VŠB TECHNICAL UNIVERSITY OF OSTRAVA

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2 VŠB TECHNICAL UNIVERSITY OF OSTRAVA Deprtment of Mthemtics nd Descriptive Geometry MATHEMATICS II Pvel Kreml Ostrv

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4 CONTENTS INDEFINITE INTEGRALS The Indefinite Integrl Computtion of Integrls Some Properties of the Indefinite Integrl 5 4 Sustitution 5 4 Integrtion y Prts 7 DEFINITE INTEGRALS 9 The Definite Integrl 9 The Are Prolem nd the Definite Integrl 0 Rules for Definite Integrls 4 Integrtion y Prts in Definite Integrls 5 Sustitution in Definite Integrls 6 Applictions of Integrtion Finding Ares under Curves Are Between Two Curves 4 Volumes of Revolution 5 The Length of Curve 6 Are of Surfce of Revolution 7 DIFFERENTIAL EQUATIONS 8 Introduction 8 Seprle Equtions 0 Liner Differentil Equtions Homogeneous Liner Differentil Eqution Nonhomogeneous Liner Differentil Eqution Method for Solving Liner Differentil Eqution 4 Liner Differentil Equtions n th order 5 5 Homogeneous Equtions with Constnt Coefficients 6 6 Homogeneous nd Order Equtions with Constnt Coefficients 7 I Rel Distinct Roots: 7 II Repeted Roots: 7 III Comple Conjugte Roots: 8 6 Nonhomogeneous Equtions with Constnt Coefficients 9 Method of Vrition of Prmeters 0 Method of Undetermined Coefficients LITERATURE 5 - -

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6 Indefinite Integrls INDEFINITE INTEGRALS We will turn our ttention to reversing the opertion of differentition Given the derivte of function, we cn find the function This process is clled ntidifferentition The Indefinite Integrl The derivtive of function f() = 5 4 t ny vlue is f () 0 We put question now If F () = 5 4, wht is F()? We know the function could e F () = 5 ecuse F ()=( 5 ) = 5 4 = f() But the function could lso e F () = 5 + C, ecuse F () = ( 5 + C) = 5 4 Thus we sy the ntiderivtive of f() = 5 4 is F () = 5 + C, where C is n ritrry constnt The process of finding n ntiderivtive is clled integrtion The function tht results when integrtion tkes plce is clled n indefinite integrl or more simply n integrl We cn denote the indefinite integrl (tht is, the ntiderivtive) of function f() y f ( d ) 4 Thus we cn write 5 d to indicte the ntiderivtive of the function f() = 5 4 The epression is red s the integrl of 5 4 with respect to In this cse, 5 4 is clled the integrnd, the integrl sign indictes the process of integrtion, nd the d indictes tht the integrl is to e tken with respect to We cn write d C Definition: A function F() is n ntiderivtive of function f() if F () = f() for ll in the domin of f The set of ll ntiderivtives of f is the indefinite integrl of f with respect to, denoted y f ( d ) F( ) C The symol is n integrl sign The function f is the integrnd of the integrl, nd is the vrile of integrtion Computtion of Integrls Now tht we know wht re integrls The net prolem is to find out how to do them For differentition we hd tidy set of rules tht llowed us to work out the derivtive of just out ny function tht we cred to write down The procedure is siclly mechnicl nd cn e done quite well y computers There is nothing like this for integrtion Integrtion is more of skill thn routine - -

7 Indefinite Integrls If you cn spot function which differentites to give your function then you hve found n integrl Look t the following simple emples Emple: Wht is the ntiderivtive of f() for (, ) : ) f() = 0, F () = C, ecuse F( ) C 0 f() ) f() =, F () = +C, ecuse F( ) ( C) 0 f() c) f() =, F () = C, ecuse F( ) ( C) 0 f( ) d) f ( ) sin, F () = cos C, ecuse F( ) ( cos C) ( sin ) 0 sin f( ) Tle : Integrtion of some common functions [] 0 d C [] d C n n [] d C n [4] d ln C [5] sin d cos C where 0, n where 0 [6] cos d sin C [7] d tg C where ( k ), k Z cos [8] d cotg C where k, k Z sin [9] d rcsin C where (, ) [0] d rctg C [] d C ln [] e d e C where 0, - 4 -

