5. DIFFERENTIAL EQUATIONS

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1 5-5. DIFFERENTIAL EQUATIONS The most commo mathematical structure emploed i mathematical models of chemical egieerig professio ivolve differetial equatios. These equatios describe the rate of chage of a depedet variable with respect to a idepedet variable. The ca be expressed i a of the three coordiate sstems metioed earlier i Sectio... The idepedet variables ma also iclude time as the fourth elemet i the descriptio of a problem. Let s first discuss the mathematical origi of differetial equatios. 5.. Mathematical origi of differetial equatios A algebraic equatio that ivolves a parameter describes a famil of curves. For example, equatio 5. describes the famil of curves show i Figure. cx (5.) 4 3 c c Figure Famil of curves for the algebraic equatio cx (Mickle, Sherwood & Reed) Usig differetiatio ad algebraic maipulatio equatio 5. ca be expressed as a differetial equaito free of the parameter c. Let s take the derivative of both sides with respect to x i this equatio. cx (5.) Process Modelig ad Aalsis ECH 64 - RG & NG

2 5- Substitutig the value of c, expressed as c x,obtaied from the origial equatio, the famil of curves is x (5.3) Equatio 5. is kow as the primitive of the differetial equatio represeted b equatio 5.3. If there are two parameters i the primitive equatio we eed to differetiate twice to elimiate the two parameters. Solvig a differetial equatio meas fidig the primitive either b aaltical or umerical meas. So, usuall durig a modelig exercise the modeler comes up with oe or more differetial equatios that describes the sstem. These equatios are the solved to obtai the algebraic represetatio, the primitive equatio, for the model. Q.5. What is the equivalet differetial equatio describig the primitive equatio expressed b c( x a)? 5.. Defiitio of terms There are several importat terms used i the cotext of differetial equatios. These terms classif the various differetial equatios ito groups accordig to certai features. Most of the time, this classificatio eables oe to select a suitable solutio method Order The order of a differetial equatio is the order of the highest derivative i the equatio Degree The degree is the power the highest order derivative is raised after the equatio is cleared of fractios ad ratioalized Geeral solutio The solutio of a differetial equatio that ivolves costats is the geeral solutio. I the previous sectio, equatio 5. is the geeral solutio or the primitive to differetial equatio 5.3. Applig various boudar ad/or iitial coditios the particular curve out of the famil of curves ca be specified. The order of a differetial equatio determies how ma separate iitial ad boudar values ca be specified. For example, a secod order differetial equatio eeds either oe iitial ad oe boudar value or two boudar values. If there are more tha that the problem might be overspecified Particular solutio Whe all the iitial ad boudar values are applied to a geeral solutio, the costats are specified ad a particular solutio is obtaied. For example, ( ) the differetial equatio 5.3. x is a particular solutio to 5.3. Classificatio of differetial equatios The two top most classes for differetial equatios are Ordiar Differetial Equatios (ODE) ad Partial Differetial Equatios (PDE). A ODE is a differetial equatio that has ol oe idepedet variable i its defiitio, whereas a PDE has more tha oe. Examiatio of the phsical settig ad

3 5-3 problem descriptio usuall gives clues about the sstem ad hit if the sstem should be modeled b ODEs or PDEs. I this chapter we will cocetrate o ODEs. The most geeral form of a ODE with oe idepedet variable x ad oe depedet variable ca be expressed as d d f x,,,, L, (5.4) A value that satisfies equatio 5.4 is said to be the solutio of the differetial equatio. The most geeral algebraic solutio to this differetial equatio that ties ad x is ( x, c, c,, ) F L (5.5), c It is mathematicall prove that each equatio of the form 5.4 must have a solutio of the form 5.5, whether we are able to fid it or ot. Let s examie the solutio methods for first order ODEs First order ad first degree ODEs A first oder ad first degree ODE ca be expressed as ( x, ) + N( x, ) M (5.6) Depedig o the fuctioal form of M(x,) ad N(x,) i equatio 5.6, there are various solutio methods Separable equatios as If M ad N i equatio 5.6 are fuctios of ol x ad, respectivel, the equatio ca be restated ( x) + f ( ) f (5.7) The solutio to equatio 5.7 is obtaied simpl b itegratig both sides of the equatio. ( x) f ( ) C f + (5.8) Equatios that become separable after chage of variable There is o uiversal method that ca fid the right variable chage ad trasforms the problem to a simpler form. However, let s illustrate the idea with a example. Cosider the equatio below ( + ) + x x (5.9) Equatio 5.9 ca be easil solved b the use of variable chage xv. Q.5. What is the solutio to equatio 5.9 after the variable chage xv? Homogeeous equatios

