 4, y(1) 11. y 6 y y, y(0) 8 1. y y 4, y(0) 13. y (1 + y ) sec x, y(0) 1 14. y e y sin(4), y(π) 1 x 3 x+ 1 15. y, (a) y(0) 1, (b) y(1) 4. y+4 16. y y (y + 5), (a) y(0), (b) y( 1) 4. 17. y xy, (a) y(4) 1, (b) y( ) e. ln(y) 18. y y y, (a) y(0), (b) y( ) 0. 008, 01 Zachary S Tseng A-1-40
Answers A-1.3:. y C 3. y sin 1 ( cos + C) 1 1 1 x 4. y sin an + C 1 y 5. ln + C y (in implici form), and y 0. 8 + 4 6. y cos 1 (e + C ) 7. y 1 8. y 3 x 1 x + 1 3 9. y 3 ln x + 7 10. 6 1/ 3 3 / 3 y e e 4 11. y 8 exp( 3 ) 1. y an(x π/4) 13. y an(an(x) π/4) 1 1 14. y 1 ln cos(4) sin(4) + e π 8 4 4 15. (a) y + x 4x + x+ 9, (b) y x 4x + x+ 5 16. (a) y 9exp( ) 5, (b) y 1exp( ) 5 17. (a) ln ( ) x y 8, (b) ln ( y ) x 1 (in implici form). 18. (a) ln y ln() y+ 1 + (in implici form), (b) y 0 008, 01 Zachary S Tseng A-1-41
Applicaions of Firs Order Equaions I. Mixing soluion A mixing ank iniially conains Q 0 amoun sal (solue) dissolved in S 0 amoun of waer (solven). Addiional sal waer (soluion) of concenraion c i flows ino he ank a a rae r i. Assume he conen of he mixing ank is sirred very rapidly such ha he soluion wihin is always of uniform concenraion. The mixed conen is hen pumped ou of he ank for use elsewhere a a rae r o. Find he amoun (mass) or concenraion (mass/volume) of sal conained in he ank a any ime > 0. Denoe: Q() amoun of solue in ank a ime S() volume of soluion in ank a ime An expression for S() is simple o derive: Since here are iniially S 0 amoun in he ank; and during each uni of ime r i amoun flows ino and r o amoun flows ou of he ank, for a ne change of (r i r o ) per uni ime. Therefore, S() S 0 + (r i r o ) Nex, we wan o come up wih an equaion ha governs Q. The general form of he differenial equaion ha governs he amoun of solue in he mixing ank, Q(), a any ime > 0 is: Q (rae of solue flowing in) (rae of solue flowing ou) The rae of solue in/ou is equal o (rae of soluion in/ou) (concenraion of soluion in/ou) 008, 01 Zachary S Tseng A-1-4
Therefore, he necessary iniial value problem is Q Q r i c i r o S(), Q(0) Q 0 The equaion is a firs order linear equaion wih he sandard form r0 Q + S( ) Q r i c i.. Consequenly, i can always be solved using he inegraing facor mehod we have already seen. 008, 01 Zachary S Tseng A-1-43
The consan-volume mixing problem (w/ consan raes, r i r o r) In his case, S() S 0. The mixing problem becomes Q + r Q r c i., Q(0) Q 0 S 0 Firs idenify p() The inegraing facor is r, and g() r c i. S o r S r S d 0 0 µ ( ) e e 0 e r r r r 1 S S r ci S 0 0 0 S0 S0 Q( ) r c r S i e d e e + C ci S0 + Ce r Q(0) Q 0 Therefore, 0 ci S0 + Ce ci S0+ C C Q 0 c i S 0 ( ) ci S0+ ( Q0 ci S0 r S 0 Q ) e. The concenraion as a funcion of ime is Q() / S() Q() / S 0. Q( ) ci S0 The limiing concenraion is lim ci. Tha is, afer a S( ) S0 very long ime, he concenraion of he conen of he ank will approach he concenraion of he new inflow. (Since evenually every las drop of he original conen will be flushed ou of he ank and be replaced by he inflow soluion.) 008, 01 Zachary S Tseng A-1-44