8 Indefinite Integrls This is short tle of some stndrd integrls You cn dd in constnt of integrtion if you wnt to Some Properties of the Indefinite Integrl These re immedite consequences of the corresponding properties of derivtives In ech eqution there is relly n ritrry constnt of integrtion hnging round Let f () nd g() e functions nd k constnt Then ) g( ) d f ( ) d ) g( ) d f ( ) d kf ) d k f ( ) d f ( g( ) d, f ( g( ) d, (, kr These rules my llow us to reduce n integrl to the point where we cn spot the nswer Emple: e d cos 7 6 sin = cos d d 7 e d 6 d sin sin cot g 7e 6 C = sin cotg 7e 6 C = = 4 Sustitution Mny integrls re hrd to perform t first hnd A smrt ide consists in clening ' them through n lgeric sustitution which trnsforms the given integrls into esier ones Theorem: Let f e continuous function defined on intervl (, ) nd let :(, ) ( cd, ) e differentile function Then the following sttement hold f ( d ) f( t) ( tdt ) Remrk: To epress integrtion y sustitution, we use the nottion () t, then d () t dt nd f ( d ) f( t) ( tdt ) Emple: Find d Solution: It is esy to see tht sine-sustitution is the one to use Set () t sint The function is continuous on 0, Indeed, we hve d ( t) dt cos t dt nd - 5 -

9 Indefinite Integrls therefore sin t cos t d costdt dt dt cotg t t C sin t sin t sin t This will not nswer fully the prolem ecuse the nswer should e given s function of Since t rcsin we get fter esy simplifictions sin cotg t d t t C t C rcsin C sin t Emple: Find 55 ( 7) d 55 Solution: It is cler tht once we develop the ( 7) through the inomil formul, we will get polynomil function esy to integrte But it is cler tht this will tke lot of time with ig possiility of doing mistkes! Let us consider the sustitution t 7 (the reson ehind is the presence of in the integrl since the derivtive of ( 7) is ) Indeed, we hve dt d nd therefore dt 56 ( 7) d t t C 55 Indefinite integrl ( 7) d is function of not of t Therefore, we hve to go ck nd replce t y t(): ( 7) d t C ( 7) C Emple: Let us evlute k e d Solution: If you sustitute t k, then dt k d kd, d dt nd we otin: k t t t k e dt e dt e C e C k k k k Similrly we cn etend Tle Tle (dditionl): Integrtion of some functions [] e k d= k e k C [4] sin kd = cos k C k [5] cos kd = sin k C k f ( d ) [6] = ln f ( ) C f ( ) - 6 -

10 4 Integrtion y Prts Indefinite Integrls One of very common mistke students usully do is f ( gd ) ( ) f( d ) gd ( ) To convince yourself tht it is wrong formul, tke f() = nd g()= Therefore, one my wonder wht to do in this cse A prtil nswer is given y wht is clled Integrtion y Prts In order to understnd this technique, recll the formul ( uv ( ) ( )) u( v ) ( ) uv ( ) ( ) which implies uv ( ) ( ) u( vd ) ( ) uv ( ) ( d ) Theorem: Let u() nd v() hve continuous derivtives on (,) u v du v u v d Then Remrk: We usully use the nottion uvd uv u vd u vd u v u v d lterntively Therefore if one of the two integrls u v d nd u v dis esy to evlute, we cn use it to get the other one This is the min ide ehind integrtion y prts Emple: Evlute ed Solution: Since the derivtive or the integrl of e led to the sme function, it will not mtter whether we do one opertion or the other Therefore, we concentrte on the other function Clerly, if we integrte we will increse the power This suggests tht we should differentite nd integrte e Hence u e, v, fter integrtion nd differentition, we get u ud e de, v( ) The integrtion y prts formul gives ed e ed It is cler tht the new integrl ed is not esily otinle Due to its similrity with the initil integrl, we will use integrtion y prts for second time The sme discussion s efore leds to u e, v, fter integrtion nd differentition, we get - 7 -

11 u ud e de, v( ) The integrtion y prts formul gives ed e ed e e ed e e e C Indefinite Integrls From this emple, try to rememer tht most of the time the integrtion y prts will not e enough to give you the nswer fter one shot You my need to do some etr work: nother integrtion y prts or use other techniques Emple: Evlute ln d Solution: This is n indefinite integrl involving one function The second needed function is g() = Since the derivtive of this function is 0, the only choice left is to differentite the other function f ( ) ln : u, v ln, fter integrtion nd differentition, we get u d, v ln The integrtion y prts formul gives ln d = ln d ln d ln C ln C Remrk: Since the derivtive of ln is /, it is very common tht whenever n integrl involves function which is product of ln with nother function, to differentite ln nd integrte the other function Emple: Evlute I = e cos d Solution: The two functions involved in this emple do not ehiit ny specil ehvior when it comes to differentiting or integrting Therefore, we choose one function to e differentited nd the other one to e integrted We hve u e, v cos, which implies u ud e d e, v(cos ) sin The integrtion y prts formul gives I = e cos d= e cos e sind = e cos e sin d The new integrl e sin d is similr in nture to the initil one One of the common mistke is to do nother integrtion y prts in which we integrte sin nd differentite e This will simply tke you ck to your originl integrl with nothing done In fct, wht you would hve done is simply the reverse pth of the integrtion y prts (Do the clcultions to convince yourself) Therefore we continue doing nother integrtion y prts s u e, v sin, which implies u ud e d e, v(sin ) cos The integrtion y prts formul