4 5-4 A fuctio f(x) is said to be homogeeous of the -th degree if whe x multiplied b a value t results i a futioal value t f ( x) equatio should hold. f ( tx t) t f ( x, ). Simillarl, If a fuctio is homogeeous i x ad the below, (5.) For the differetial equatio 5.6 where M(x,) ad N(x,) are both homogeeous fuctios i x ad of the same degree the chage of variable ux leads to a solutio as show. f (, u) (, u) + uf ( u) l x + du C f, (5.) Q.5.3 Show the derivatio of equatio 5. with the substitutio ux First order first degree equatios with liear coefficiets If M(x,) ad N(x,) are liear fuctios of x ad so that equatio 5.6 ca expressed as ( + b + c) + ( gx + h + k) ax (5.) the followig substitutios usuall trasforms equatio 5. to a homogeeous oe. x w + m v + dw dv (5.3) The costats m ad are foud solvig the followig equatios am + b + c gm + h + k (5.4) Q.5.4 This method would fail if a g b h. If this is the case oe ca elimiate from equatio 5. b subtitutig w ax + b. The resultig equaio would be separable i x ad w. Proove this method ad suggest a geeral solutio Exact equatios If the partial deriatives of M(x,) ad N(x,) with respect to ad x, respectivel, are the same, the left had side of equatio 5.6 is said to be a exact differetial. The solutio ca be expressed b or ( x ) M ( x, ) + N( x, ) M ( x ) F,, (5.5) F ( x ) N ( x, ) + M ( x, ) N ( x, ) x, (5.6)

5 5-5 Q.5.4 Cosider the followig differetial equatio x + a) Proove that the left had side forms a exact differetial. b) What is the geeral solutio? Equatios solvable b itegratig factors A itegratig factor trasforms a differetial equatio ito a equatio where the left had side is a exact differetial. Cosider the below differetial equatio 3 ( 6 x + ) + ( 6x + 6 x + ) (5.7) The left had side of Equatio 5.7 forms a exact differetial. However, the same equatio ca also be simplified b dividid both sides b to give ( 3 + ) + ( 3x + 3x + ) x (5.8) Equatio 5.8 does ot have a exact differetial o its left. To solve it we eed to realize that is a itegratig factor that trasforms the left had side to a exact differetial First order liear equatios The form of these tpe of differetial equatios ca be rewritte as + P Q (5.9) where P ad Q are either costats or fuctios of x ol. differetial equatioad the solutio is P e is a itegratig factor for this e P P e + C (5.) Beroulli s equatio The form of these tpe of differetial equatios ca be expressed as + P Q (5.) where P ad Q are either costats or fuctios of x ol. Whe equatio 5. is divided b the substitutio described i sectio Other itegratig factors z is used it becomes a liear first order equatio ad ca be solved as Fidig a itegratig factor is usuall a difficult task. However, there several stadard forms which have kow itegratig factors. Three of these forms are below. ad

6 5-6 M N x ( ) If f ( x) N is true, the f ( x) e is a itegratig factor. M N x ( ) If f ( ) M is true, the ( ) e f is a itegratig factor. If M f ( x) ad N xf( x) hold, the ( xm N) is a itegratig factor Equatios of the first order ad higher degree Equatios of the first order ad higher degree have the followig form x,, f (5.) Equatio 5. ca be solvable for /, or x. Depedig o this criterio there 3 cases to be cosidered for solutio. Case I. Eqatios solvable for /. Let s ivestigate this case with a example. A tpical equatio is as follows + (5.3) Equatio 5.3 ca be rewritte as (5.4) Equatio 5.4 has the followig two solutios 5 4 leads to 5x + c leads to 4x + c (5.5) These two solutios ca be combied b multiplicatio to give the fial solutio as ( + x c )( 4x c ) (5.6) 5 I differetial equatios of first order ad first degree for ever value of the idepedet variable there is ol oe correspodig slope at that poit. However, for equatios of -th degree there are slopes specified at ever poit. Case II. Eqatios solvable for.

