# Differential Equations and Linear Superposition

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1 Differenial Equaions and Linear Superposiion Basic Idea: Provide soluion in closed form Like Inegraion, no general soluions in closed form Order of equaion: highes derivaive in equaion e.g. dy d dy 2 y 0 d is a rd order, non-linear equaion. Firs Order Equaions: (separable, eac, linear, ricks) A separable equaion can be wrien: e.g. dy A( ) B( y) A( )d d A ( )d B ( y)dy C dy y 2 d 2 0 B( y)dy dy y 2 d y 2 y 2 C 0 2

2 More Generally, an equaion can be an eac differenial: A(, y)d B (, y)dy 0 if he lef side above is he eac differenial of some u(,y) hen he soluion is: u(, y) C A necessary and sufficien condiion for his is: A(, y) y B(, y) e.g. ( 2 2y) ( 2 y 2 )y is eac. We can hen parially inegrae A(,y) in and B(,y) in y o guess a from for u(,y): ( 2 2y)d ( 2 y 2 )dy u(, y) 2y 2y y y C ( y) C( ) 2y C For firs order equaions, i is always possible o find a facor λ(,y) which makes he equaion eac. Unforunaely, here is no general echnique o find λ(,y)...

3 Consider he general linear firs-order equaion: dy d f( )y g( ) Les ry o find an inegraing facor for his equaion... λ ( y, ) We wan λ ( )dy λ ( ) ( f( )y g( ) )d 0 o be eac. Equaing he parial derivaives: A(, y) y B(, y) d d λ ( ) λ( )f( ) This new equaion for lambda is separable wih soluion: λ ( ) e f( ) d So we can always find an inegraing facor for a firs order linear differenial equaion, provided ha we can inegrae f(). Then we can solve he original equaion since i is now eac. e.g. The inegraing facor is: y ( )y e y y ( ) e λ ( ) e f( ) d e So we ge he eac equaion: e y e y( ) e 2 Inegraing boh sides: e y e 2 C y 2 e 2 Ce

4 Why do we call he equaion below linear? Consider a se of soluions: dy d f( )y 0 y { y ( ), y 2 ( ),... } Any linear combinaion of he above soluions is anoher valid soluion of he original equaion. If we se: y ay ( ) by 2 ( ) we ge: dy a d dy 2 b af ( )y d ( ) bf ( )y 2 ( ) 0 which is clearly saisfied if y and y 2 are indeed soluions. Linear equaions are saisfied by any linear superposiion of soluions. We divide he se of soluions ino a se of linearly independen soluions saisfying he linear operaor, and a paricular soluion saisfying he forcing funcion g(). In general, i can be shown ha over a coninuous inerval, an equaion of order k will have k linearly independen soluions o he homogenous equaion (he linear operaor), and one or more paricular soluions saisfying he general (inhomogeneous) equaion. A general soluion is he superposiion of a linear combinaion of homogenous soluions and a paricular soluion. A wide variey of equaions of ineres o an elecrical engineer are in fac linear...

5 Consider he circui below: I R V C Assuming ha C is iniially uncharged, when he swich is closed, a curren flows across he resisor, charging he capacior. We know ha he oal charge Q across a capacior is jus C imes he volage across i: Q() CV C () Also, he charge on he capacior mus be he inegral of he curren I hrough he resisor: Q() I C () d The curren hrough he resisor is deermined by he volage across i: I R () V R () R From Kirchoff s laws, we have ha he oal volage drop around a loop is zero and ha he oal curren a each node is zero. So we can find V C from he consan baery volage V and he volage across he resisor: V V R () V C (), I R () I C () So we ge: V C () d d V C () C ( V V C () ) d R ( V V C () )

6 Given ha V C (0) 0, find he volage across he capacior as a funcion of ime. We have: The homogenous soluion is separable: d d V C () And he paricular soluion is rivial since d/d (V) 0: So he general soluion is: d d V C () Since V C (0) 0, we have: V C () V V C () 0 V C () e V C () V V C () V ae V C () V Ve V e I is cusomary o se o ime scale of such a circui by he quaniy which has he unis of ime. A, V C V(-e - ) 0.62V

7 Now consider he case in which he driving volage is a sine wave such as power grid disribuion curren. We hen have: d d V C () This paricular soluion is a bi harder han he las, bu he equaion sill falls ino he firs-order linear mold: f() / and g() A/*Sin(w). The inegraing facor is: so he equaion below should be eac: e d d V C () V C () Asin ( w) λ () e Carrying hrough he inegraion as before: e V C () Asin ( w)e V C sin ( w) wcos ( w) A R 2 C 2 w 2 A ( sin ( w) wcos ( w) ) w2 is he paricular soluion. Since we can superpose soluions, if we have a consan volage and a sinewave drive, he general soluion is: V C () ae A ( sin ( w) wcos ( w) ) w2 An eample of such a soluion is he circui wih a consan volage and a small amoun of AC noise. Alhough he soluion above looks dauning, remember ha any linear combinaion of sin and cos funcions of he same frequency can always be wrien in he form: Asin(wθ). Thus he capacior is changing he phase of he impressed volage. The morass of algebra above is why we ofen choose o wrie such funcions as he real par of e i(wθ). (Phasor noaion).