12 - 9 - Indefinite Integrls e sin d= e sin e cosd = e sin e cos d Comining oth formuls we get I = e cos d= e cos e sin e cosd Esy clcultions give I = e cos 4 d= e cos e sin I 9 9 After two integrtion y prts, we get n integrl identicl to the initil one You my wonder why nd simply ecuse the derivtive nd integrtion of e re the sme while you need two derivtives of the cosine function to generte the sme function Finlly esy lgeric mnipultion gives I = e cos d= e cos e sin C DEFINITE INTEGRALS Integrtion is vitl to mny scientific res Mny powerful mthemticl tools re sed on integrtion Differentil equtions for instnce re the direct consequence of the development of integrtion So wht is integrtion? Integrtion stems from two different prolems The more immedite prolem is to find the inverse trnsform of the derivtive This concept is known s finding the ntiderivtive The other prolem dels with res nd how to find them The ridge etween these two different prolems is the Fundmentl Theorem of Clculus The Definite Integrl Definition: Suppose tht F() is n indefinite integrl of f (), ie F'() = f () The Definite Integrl f ( ) d where nd re numers, is defined to e the numer F( ) F( ) F( ) f ( ) d nd re clled the Limits of Integrtion, is clled the Lower Limit nd is clled the Upper Limit Remrk: The choice of indefinite integrl (choice of constnt of integrtion) does not mtter- -the constnt of integrtion cncels out f ( d ) F ( ) C F ( ) C F ( ) C F ( ) CF ( ) C F ( ) F ( )

13 Definite Integrls Emple: Evlute 4 d Solution: (4 4) d4 (4 4 4) (4 ) The Are Prolem nd the Definite Integrl Consider the intervl <, > Let f () e continuous function defined on this intervl Let us think once more of the prolem of finding the re under the grph of f () etween = nd = Figure : Dividing up the re under curve We will dopt n pproch tht is much more elementry (nd much older) thn our previous method Divide the intervl <, > up into lrge numer of smll prts For convenience we will tke them ll to e of the sme width, ut tht is not very importnt Now use this sudivision to rek up the re into thin strips s shown Denote the sudivision points y 0,,, n where k = + k nd is the width of ech strip = ( - )/n We cn get n pproimtion to the re under the grph y dding up the res of the n rectngles shown in the picture The rectngle on the se < k, k + > hs re f( k ) So pproimte re = n k 0 f ( ) k Figure : One of the rectngles - 0 -

14 Definite Integrls Now we use the sme kind of rgument tht we used when inventing the derivtive As n gets igger nd igger we epect the sum of the res of the rectngles to get closer nd closer to the true re under the grph We would hope tht if we took the limit s n the sum would tend to the true re s its limiting vlue We will ssume tht this is true So A = re under grph = If there eists numer A such tht n n k 0 n lim f ( ) n k 0 k A lim f( ), then f is integrle on <, > nd A is the definite integrl of f over <, > This is denoted f ( ) d This interprettion of the definite integrl is the one tht is most useful in pplictions, s we will soon see Rules for Definite Integrls Order of Integrtion: f ( ) d f ( ) d Zero: f ( ) d 0 Constnt Multiple: kf ( ) d k f ( ) d (F() - F() = - (F() - F())) 4 Sum nd Difference: ( f ( ) g( )) d f ( ) d g ( ) d c c 5 Additivity: f ( ) d + f ( ) d f ( ) d 6 Domintion: f ( ) g( ) on, f ( ) d g ( ) d k - -