7 5-7 Differetial equatios solvable for has the geeral form of f x, (5.7) For solutio first use the substitutio /p ad the differetiate with respect to x. The result would have the geeral form dp p f x, p, (5.8) If equatio 5.8 ca be itegrated to give a algebraic relatio i x, p ad c as ( x, p, c) f (5.9) 3 the fial solutio ca be expressed after elimiatig p betwee equatios 5.7 ad 5.9. If equatio 5.8 is difficult to itegrate, it is also possible to express the fial solutio as a parametric oe b equatios 5.7 ad 5.9 where p is the parameter that specifies x ad. Case III. Eqatios solvable for x. Differetial equatios solvable for x has the geeral form of x f, 4 (5.3) A more coveiet versio of equatio 5.3 is obtaied after differetiatig with respect to ad substitutig //p. p dp f, p, 5 (5.3) After the maipulatio the solutio method outlied i the secod case should be followed to obtai either a parametric or algebraic solutio. Q.5.5 Cosider the followig differetial equatio of first order ad secod degree x 4 a) Which of the three cases are applicable for this ODE? b) Which case results ito a easier solutio path? Clairaut s equatio These ODEs have the geeral form of x + f (5.3)

8 5-8 Equatio 5.3 resembles to Case II studied i Sectio Followig a similar approach let s differetiate with respect to x ad make the substitutio /p. dp p p + x + f dp ( p) (5.33) Equatio 5.33 rearraged to dp [ + f ( p) ] x (5.34) For equatio 5.34 to hold either x+f (p) or dp/ should equal. The case dp/ leads to the fact pcostat. I that case the geeral case becomes ( c) cx + f (5.35) The secod solutio ca be obtaied b elimiatig p from x+f (p) ad equatio 5.3. This solutio will satisf equatio 5.3. So, it is a solutio to the origial ODE. However, as it does ot cotai a arbitrar costats, it caot be the geeral solutio. This tpe of solutio is called the sigular solutio Secod order ad first degree ODEs The geeral form of the this tpe of equatios is x d,, f (5.36), It is impossible to discuss all the possible equatios that coform to equatio 5.36 ad suggest a solutio method. However, there are several stadard forms of equatio 5.36 that are importat i chemical egieerig. These stadard forms have well established solutio methods. We will first cosider equatios with missig terms ad the secod order liear ODEs with costat coefficiets. Case I. Equatios ot cotaiig. ) If / is ot preset, the form has to be d f ( x) (5.37) The solutio ca be foud b itegratig f(x) twice with respect to x. ) If / is preset, the form has to be d f x, (5.38) The substittio /p will create a ew first order ODE i p ad x that ca be solved b previousl discussed methods. Please ote that d / dp/.

9 5-9 Q.5.6 Cosider the followig ODE of secod order ad first degree d + x x What is the solutio for this ODE? Case II. Equatios ot cotaiig x. These ODEs would have the followig geeral form d, f (5.39), Usig the substitutio p/ ODE should be first reduced to oe of the first order. However, the ODE will be easier to solve if d / is expressed as follows dp dp dp p (5.4) Q.5.6 Cosider the followig ODE of secod order ad first degree d + What is the solutio for this ODE? Case III. Liear ODEs of secod order ad first degree with costat coefficiets. Actuall, i the ext sectio we will discuss the geeralized liear ODEs. However, as ODEs of secod order ad fi rst degree with costat coefficiets pla a importat role i the chemical egieerig cotrol theor, we foud it appropriae to discuss this class here i greater detail. Followig the otatio we used util ow the ODE s geeral form ca be expressed as d a + b + f ( x) (5.4) However, it is a better idea to adopt time, t, as the idepedet variable ad to follow the otatio used i the cotrol theor. The geeral form of the ODE is ow expressed b d + ζ dt + m( t) dt (5.4) The parameters? ad are called the dampig coefficiet ad time costat, respectivel. The fuctio m(t) is said to be the forcig fuctio. It is better to express the coefficiets this wa as the both relate to some specific phsical facts i the cotrol theor. I this sectio the details of solutio methods are ot give. The details will be pieced together whe the geeral liear differetial equatios are discussed i the ext sectio. For ow let s just cocetrate o the tabulated solutios for the five critical regios of? aroud zero ad uit. The full solutio to the ODE i equatio 5.4 requires the complemetar solutio first. Complemetar solutio is the solutio for the case where the forcig fuctio m(t) is assumed to be zero as show below.