Example: A swimming pool iniially conains 1000 m 3 of sale, unchlorinaed waer. Waer conaining grams per m 3 of chlorine flows ino he pool a a rae of 4 m 3 per minue. The well-mixed conen of he pool is drained a he same rae. Find he ime when he chlorine concenraion in he pool reaches 1 gram per m 3. In his problem, r i r o 4. Therefore, i is a consan volume problem, and he iniial volume S 0 1000 S(). There is iniially no solue (chlorine, in his case) in he pool, hence Q(0) 0. The inflow concenraion c i. We can hen se up he following iniial value problem 4 1 Q (4)() Q 8 Q, Q(0) 0. 1000 50 1 In sandard form: Q + Q 8, Q(0) 0. 50 1 Where p() 50, and g() 8. The inegraing facor is 1 1 d 50 50 ) e e µ (. The general soluion is, hen, Q 1 50 50 50 50 8e d e 000e C 000+ Ce / 50 e +. Applying he iniial condiion Q(0) 0, Q(0) 0 000 + C C 000. 008, 01 Zachary S Tseng A-1-45
Consequenly, he paricular soluion for his problem is Q( ) 50 000 000e. The concenraion of chlorine in he pool is given by he expression Q( ) S( ) 50 000 000e 50 e. 1000 Se he expression o equal 1, and solve for. The ime (in minue) i akes for he chlorine concenraion in he pool o reach 1 gram per m 3 is: 50 Q( ) 1 e S( ) 1 e 50 1 50 e 1 ln 50 1 50 ln 50 ln(). 008, 01 Zachary S Tseng A-1-46
Non-consan-volume mixing problem (w/ consan raes, bu r i r o ) Example (Exam 1, summer 00): A 400-lier ank is iniially filled wih 100 liers of dye soluion wih a dye concenraion of 5 grams/lier. Pure waer flows ino he ank a a rae of 3 liers per minue. The well-sirred soluion is drained a a rae of liers per minue. Find he concenraion of dye in he ank a he ime ha he ank is compleely filled. In his problem, r i 3, r o, and he iniial volume is S 0 100. So he soluion s volume is S() S 0 + (r i r o ) 100 +. The iniial concenraion of he solue is 5 grams per lier. Muliplying i by he iniial volume gives us he iniial condiion of Q(0) 500 (grams of dye). No number is given for he inflow concenraion c i, bu i can be seen ha c i 0 (why?). A he sar, he 400-lier ank sill has 300 liers of spare capaciy lef. A he rae of 1 lier ne gain of conen per minue, i can las 300 minues unil i is fully filled, so overflow 300. Therefore, we can se up he iniial value problem, for 300 (beyond ha ime, he mixing process will be of a differen naure!): Q + Q 0, Q(0) 500. 100+ Where p() 100+, and g() 0. The inegraing facor is µ ( ) e d 100 ln 100 + + ln(100+ ) e e (100 + ). The general soluion is Q 1 (100+ ) C (100+ ) 0 d. 008, 01 Zachary S Tseng A-1-47
Applying he iniial condiion Q(0) 500, C C Q ( 0) 500 (100+ 0) 10000 C 5000000. 5000000 Hence, he paricular soluion is Q( ) (100+ ). The problem asks for he concenraion of dye a overflow 300 minues. Wrie down he formula for he solue s concenraion and hen se 300 o obain Q(300) S(300) 5000000 /(100+ 300) (100+ 300) 5000000 3 400 5000000 64000000 5 64 0.07815. 008, 01 Zachary S Tseng A-1-48
Exercises A-1.4: 1. A ank is iniially filled wih 600 liers of a soluion conaining 100 grams of sugar. Soluion conaining a concenraion of g/lier sugar eners he ank a he rae 4 liers/minue and he well-sirred mixure leaves he ank a he same rae. Find he amoun of sugar in he ank a ime, and find he limiing amoun of sugar in he ank as.. A swimming pool holds 100 m 3 of pure waer. Soluion conaining kg/m 3 of chlorine eners he pool a a rae of 3 m 3 /min. A drain is opened a he boom of he pool so ha he volume of soluion in he pool remains consan. Find : (i) he amoun of chlorine in he pool a ime, (ii) he amoun of chlorine in he pool afer one hour, and (iii) find he maximum amoun of chlorine in he pool if he process is o coninue indefiniely. 3. A 00-lier ank is filled o capaciy wih brine conaining 1 g/lier of sal. Addiional brine conaining 5 g/lier of sal eners he ank a he rae liers/min and he well-sirred mixure leaves he ank a he rae of 4 liers/min. Find he amoun of sal in he ank a any ime, unil he ank is compleely drained (0 < < 100). Wha is he maximum amoun of sal presen in he ank during his period? 