15 4 Integrtion y Prts in Definite Integrls Definite Integrls Theorem: Let u() nd v() hve continuous derivtives on <,> u v d u v u v d Then Remrk: We usully use the nottion u vd u v u vd lterntively uvd uv u vd Emple: Evlute ( )sind 0 Solution: We choose u sin, v, fter integrtion nd differentition, we get u ud sin dcos, v( ) 0 The integrtion y prts formul gives ( )sind cos cos d cos sin = = cos sin ( 0)cos0 sin Sustitution in Definite Integrls In definite integrl f ( ) d it is lwys understood tht is independent vrile nd we re integrting etween the limits nd Thus when we chnge to new independent vrile t, we must lso chnge limits of integrtion Theorem: Suppose f is continuous function nd hs ntiderivtive on intervl, nd let function ( t) hs continuous derivtive on, nd mps, into, ( ( ) nd ( ) ) Then f ( d ) f( ( t)) ( tdt ) In this cse, you will never hve to go ck to the initil vrile Emple: Evlute 0 cos 4 sin d - -

16 Solution: Put cos t, so -sin d Definite Integrls dt, sin d -dt As goes from 0 to the vlue of t goes from = cos 0 = to = 0 cos 4 0 cos = 0 So the integrl trnsforms into : t sin d = t dt t dt Applictions of Integrtion Finding Ares under Curves 0 0 y y = f() We wnt to find the re of the region ounded ove y the grph of positive function f (), ounded elow y the - is, ounded to the left y the verticl line =, nd to the right y the verticl line = (see Figure ) Figure If f is continuous function on <, >, then the re is given y A = Are (shdet) = f ( ) d Remrk: I drew the picture conveniently with the grph ove the is If f () goes negtive then the re clculted y the integrl lso goes negtive Emple: Find the re ounded y the curve y = 4 nd -es Solution: The re to integrte must e n enclosed re This time the upper ound is the - is, the lower ound is the curve (see Figure 4) y We first find nd y finding the -coordintes of points of intersection of the function nd -is We solve eqution y = for 0 4 ( 4) 0 so 0, 4 Figure 4 The functionl vlues over the intervl <0, > re negtive Thus the vlue of the definite integrl over this intervl will e negtive too The re is given y - -

17 Definite Integrls d 4 d A = Are Between Two Curves We hve used the definite integrl to find the re of the region etween curve nd the -is over n intervl where the curve lies ove the -is We cn esily etend this technique to finding the re etween two curves over n intervl where one curve lies ove the other (see Figure 5) Suppose tht the grphs of oth y f( ) nd y g( ) lie ove the -is, nd tht the grph of y f( ) lies ove y g( ) throughout the intervl Figure 5 from to ; tht is f ( ) g( ) on <, > Figure 6 Then f ( d ) gives the re etween the grph of y f( ) nd the -is (see Figure 6 ()), nd gd ( ) gives the re etween the grph of y g ( ) nd the -is (see Figure 6 ()) As Figure 69 (c) shows, the re of the region etween the grphs of y f( ) nd y g( ) is the difference of these two res Tht is f d d f g d A = Are etween the curves = ( ) g ( ) ( ) ( ) - 4 -

18 Definite Integrls Emple: Find the re of the region ounded y y nd y Solution: The grph of the region is shown in Figure 7 We first find nd y finding -coordintes of the points of intersection of the grphs Setting the y-vlues equl gives y y = y = + - Figure 7 = +, - - = 0, ( )( + ) = 0, =, = -, thus =- nd = The re of enclosed region is: A = d Volumes of Revolution Tke the grph of y = f () on the intervl <, > nd spin it round the -is so s to produce wht is known s solid of revolution s shown in Figure 8 We wnt to get formul for the volume of this solid The method is lmost ectly the sme s in the previous emples Think of the intervl <, > eing sudivided into lots of little its Now look t one of the its nd try to get n pproimtion for the volume of the thin slice of the solid otined y rotting the piece of the grph on this intervl Figure 8 In the nottion of the digrm, the thin slice of the solid is virtully cylinder of rdius y nd thickness Figure 9 The volume of cylinder is the product of its height nd the re of its se So we get the pproimtion V y for the volume of the slice The pproimtion to the totl volume cn then e written s Figure 9-5 -

19 Definite Integrls Now tke the limit s n nd get V y y d f ( ) d V=Volume = Emple: Find the volume V of solid generted when the grph of function y revolves round the -is on the intervl <0, > y y = Solution: In Figure 0 we see tht 0, The volume of the solid is 0 y = - V = 0 Figure d d The Length of Curve Suppose we wnt to clculte the length of the grph of y f( ) etween nd Sudivide <, > s efore into n smll prts Look t the grph on one of these prts The ide is to get n pproimtion to the length of the grph on this prt, in the sme wy tht we used the rectngle pproimtion when finding the re Then we dd up the pproimtions nd tke the limit s n so s to produce n integrl which gives the true length The ovious pproch is to use the length of the chord PQ s n pproimtion to the length of the grph etween P nd Q In the nottion of Figure this length is s y which we cn write more conveniently s y s Figure We now hve the pproimtion to the length which I will write crudely s - 6 -