8 Eample If he Baery was in fac a source of alernaing curren, wih uni frequency and ampliude, and if he inial charge on he capacior was zero, we ge he following plo for he volage across he capacior: Noe ha here is a ransien adjusmen o he phase and poenial. A wice he frequency, he phase and he ampliued are boh changed: However, noe ha he ime scale of he ransien is unchanged.

9 We can eend his noion of superposiion of soluions o he case of muliple sources and muliple simulaneous equaions. To solve, we simply consruc soluions for each of he sources acing independenly, and hen superpose he (add) hem o find he ne behavior. For eample, if he inpu o he previous circui was wo waves, one a uni frequency and zero phase and anoher a wice he frequency and zero phase, we ge he following superposed soluion: V C () 2π 4π 2 e sin ( 2π) 2πcos ( 2π) 4π 2 4π 6π 2 e sin ( 2π) 2πcos ( 2π) 4π 2 Noe, ha he area of he wave ends o balance across he capacior:

10 Phasors Represen an alernaive curren by a single comple number Trick -- use o represen boh he ampliude and phase Ae iϕ The idea here is ha i is much easier o manipulae polar form comple numbers han rigonomeric funcions. We shall represen: Acos ( ω ϕ) Re Ae iϕ e iw ( ) So far -- i doesn look much beer. However, in some applicaions all pars of he circui have he same frequency of operaion. In his case, we only need o carry he comple phasor Ae iϕ o keep rack of boh. A phasor of he sum of wo alernaing signals is he same as he sum of he phasors So if we have wo volage sources A and A2, represened by A e iϕ and bya 2 e iϕ 2 heir sum is jus A e iϕ A 2 e iϕ 2. (Noe ha his doe no hold for he produc of wo phasors. I is herefore safe o represen soluions o linear differenial equaions as phasors and do calculaions in shorand, as long as one realizes ha he real-par is he desired soluion. A second applicaion will be he calculaion of branch volages and currens in alernaing curren (A.C.) circuis. In a general nework of resisors, inducors and capaciors, he general behavior can be shown o be linear. Thus, he volage beween any wo nodes or he curren along any branch will have he same frequency as he source. Thus only wo numbers, (a phasor) sufices o deermine he circui behavior--- he ampliude and he phase. (We shall come back o his opic in a laer lecure).

11 Oscillaing (Resonan) Circuis Consider he circui below: I C L R If we ake posiive curren as shown, we have: dq di I Q CV V L RI d d To solve his, we will eliminae boh Q and I -- o ge a differenial equaion in V: di d I d d ( CV) dv C d 2 2 dv dv C V LC 2 d 2 d 2 dv R d 2 dv Ld LC V 0 dv d This is a linear differenial equaion of second order (noe ha solve for I would also have made a second order equaion!). Such equaions have wo indepeden soluions, and a general soluion is jus a superposiion of he wo soluions. We shall assume ha inially, he volage sored in C is A.

12 Suppose we ry: V Âe iw for a comple phasor Â. We have: ω 2 Âe iw iω R L Âeiw LC Âeiw 0 This equaion holds for all, w, A only if he following quadraic equaion is idenically zero: We ge: ω 2 i R L ω 0 LC ω i R 2L ± LC R 2 4L 2 So w is in general comple. This equaion has several special cases: R 0 (lossless case) hen w is pure real and he soluion is an alernaing volage of frequency:. V LC Re( Âe iw ) A cos ( LC ϕ) Here he phase is zero for he iniial condiions. R posiive, bu w sill has a real par: LC R 2 > 4L 2 R < 2 L C V R 2L Ae cos LC R 2 4L 2 ϕ In his case, he real par of w defines he frequency of oscillaion, while he

13 imaginary par deermines he energy loss. Noe ha he ampliude decreases eponenially for increasing ime. R large -- w is pure imaginary: V R 2L A e R 2 2L 2 LC R 2L e R 2 2L 2 LC In his case, he volage is no oscillaory -- in fac, i simply falls eponenially o zero. Finally, here is a special criical value: R 2 L C In his case, he equaion above is singluar -- or equivalenly, he quadraic equaion for w has a repeaed roo. ω i R 2L V ( A B)e i LC LC In each of he above cases, we have consisanly used he noion ha he value was represened by he real par of he comple polar form. This rick has saved an enormous amoun of work -- compared o using rigonomeric funcions.