4. A 150-lier mixing va is iniially filled wih 60 liers of waer conaining g/lier of dissolved poassium chloride. Saring a 0, 5 g/lier soluion of poassium chloride flows ino he va a a rae of 6 liers/minue. The well-mixed soluion leaves he va a a rae of 3 liers/minue. (i) Se up an iniial value problem describing he amoun of poassium chloride in he va a any ime prior o overflow, 0 < < 30. (ii) Solve his problem. (iii) Find he amoun and concenraion of poassium chloride in he va a he ime of overflow. (iv) Suppose he inake pipe coninues o supply 6 liers/minue of soluion pas he ime of overflow, and he excessive soluion spills over he open op of he va. Therefore, he well-mixed soluion would leave a a rae of 6 liers/minue by means of boh he oupu pipe and spill-over. Se up an iniial value problem describing he amoun of poassium chloride in he va a any ime from he ime of overflow onward, > 30. (v) Solve his second iniial value problem. (vi) Find he limiing concenraion of poassium chloride in he va as. 008, 01 Zachary S Tseng A-1-49
5. A reenion pond iniially conains 000 m 3 of waer having a polluans concenraion of 0.5 kg/ m 3. Each hour, 10 m 3 of waer conaining polluans of variable concenraion + sin() kg/ m 3 flows ino he pond. Thoroughly mixed waer flows ou of he pond a he same rae. (i) Se up an iniial value problem modeling his process. (ii) Solve his problem. Answers A-1.4: 1. Q() 100 1100e / 150, 100 grams. (i) Q() 00 00e 3 / 100, (ii) Q(60) 00 00e 9 / 5 166.94 kg, (iii) Q max approaches 00 kg, occurs as. 3. Q() 0.08 + 6 + 00, Q max 31.5 grams (a 37.5) 1 4. (i) Q + Q 30, Q(0) 10 0+ 15 + 600+ 400 (ii) Q( ) 0+ (iii) Amoun of KCl 678 grams, concenraion 4.5 g/lier 1 (iv) Q + Q 30, > 30, Q(30) 678 5 (30 ) / 5 (v) Q() 750 7e (vi) 5 g/lier 1 5. (i) Q + Q 0 + 10sin(), Q(0) 1000 00 (ii) Q( ) 4000+ 40000 40001 1 00 sin( ) cos( ) 119963000 40001 e 00 008, 01 Zachary S Tseng A-1-50
II. Air-resisance / Moion of an objec in a resisive fluid medium Freefall and Air-resisance An objec of mass m is freefalling near sea level (herefore, assume consan graviy). Unlike in he calculus class earlier, we will include he effec of air-resisance in our consideraion. For he ime being, we shall assume ha he resisive force (drag force) is proporional o he insananeous speed * of he objec in moion (e.g., when he resisance is due o fricion only). Find he velociy of he freefalling objec as a funcion of ime. Noe: The exbook s convenion is ha he downwards direcion is posiive. Forces acing on he objec undergoing freefall: Graviaional force w mg (always downwards) Resisive force (drag) F r k v (agains he direcion of v) The graviy/weigh is always downward, so w is always posiive. The drag force always opposes he direcion of he moion (given by he sign of velociy funcion v()). Therefore, F r k v, which is always opposie of v bu whose magniude equals k v. The proporionaliy consan k is he drag coefficien. * Speed he magniude of velociy v 008, 01 Zachary S Tseng A-1-51