20 Definite Integrls y L Our new interprettion of the definite integrl tells us tht, s n this tends to the vlue of the definite integrl Remrk: The y tends to dy d L = Length in the limit dy d d Note: If the curve is given prmetriclly y = (t) nd y = y(t) then very similr rgument gives us the formul t y t dt L = Length () () for the length of the curve etween t nd t Emple: Find the length L of the steroid y, where 0 Solution: Let us employ the prmetric epression cos t, y sin t, where t 0, We clculte cos tsint, y sin tcost Figure In this cse 4 4 () t y() t 9 cos tsin t9 sin tcos t 9 cos tsin t By the symmetry (see Figure ), we hve cos t L 4 9 cos t sin tdt cost sin tdt 6 sin tdt Are of Surfce of Revolution I'm going to e rief here nd just give you the formul - 7 -

21 Definite Integrls Suppose we tke the grph y = f () on the rnge <, > nd rotte it round the -is s efore Then the Surfce Are of the surfce formed y this is given y S = Surfce Are dy y d d Emple: Find the surfce re of the sphere Solution: We otin our sphere y rotting the semicircle Tke the semicircle y R on <- R, R> nd spin it round the -is We get sphere of rdius R For this curve dy d R So dy R d R R So the re is given y R R R R S=Surfce of sphere R d RdR 4R R R R So sphere of rdius R hs re 4 R DIFFERENTIAL EQUATIONS Introduction A differentil eqution is n eqution for n unknown function, sy y(), which involves derivtives of the function dy For emple: d, y y y cos, y y y 4 y The order of differentil eqution is the order of the highest derivtive occurring in it In the ove emples the orders re, nd Remrk: Techniclly, these re known s Ordinry Differentil Equtions (ODE) ecuse the unknown function is function of one vrile Differentil equtions involving functions of severl vriles nd their prtil derivtives re clled Prtil Differentil Equtions (PDE) R

22 Differentil Equtions Mny lws in science nd engineering re sttements out the reltionship etween quntity nd the wy in which it chnges The chnge is often mesured y derivtive nd therefore the mthemticl epression of these lws tends to e in terms of differentil equtions Given differentil eqution the ovious rection is to try to solve it for the unknown function As with integrls, nd for much the sme reson, this is esier sid thn done Consider the differentil eqution y''() =, where the dsh denotes differentition with respect to Integrting oth sides of this eqution with respect to we get y d C, where C is n ritrry constnt of integrtion This is one plce where it is crucilly importnt to include the constnt of integrtion! Now integrte once more nd get y ( C) d CD, where D is further constnt of integrtion We now hve the Generl Solution of the differentil eqution with ritrry constnts C nd D Note tht this is relly n infinite clss of solutions If we give C nd D prticulr vlues then we get Prticulr Solution For emple, y nd y 5 re prticulr solutions of the eqution The vlues of the ritrry constnts tht we lmost invrily cquire when solving differentil eqution re usully determined y giving conditions tht the solution is required to stisfy The most common kind of conditions re Initil Conditions, where the vlues of y nd some of its derivtives re given for specific vlue of Emple: Find the solution to y''() = then stisfies y(0) nd y(0) 0 Solution: We know tht the generl solution is y C D The condition y(0) sys tht = D, so D = The condition y(0) 0 sys tht 0 = 0 + C, so C = 0 So the required solution is y - 9 -

23 Differentil Equtions Note: An eqution of order n generlly requires n integrtions to get the generl solution, so the generl solution cn e epected to contin n unknown constnts nd you would epect to hve to give n conditions to fi these constnts Seprle Equtions It is frequently necessry to chnge the form of differentil eqution efore it cn e solved with the techniques of the previous section For emple, the eqution y y cnnot e solved y simply integrting oth sides of the eqution with respect to We cnnot evlute solution we seek y d unless we cn write y s function of, ut y = f() is the Becuse dy y, we cn multiply oth sides of d dy y d y d y to otin n eqution tht hs ll terms contining y on one side of the eqution nd ll terms contining on the other side Tht is, we otin dy d y A differentil eqution is sid to e seprle if it cn e mnipulted into the form f ( ydy ) gd ( ) The solution of seprle differentil eqution is otined y integrting oth sides of the eqution fter the vrile hve een seprted f ( ydy ) gd ( ) C It my not e possile to epress y simply in terms of Emple: Solve the differentil eqution ( y ) y Solution: To write the eqution in seprle form, we first fctor dy epress y d ( y) dy d ( y ) dy d from the left side nd