10 5- d + ζ + dt dt (5.43) This equatio, hece the complemetar solutio, describes the sstem s respose whe there are o exteral disturbaces. Whe the differetial operator D is substituted for the derivative of with respect to t, the characteristic equatio for the ODE is obtaied as D + ζd + (5.44) The roots of this characteristic equatio determies what the complemetar solutio will be. Table Table tabulates the five possible cases. For a sstem to be stable the dampig coefficiet has to be greater tha zero. If it is equal to zero the the sstem is said to be udamped. It will sustai cotiuous oscillatios. If the dampig coefficiet is less tha oe but greater tha zero, the the sstem is uderdamped. A uderdamped sstem will go to its stable poit through some oscillatios. If the dampig coefficiet is greater tha uit the sstem is called overdamped. A overdamped sstem will go to its stable poit with o oscillatios. Table Solutios for secod order liear ODEs with costat coefficiets Name? Roots, r ad r Solutio, (t) Overdamped sstem? > r r ζ ζ + ζ ζ rt r t c e + ce Criticall damped sstem? t r ( ) c + c t e Uderdamped sstem? < r r ζ ζ + i i ζ ζ e ζt R ( ) ζ I c cos t + i( c ) si ζ t Udamped sstem? i r i r R I ( c ) cos t + i( c ) si t ζ ζ Ustable sstem? < r r ζ ζ + i i ζ ζ e ζt R ( ) ζ I c cos t + i( c ) si ζ t

11 5- Q.5.7 Cosider two idetical, isothermal, costat volume CSTRs i series with a first order irreversible reactio. Develop ad solve the sstem of simultaeous ODEs describig the sstem. (Hit: You ca express simultaeous ODEs as a higher order ODE)

12 Liear ordiar differetial equatios A geeral -th order liear ODE ca be writte as a d d + a a a h ( x) L (5.45) where a, a,,a are cosat coefficiets ad h(x) is a arbitrar fuctio of x. If D is defied as the differetial operator Dd/, equatio 5.45 ca be rewritte as follows ( a D a D + + a D + a ) h( x) + L (5.46) If p is a particular solutio to equatio 5.46 the the geeral solutio ca be expressed as the sum of the particular solutio p ad a complemetar solutio c. p + c (5.47) Let s combie the equatios 5.46 ad ( a D a D + L+ a D + a ) + ( a D + a D + L+ a D + a ) h( x) + p c (5.48) term Now, ote that p was a solutio ad therefore term must equal to h(x). So, term must equal. ( D a D + + a D + a ) a L (5.49) + c I other words, the complemetar solutio of a liear ODE is the solutio to the ODE where h(x) is replaced b, the homogeeous form of the origial ODE. This solutio summed with a particular solutio gives the full solutio to the ODE Determiig the complemetar fuctio Equatio 5.49 is a polomial i D with roots r, r,, r ad ca be represeted as follows. ( r )( D r ) ( D r ) D L (5.5) c Depedig o the tpe of roots there are three cases: Case I. Roots are real ad distict. The solutio is term c c e r x + c e r x + + c e r x j c j e r x i L (5.5) Case II. Two or more roots are equal. If there are m equal roots r, the part of the complemetar solutio due to these roots is