By Newon s second law of moion ma forces Tha is ma mv mg + ( kv). Hence he required equaion of moion governs he velociy of he objec is mv mg kv. I is a simple firs order linear equaion, wih consan coefficiens. Is soluion is v ) mg + Ce k k m (. The posiion funcion of he moion can be found, as usual, by inegraion: x() v() d. Limiing velociy: v L lim v( ) The limiing velociy is he maximum velociy achievable by he objec, in his model, given infinie amoun of ime o accelerae. Take he limi of he soluion found above, we obain v L mg / k. More easily, i could also be found, wihou having o find v() firs, by seing v 0 in he original moion equaion and solve for v. (Since v L is he maximum velociy, i occurs a a criical poin of v()! Hence, v 0. Physically, his happens when he graviaional force and drag cancel each oher, leaving zero ne force in he moion equaion.) 008, 01 Zachary S Tseng A-1-5
Tha is 0 mg k v L mg k v L Therefore, v L mg lim v( ). k Noe ha v L is independen of any iniial condiion. Wha happens if he iniial velociy, for whaever reason, is larger han v L? In ha case he righ hand side of he moion equaion, mg kv, is negaive. The process modelled becomes a gradually deceleraing moion whose velociy would evenually slow down o v L, which would be he minimum achievable velociy. 008, 01 Zachary S Tseng A-1-53
Anoher (more realisic) air-resisance model According o fluid dynamics, he drag force exered on an objec moving, sub-sonically, in a fluid (liquid or gas) medium is acually proporional o he square of is speed. Therefore, a more realisic equaion ha models he sub-sonic moion of an objec in a resisive fluid medium is mv [propulsive force] kv, v 0 This equaion is a nonlinear firs order differenial equaion. Forunaely, i is a separable equaion. Therefore i is well wihin our capabiliy o solve i. Le p > 0 denoes he propulsive force (graviy, or he hrus of an engine, for examples) and v ± p / k : m mv p kv v 1 p kv Inegrae boh sides (he lef can be inegraed by parial fracions) o obain he implici soluion: m dv + C p kv Even wihou an explici funcion, he limiing velociy can neverheless be found easily by seing v 0 in he moion equaion. v L p k 008, 01 Zachary S Tseng A-1-54
Example: A 100 kg Unmanned Aerial Vehicle (UAV) possesses propulsive force of 10000 N and has drag coefficien k 4. Find he velociy funcion of is fligh. m 100, k 4, and ake v(0) 0 as he iniial condiion: 100v 10000 4v, v(0) 0 Simplify he equaion, separae he variables, and inegrae boh sides. (Bu firs noing ha v ± 50 are boh also soluions of his equaion.) dv v 100 0.04v, d 100 0.04v dv 100 0.04v d. The lef-hand side could be simplified by parial fracions ino: Hence, 1 1/ 0 1/ 0 +. 100 0.04v 10+ 0.v 10 0. v 1 0 1 1 + dv + C 10+ 0.v 10 0. v 1 0 0. dv 0. dv 5 10+ 0.v 10 0. v + C [ ln 10+ 0.v ln 10 0. v ] + C 1. 4 Now use he iniial value v(0) 0 o find c 0. Therefore, [ ln 10+ 0.v ln 10 0.v ] 1 1 10+ 0.v ln. 4 4 10 0.v The limiing velociy (forward) is v L 50 m/sec, which is found by seing v 0 in he original equaion and solve for v. 008, 01 Zachary S Tseng A-1-55
Wih some algebra (and a lile paien) we can also find he explici soluion for his problem wihou much difficuly. For reasons we shall see very shorly (in he secion on auonomous equaions), and given he condiion ha v 0, here are 3 families of soluions, depending on he iniial v-value. Since v 0 0 in his example, we will only find he relevan soluion, which exiss on he inerval 50 > v 0. From he implici soluion 1 10+ 0.v ln, 50 > v 0. 4 10 0.v 4 ln 10+ 0.v 10 0.v e 4 10+ 0.v 10 0. v Since 50 > v 0, we can drop he absolue value: (10 0.v) e 4 10+ 0.v 10e 4 10 0.e 4 v+ 0.v 10( e 4 1) 0.( e 4 + 1) v 50( e 4 1) ( e 4 + 1) v v 50( e 4 e 4 1) + 1 Verify ha v(0) 0 and lim v( ) 50 v L. The consan funcion v 50, by he way, is also a soluion o he equaion (verify his). I does no come from he implici general soluion found earlier, bu raher comes abou by merely seing v 0 and solve for v. 008, 01 Zachary S Tseng A-1-56