14 More Techniques for Differenial Equaions A paricularly powerful echnique is ha of making an appropriae change of variable in an equaion. For eample, y' f( a by c) is easily solved if we make he subsiuion: u a by c A variable change is performed by wriing he form o be subsiued, find each (oal) derivaive and hen subsiuing. e.g. u a by c, u' a by' So he equaion above becomes: u' a b f( u) du a bf( u) d which is separable. Noe ha afer he inegraion, we mus replace u by abbyc o ge he final soluion. In general, a appropriae subsiuion is a very powerful echnique o simplify or solve a ypical differenial equaion. Recognizing ha a paricular subsiuion is useful is ofen a ricky proposiion. Some special cases are lised below: y' f( )y g( )y n Bernoulli Equaion can be made linear by seing u y n. Noe ha: u' ( n)y n y'

15 and: y' y n f( ) y n g( ) So subsiuing: u' n uf ( ) g( ) which is linear. Homogenous firs-order equaions: denoes a homogenous funcion of degree r. A general firs order equaion is homogenous if boh A and B are homogenous funcions. Then can be solved by subsiuing: A( a, ay) a r A(, y) A(, y)d B (, y)dy 0 y( ) u ( ), dy u( )d du E.g. yd ( 2 y )dy 0 Here r. If we subsiue: u ( )d ( 2 2 u ) ( u( )d du) 2 2u d ( 2 u )du 0 which is separable. 2d ( 2 u ) u 2 du

16 If he dependen variable y is missing, subsiue y u, lowering he order by one. e.g. y'' f( )y' g( ) 0 u' f( )u g( ) 0 Then solve and inegrae when done... If he independen variable is missing, le y be he new independen variable, and le u y, u y ec. This also lowers he order by one. e.g. e.e.g. y'' f( y' ) g( y) 0 u' f( u) g( y) 0 y'' y' y 2 0 u' u y 2 0 The homogenous soluion: u' u 0 u ae y The paricular soluion: u y 2 2y 2 In general: u y' ae y y 2 2y 2 So finally: ae y y 2 dy 2y 2

17 Always wach for an eac equaion or for an appropriae inegraing facor: e.g. y'' f( y) can be inegraed immediaely if boh sides are muliplied by y... y''y' f( y)y' 2 ( y' )2 f ( y)dy e.e.g. y'' y 2 2 ( y' )2 y y 6 2 y' y 2 ± 6 ( C) 2 A paricularly useful subsiuion is one of he simples: for a linear second order equaion. Given he equaion: Subsiuing, we obain: y u( )v( ) y f( )y g( )y 0 2u u fu gu v ( f( ) )v ( )v 0 u u Now, i is clear ha special chioces for u will poenially give us simplified equaions for v... For eample, if u is a soluion o he original equaion, he hrid erm drops ou, leaving an equaion han can be reduced o firs order, linear. If we ry: u hen u saisfies he second erm and i drops ou, leaving a simpler second order equaion. ( ) f( )d 2 e

18 e.g. If our equaion is: y 2 y 4 y 0 and we happen o guess ha one soluion is: y e we can use he previous subsiuion o find he oher soluion... So we subsiue: y( ) v( )e v( ), y ( ) ( v ( ) )e 2 y ( ) v 4 v v v e 4 And we ge: v v v 0 2 This equaion can be simplified by leing w v : Which is separable-- we ge: w w ( ) 0 2 log ( w) C 2 log ( ) 0 2 w ae 2 v So inegraing, one las ime: v ( c)e 2 y is he oher soluion... ( c)e 2 e ce

19 We can use he preceeding rick o help find paricular soluions as well. Consider he following 2nd order linear equaion: If we know a soluion of he homogenous equaion, u(), we can subsiue as before o ge a new linear s order equaion in v (). Solving his equaion and hen inegraing again will allow he paricular soluion o be found. This migh be an easier procedure han variaion of parameers... e.g. y f( )y g( )y h( ) y ( 2 4)y 4 2 y 2e Noe ha a soluion o he homogenous equaion is simply: y ae 4 So we subsiue: y v( )e 4, y v e 4 4ve 4, y 6e 4 v 8e 4 v e 4 v To obain: v ( 2 4)v 2e This equaion is firs order, linear in v -- we ge: v ( ) 4 8e 4 ae

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