13 5-3 ( m c c x + + c x ) e rx + L (5.5) m The other (-m) distict roots follow the patter i equatio 5.5. Case III. Oe or more cojugate pairs of complex roots. Let s assume that cojugate pairs of complex roots have the followig form a±ßi. If there are m cojugate pairs of complex roots, the complemetar solutio would be. c m ( α+ iβ ) x ( α iβ ) x ( c +, je c, je ) + j c j j m e r x i (5.53) Usig the followig trigoometric relatios e e iβx iβx cos cos ( βx) + isi ( βx) ( βx) i si ( βx) (5.54) equatio 5.53 is usuall writte as c e αx m ( c ( ) ( )), j cos βx + c, j si βx + j c e j j m r x i (5.55) For other trigoometric relatios that might be helpful i this tpe of ODE problems please refer to page 5 i Mickle, Sherwood & Reed, 957. Q.5.6 Cosider the followig geeral secod order ODE with real costat coefficiets. d + a + b What are the possible complemetar fuctios? Determiig the particular solutio There are two commo methods for fidig the particular solutio p. The first oe is method of udetermied coefficiets. This method is eas to use, however, it requires h(x) to be of certai fuctioal form. It also applies ol to the costat coefficiet liear ODEs. The secod method is called variatio of parameters. This method ca be applied to cases where the coefficiets are fuctios of x ad h(x) is complex. I. Method of udetermied coefficiets. I chemical egieerig applicatios, the fuctio h(x) i equatio 5.45 is usuall the sum of oe more of five simple fuctioal relatios. These fuctioal relatios are: costat, x, e rx, cos kx, si kx. It is possible to show mathematicall that accordig to the fuctioal form of h(x) oe ca predict the geeral form of the particular solutio it should have. As this particular solutio has to satisf the origial differetial equatio, oe ca substitute the predicted form of the particular solutio ito the ODE ad fid out what the coefficiets should be. Table shows what the geeral form of the particular solutio should be accordig to the above metioed commo h(x) terms.

14 5-4 Table Form of particular solutio (adapted from Mickle, Sherwood & Reed, 957) h(x) form of p costat a costat A ax A x + A x + L + A x + A rx be ccos ( kx) d si ( kx) gx hx e e rx rx cos si ( kx) ( kx) rx Be ( kx) Dsi ( kx) Ccos + rx ( Gx + G x + L+ Gx + G) e cos( kx) rx + ( H x + H x + L+ H x + H ) e si ( kx) The followig two rules appl for buildig the predicted form of the particular soulutio.. Whe h(x) is the sum of two or more terms, the geeral form of the particular solutio is the sum of the correspodig terms matchig the idividual forms from Table.. If a of the suggested terms b Table is alrea preset i the complemetar solutio, the the form eeds to be multiplied b x to form the fial form of the particular solutio. Example. Let s solve a costat coefficiet liear ODE of fourth order to illustrate the method of udetermied coefficiets. Cosider the followig ODE 4 d 4 3 d d + x 3 3 (5.56) I differetial operator D, the characteristic equatio is 4 3 ( 3D + D ) D (5.57) After factorig the characteristic equatio ca writte as ( D )( D ) D (5.58) With the roots r r, r 3 ad r 4, the complemetar solutio, c, is foud as c x x ( c + c x) + c e c e (5.59) As h(x) is x, from Table, the particular solutio, p, is predicted as

15 5-5 p A x + (5.6) A However, as there are alrea a costat term ad a term with first power of x i the complemetar solutio, we eed to multipl equatio 5.6 with x to get fial form of the geeral particular solutio as 3 ( A x + A ) A x A x p x + (5.6) Let s substitute equatio 5.6 ito equatio 5.56 ( 4) ( 3) ( ) + ( ) 3( 6A ) + ( 6A x + A ) x p p p 3 (5.6) Some basic maipulatio ields 8 A A x + A x (5.63) + 4 After equatig the like powers, the coefficiets are foud as A / ad A -3/8 so the particular solutio is p 3 3 x x (5.64) 8 As the fial solutio is the sum of complemetar ad particular solutios, the results is x x 3 ( c + c x) + c e + c e + x x (5.65) II. Variatio of parameters. The method of edetermied coefficiets is applicable to liear ODEs where the coefficiets are costat. This secod method is more relaxed i that sese ad allows oe to solve higher order liear ODEs with variable coefficiets. The procedure will be discussed for a geeral secod order ODE with the followig form d + a ( x) + a ( x) h( x) (5.66) Let s assume that the complemetar solutio to equatio 5.66 is kow ad has the followig form c c + (5.67) c The the particular solutio ca be will have the followig geeras form. ( x) v( x) p u + (5.68) We have ow oe equatio ad three ukows, p, u(x) ad v(x). We eed two more equatios. The first oe results from the fact the particular solutio must satisf the origial ODE.