24 The eqution is now seprted, so we integrte oth sides ( y) dy d y y C This eqution, s well s the eqution Differentil Equtions y y C gives the solution implicitly Liner Differentil Equtions Some very importnt pplictions re modeled y specil clss of differentil equtions, clled Liner Differentil Equtions We will solve liner differentil equtions in which the highest derivtive is the first derivtive; these re clled First-Order Liner Differentil Equtions A first-order liner differentil eqution is n eqution of the form y py ( ) q ( ), where p nd q re functions of y Emple: Differentil eqution y e is clled first-order, liner, nonhomogeneous differentil eqution First-order no derivtive higher thn the first derivtive y, liner no powers in y higher thn, nonhomogeneous q ( ) is not zero To see how to solve nonhomogeneous differentil eqution for y(), let's first consider the simpler eqution when q ( ) 0, which is clled Homogeneous Eqution: Homogeneous Liner Differentil Eqution A differentil eqution tht cn e written in the form y p( ) y 0 is clled first-order homogeneous liner differentil eqution It is understood tht vries over some intervl in the rel line, nd p() is continuous function of in the intervl The eqution is clled liner ecuse y nd y' occur only linerly nd homogeneous ecuse the right side of the eqution is zero - -

25 Differentil Equtions The first order homogeneous liner differentil eqution hs seprle vriles, ecuse it cn e written s dy p( y ) d This forms llows us to seprte ll the y-terms on the left nd the -terms on the right: dy p( d ) y,, which gives: ln y p( d ) c where c is the constnt of integrtion Now solve for y pd ( ) c y e, pd ( ) y ( ) Ce, where c C e if y > 0, nd c C e if y < 0 Emple: ) Find the generl solution of the eqution y y 0 for > 0 ) Find the prticulr solution with the initil vlue y() Solution: ) We first put the eqution into the homogeneous liner form y dividing y : y y 0, dy d y ln y ln cln c The constnt of integrtion c is sored into the constnt C, nd the generl solution is C y ( ) ) The prticulr solution with initil vlue y() is C, C, y ( ), - -

26 Nonhomogeneous Liner Differentil Eqution Differentil Equtions First-order nonhomogeneous liner differentil equtions re those in which, fter isolting the liner terms contining y( ) nd y ( ) on the left side of the eqution, the right side is not identiclly zero In these cses the right hnd side of the eqution is usully represented s one function q, ( ) nd the stndrd form looks like y py ( ) q ( ) Remrk: When the right hnd side is ctully constnt k, it is still vlid to think of it s function; it's merely the constnt function q ( ) kfor ll Method for Solving Liner Differentil Eqution A generl solution of liner eqution cn e found y the method clled Vrition of Constnts: Strt with solution y ( ) of the corresponding homogeneous eqution where y p( ) y 0, pd ( ) y( ) Ce Cu( ), pd ( ) u ( ) e We egin y ssuming tht the prticulr solution hs the form y ( ) Cu ( ) ( ), where C ( ) is n unknown function (we replced constnt C y function C) ( ) We sustitute this into the differentil eqution d y p ( ) y q ( ), d C( u ) ( ) Cu ( ) ( ) pcu ( ) ( ) ( ) q ( ), C( ) u( ) C( ) u( ) p( ) u( ) q( ) Since uis ( ) solution of homogeneous eqution, u( ) p( ) u( ) 0 We otin q ( ) C( ), u ( ) q ( ) C ( ) d K, u ( ) where K is gin the constnt of integrtion - -

27 Thus, the finl epression for y( ) is: q ( ) y( ) Cu ( ) ( ) Ku ( ) u ( ) d u ( ) Differentil Equtions Note tht the generl solution y( ) y ( ) Y( ) is the sum of n ritrry constnt times homogeneous solution, y ( ) Ku( ), tht stisfies y p( ) y 0 nd prticulr solution, q ( ) Y( ) u( ) d, tht stisfies Y p( ) Y q( ) u ( ) Emple: ) Find the generl solution of the eqution y y for > 0 ) Find the prticulr solution with the initil vlue y() Solution: ) We first put the eqution into the homogeneous liner form y dividing y : y y The solution y ( ) of the corresponding homogeneous eqution is (see pge ) y y 0 C y ( ) C ( ) We ssume tht the generl solution hs the form y ( ) unknown function We sustitute this into the differentil eqution C ( ) C( ) C( ), 6 C( ) C( ) C( ), C( ), where C ( ) is n The generl solution is 6 5 C ( ) d K 6 K C ( ) ( ) 6 K y