III. Coninuous compound ineres wih addiional ransacions From calculus: Saring wih a fixed principal amoun A 0, he balance of an accoun garnering a fixed ineres rae r (per year, usually) compounding a a frequency of m (per year) is given by he formula r A ( ) A0 1+. m If he ineres is compounded coninuously, i.e. as he frequency m, hen m m r A ( ) lim A 1 0 + 0 m m Indeed, A 0 e r is he acual soluion of he differenial equaion A ra, subjec o he iniial condiion A(0) A 0. (Exercise: verify his claim.) A e r. Noe: The simple equaion above, A ra, means simply ha he rae of change of he accoun balance is (coninuously) proporional o is presen size. The same equaion (where he rae of change of some quaniy is direcly proporional o he curren size of he said quaniy) also governs exponenial growh and radio-acive decay (when r is negaive) behaviors. Now, insead jus le he principal si unouched and allowed o grow exponenially during he lifeime of he deposi (as in a bank CD), we will consider he effec of furher deposi/wihdraw ransacions afer he iniial deposi. One cavea: since we do no have he necessary ool (he Laplace Transform, chaper 6) o deal wih discree (one-ime) evens, we have o assume ha he ransacions occur coninuously, or a leas occur regularly and frequenly enough ha hey can be hough of as o be occurring coninuously. While such an assumpion does no model well he accoun balance of a ypical checking accoun, i does give a good approximaion of accouns wih fixed insallmen paymens such as annuiies, morgage or suden loan repaymen, ec. 008, 01 Zachary S Tseng A-1-57
Hence, assume an accoun sars wih a principal of A 0, ha gains ineres a a rae of r per uni ime compounded coninuously. In addiion, ransacions occurring coninuously and neing k amoun per uni ime (k > 0 means a ne deposi ino he accoun; k < 0 means a ne wihdraw from i) are applied o he accoun. Then, he accoun balance is described by he following iniial value problem: y r y + k, y(0) A 0. Commen: The above equaion says ha a any momen in ime he accoun balance y is increased by an amoun proporional o is curren size imes he ineres rae, and he rae of change is furher modified (up or down, depending on he sign of k) by he ne ransacional amoun. We have solved his equaion earlier in he semeser. The general soluion is C r k y e. r r Apply he iniial condiion we ge: C 0 y( 0) A0 e r k r C k r ra0 C k C ra 0 + k Therefore, k + r k r r ( e 1) r r y A0 e A0e +. k r 008, 01 Zachary S Tseng A-1-58
Example (final exam, fall 007): A college suden borrows $5000 o buy a car. The lender charges ineres a an annual rae of 10%. Assume he ineres is compounded coninuously and ha he suden makes paymens coninuously a a consan annual rae k. Deermine he paymen rae k ha is required o pay off he loan in 5 years. In his problem, he yearly ineres rae r 10% 0.1, he principal balance is A 0 5000, he yearly paymen (hink i as a wihdrawal, since we are paying down he balance) k is he unknown. The loan erm is 5 years, i.e. i needs o be paid off compleely in 5 years. Tha means besides he iniial condiion, y(0) 5000, we also have a second (erminal?) condiion of y(5) 0. We se up he required iniial value problem (noe ha k has a minus sign in fron, denoing repaymen): y r y k 0.1y k y(0) 5000. I is a firs order linear equaion (i is also a separable equaion), y 0.1y k, where p() 0.1, and g() k. The inegraing facor is, herefore, 1 d 10 /10 µ ( ) e e. The general soluion is 1 /10 y k e d /10 e /10 10 k+ C e e /10 k e /10 d e /10 /10 ( 10 k e + C) Apply he iniial condiion o find C 5000 10k. The paricular soluion is y() (5000 10k) e /10 + 10k. Lasly, apply he pay-off condiion y(5) 0, we find ha 1/ 500 e 500 k 500+ 1/ 1/ e 1 e. 1 008, 01 Zachary S Tseng A-1-59