16 5-6 p du dv u + v + + (5.69) The secod equatio is arbitrar ad ca be set b the modeler. However, if du dv + (5.7) is selected, the resultig equatios will easier. d p d d du dv u + v + + (5.7) Combiig equatios 5.66 ad ields u + a du + + ( x) + a ( x) + v + a ( x) + a ( x) dv h( x) (5.7) The two expressios i the brackets should be zero as (x) ad (x) are soultios of the homogeeous equatio ad part of the complemetar solutio. du + dv h( x) (5.73) Whe equatios 5.7 ad 5.73 are solved two itegral expressios are obtaied for u ad v as u h (5.74) v h (5.75) With the fuctioal froms of u(x) ad v(x), the particular solutio is costructed substitutig them ito equatio The Euler equatio The geeral form of the Euler equatio is x d d + a x + + a x + a h ( x) L (5.76) This equatio ca be reduced to a liear ODE with costat coefficiets usig the subtitutio xe z.

17 Sstem of liear differetial equatios with costat coefficiets The solutio of sstems of liear ODEs with costat coefficies will be discussed usig sample secod order liear ODE. Cosider the followig sstem dz z dz z 4 (5.77) (5.78) The solutio techique starts with covertig the ODEs ito the operatioal form with differetial operator D. ( + ) + ( D + ) z D (5.79) ( + ) + ( D + ) z 4 D (5.8) Now maipulate the two equatios to elimiate ad fid a ODE i terms of z. As the operators of are idetical, we eed to subtract equatio 5.8 from 5.79 to do that. The result is ( ) z 4 D (5.8) The complemetar solutio to equatio 5.8 is z c A (5.8) The form of the particular solutio from Table XXX is a costat. However, there is alrea a costat term i the complemetar solutio. Therefore, we eed to multipl the suggested costat term with x to obtai the geeral form of the particular solutio as z p Bx (5.83) Usig the method of udetermied coefficiets we substitute the resultig particular solutio ito equatio 5.8. This ields a value of 4 for B. The fial solutio for z is ow z A 4x (5.84) Let s follow a similar procedure to fid a solutio for. To elimiate z from equatios 5.79 ad 5.8 mutipl them with (D+) ad (D+), respectivel. ( + )( D + ) + ( D + )( D + ) z D (5.85) ( + )( D + ) + ( D + )( D + ) z 4( D + ) D (5.86) Whe the top equatio is subtracted from the bottom oe ( + )( D + ) ( D + )( D + ) 4 D (5.87) After maipulatio the equatio becomes

18 5-8 ( D + ) 4 D (5.88) The complemetar solutio to equatio 5.88 is c x C + De (5.89) The particular soluttio is foud as described i the previous case as p x (5.9) The geeral solutio is ow C + De x + x (5.9) We are almost there. We have the equatios for z ad as C + De x + x (5.9) z A 4x (5.93) Origiall we had two first order ODEs but we eded up havig three arbitrar costats. Oe of them is too ma ad should be stated i terms of the others. To fid it let s use oe of the origial ODEs, sa equatio 5.79, ad substitute the equatios for ad z ito it. x ( D + )( C + De + x) + ( D + )( A 4x) (5.94) Applig the differetial operators ield De x + + C + De x + 4x 8 + A 4x (5.95) After some maipulatio B is stated as C 6 A (5.96) The fial solutio becomes 6 x A + De + x (5.97) z A 4x (5.98) Now with two boudar coditios the remaiig arbitar costats ca be determied to give problem specific solutio. For example, if i a specific problem at x ad z values has to be, as well, the i the fial fial solutio A ad D would be ad 3, respectivel.

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