28 ) The prticulr solution with initil vlue y() is Differentil Equtions Required prticulr solution is K, 6 K 6 y ( ) 6 ( ) Liner Differentil Equtions n th order Mny systems cn e represented in mthemticl form using Liner Differentil Equtions This suject is the min focus of most introductory courses in differentil equtions In ddition, liner differentil equtions with constnt coefficients cn e solved using reltively simple nlyticl pproches, nd the chrcteristic solutions tht re otined re simple elementry functions (sinusoids, eponentils, etc) nd they give significnt insight into the physicl ehviour of the systems under study In the previous chpter we looked t first order differentil equtions In this chpter we will move on to n th order differentil equtions A liner differentil eqution is ny differentil eqution tht cn e wrote in the following form ( n) ( n) y pn( ) y p( ) y p0( ) y f( ), where the pi ( ) coefficients nd right hnd side forcing function, f ( ), re continuous functions For emple, the stndrd form for second order system is y p( ) y q( ) y f( ) Remrk: The importnt thing to note out liner differentil equtions is tht there re no products of the function, y, ( ) nd it s derivtives nd neither the function or it s derivtives occur to ny power other thn the first power Initil Conditions is set of conditions on the solution of the form ( ) 0 0 n 0 0 y( ), y ( ),, y ( ) n The numer of initil conditions tht re required for given differentil eqution will depend upon the order of the differentil eqution s we will see - 5 -

29 Differentil Equtions When f() = 0, the eqution is clled homogeneous, otherwise it is clled nonhomogeneous To nonhomogeneous eqution we ssocite the so clled ssocited homogeneous eqution: ( n) ( n) y pn( y ) p( y ) p0( y ) 0 The generl solution to prolem of this type (ie liner ODE) cn e written s the sum of homogeneous solution nd prticulr solution, y ( ) y ( ) Y ( ), where y ( ) is the generl solution, y ( ) is the homogeneous solution contining n ritrry constnts, nd Y( ) is the prticulr solution contining no ritrry constnts 5 Homogeneous Equtions with Constnt Coefficients When pi( ) i constnt, the eqution is Equtions with Constnt Coefficients A generl constnt coefficient homogeneous liner ODE cn e written s ( n) ( n) y ny y 0y 0 One usully ssumes solution of the form y( ) e (for constnt ) Sustitution of this epression into the originl eqution gives Dividing y n n e n e e 0e 0 e gives the chrcteristic eqution n n n 0 0 nd the roots of the chrcteristic polynomil represent vlues of tht stisfy the ssumed form for y( ) For n n th order system, there will e n roots (not necessrily distinct) to the n th order chrcteristic eqution The generl solution ecomes liner comintion of the individul solutions to the homogeneous eqution with the restriction tht the n solutions must e linerly independent For n distinct roots, the homogeneous solution cn e written s n n ( ) ( ) i y C y Cie i i i i - 6 -

30 Differentil Equtions The n linerly independent solutions form the sis of solutions on the intervl of interest 6 Homogeneous nd Order Equtions with Constnt Coefficients A nd order constnt coefficient homogeneous system cn e written s The corresponding chrcteristic eqution is y y y nd the roots of this qudrtic eqution re given y, 4 0 With known (distinct) roots, the generl solution cn e written s y ( ) Ce Ce The ctul form of the solution is strongly dependent on whether the roots re rel versus comple or distinct versus repeted In fct three specil cses cn e identified sed on whether the term inside the rdicl is positive, negtive, or zero These three cses re identified in detil in the reminder of this susection I Rel Distinct Roots: If D 40 0, the roots, re rel nd distinct Therefore the generl solution is usully written s ove II Repeted Roots: If 0 y( ) C e C e D 4 0, one otins repeted roots For this sitution, the doule root is given y, Therefore, there is only one independent solution, y ( ) e The second independent solution is y ( ) e nd the generl solution is - 7 -