Example: The presen value of a loery jackpo A lucky college suden has won he loery s en million dollars jackpo. The winning is paid ou equally over 0 years. Assume he payou is made coninuously and he annual ineres rae is consan 8% over he 0-year period. How much is he jackpo worh in oday s dollar? The problem is o find he presen value (or discouned value) of his jackpo, which is no paid ou all a once, bu over a period of 0 years. Tha is, if he winning was, insead, made in a lump sum and was immediaely deposied in a bank o garner ineres for 0 years, how big a sum i mus be o equal in value, a he end, o he 0-year of seady cash sream? In accouning-speak, we are rying o discoun he fuure cash flow in order o find is presen value, or how much his fuure sream of cash paymens is worh now. There are ways o ackle his problem. The more obvious (o us non-accouns) is he indirec approach. Firs we compue he worh of his jackpo a he end of 0 years by solving he compound ineres equaion wih he yearly ineres rae r 8%, he yearly paymen k is $10,000,000 / 0 $500,000, and (he iniial condiion) he principal balance A 0 0. Se 0 in he resul o obain he erminal value afer 0 years. Then we solve a second problem of coninuous compound ineres wih an unknown iniial principal balance A 1, no addiional ransacions, and a erminal condiion y(0) equal o he amoun we have found previously. Solve his second problem o find A 1, which is how much he jackpo would be worh presenly. There is anoher, more direc, way o find he presen value. I is how accounans will approach his problem from he poin-of-view of he loery adminisraor. For he adminisraor, he problem is o se aside enough money o be deposied in a bank accoun such ha a yearly payou/wihdraw of k $500,000 can be made for 0 years (and he accoun balance becomes exacly zero a he end of he 0-h year). In his approach, we will use r 8%, he yearly wihdraw k $500,000, he iniial condiion being he unknown principal balance A 0, plus he erminal condiion y(0) 0. 008, 01 Zachary S Tseng A-1-60
Hence, we will solve he iniial value problem: y r y + k 0.08 y 500000, y(0) A 0, and such ha y(0) 0. In is sandard form, y 0.08y 500000, wih p() 0.08, and g() 500000. The inegraing facor is, herefore, 0.08d 0.08 µ ( ) e e. The general soluion is y e e 0.08 1 0.08 0.08 500000 e d e 500000 0.08 0.08 ( 650000 e + C) 650000+ C e 0.08 e 0.08 d Apply he iniial condiion o find C A 0 650000. The paricular soluion is, consequenly, y() 650000 + (A 0 650000) e 0.08. Lasly, apply he erminal condiion y(0) 0, we find ha he presen value of his nominally en million dollars jackpo is acually less han half of is saed amoun:. A 0 650000 650000 / e 1.6 A 0 $4,988,147 This example explains why ha, when a loery winner chooses (as mos of hem do, given he opion) o ake he winning in a single lump sum, raher han in periodic paymens over many years, he payou amoun becomes much smaller han he quoed jackpo, even before axes are deduced 008, 01 Zachary S Tseng A-1-61