31 Differentil Equtions y ( ) ( C Ce ) III Comple Conjugte Roots: If 0 D 4 0, the roots re comple conjugtes In this cse the roots cn e written s, i with nd D The independent solutions re ( ) y e cos nd ( ) y e sin Therefore the generl solution is y( ) e ( C cos C sin ) Emple: Solve the differentil eqution y y4y 0, y(0) 0, y(0) 7 Solution: The chrcteristic eqution is 4 0, ( 8)( ) 0 Its roots re 8 nd nd so the generl solution is nd its derivtive is 8 y( ) C e C e 8 y( ) 8C e C e Now, plug in the initil conditions to get the following system of equtions: y(0) 0 C C y(0) 7 8C C 7 7 Solving this system gives C nd C The ctul solution to the differentil 5 5 eqution is then y( ) e e 5 5 Emple: Solve the differentil eqution y y y 0 Solution: The chrcteristic eqution is 0, ( ) ( ) 0 Its roots re 0 nd the doule root, nd so the generl solution is y( ) C C e C e - 8 -

32 Differentil Equtions Emple: Solve the differentil eqution 4y 8y5y 0, y(0) 0, y(0) Solution: The chrcteristic eqution is Its roots re, i nd so the generl solution is 8i 6480, 8 y ( ) Ce cos Ce sin e Ccos Csin nd its derivtive is y ( ) e Ccos Csin e Csin Ccos Now, plug in the initil conditions to get the following system of equtions: y(0) 0 C 0C, y(0) C C Solving this system gives C 0 nd C 4 The ctul solution to the differentil eqution is then y ( ) 4e sin 6 Nonhomogeneous Equtions with Constnt Coefficients The n th order, liner nonhomogeneous differentil eqution is ( n) ( n) n 0 y y y y f( ), where f() is non-zero function We need to know set of fundmentl solutions y, y,, yn of the ssocited homogeneous eqution ( n) ( n) n 0 y y y y 0 We know tht, in this cse, the generl solution of the ssocited homogeneous eqution y ( ) C y ( ) C y ( ) C y ( ) is The generl solution to liner nonhomogeneous differentil eqution cn e written s the sum of homogeneous solution nd prticulr solution, y( ) y ( ) Y( ) n n - 9 -

33 Method of Vrition of Prmeters Differentil Equtions A more generl method for finding prticulr solutions is the vrition of prmeter technique The method cn e summrized s follows: Given the n th solution order, liner nonhomogeneous differentil eqution with homogeneous y ( ) C y ( ) C y ( ) C y ( ) n n One cn write the prticulr solution s Y( ) C ( ) y ( ) C ( ) y ( ) C ( ) y ( ), n where C( ), C( ) Cn( ) re unknown functions Second Order, Liner Nonhomogeneous Differentil Eqution We will tke look t the method tht cn e used to find prticulr solution to n eqution y y 0 y f( ) The generl solution to the ssocited homogeneous differentil eqution is y ( ) C y ( ) C y ( ), where n y ( ) nd y ( ) re fundmentl set of solutions Wht we re going to do is see if we cn find pir of functions, C ( ) nd C ( ) so tht Y( ) C ( ) y ( ) C ( ) y ( ) will e prticulr solution to nonhomogeneous differentil eqution We hve two unknowns here nd so we ll need two equtions eventully One eqution is esy Our proposed solution must stisfy the differentil eqution The second eqution cn come from vriety of plces We re going to get our second eqution simply y mking n ssumption tht will mke our work esier So, let s strt If we re going to plug our proposed solution into the differentil eqution We re going to need some derivtives so let s get those The first derivtive is Y( ) Cy CyCy C y Here s the ssumption Simply to mke the first derivtive esier to del with we re going to ssume tht - 0 -

34 Differentil Equtions Cy C y 0 Now, there is no reson hed of time to elieve tht this cn e done However, we will see tht this will work out We simply mke this ssumption on the hope tht it won t cuse prolems down the rod nd to mke the first derivtive esier so don t get ecited out it With this ssumption the first derivtive ecomes Y( ) C y Cy The second derivtive is then Y( ) Cy Cy C y C y Plug the solution nd it s derivtives into the nonhomogeneous differentil eqution Rerrnging little gives Cy C y f( ) The two equtions tht we wnt so solve for the unknown functions re Cy C y0 Cy C y f( ) Solving this system is ctully quite simple C( ) C ( ) y f( ) y y yy yf ( ) yy y y nd Net, let s notice tht W( y, y ) y y yy y y y y is the Wronskin of y nd y Finlly, ll tht we need to do is integrte C ( ) nd C ( ) in order to determine wht C ( ) nd C ( ) re Doing this gives y f( ) C ( ) d nd yy y y yf ( ) C( ) d yy y y - -

35 A prticulr solution to the differentil eqution is Differentil Equtions yf( ) yf( ) Y( ) C( ) y( ) C( ) y( ) y( ) d y( ) d yy y y yy y y Emple: Solve the differentil eqution y y cos Solution: A set of fundmentl solutions of the eqution y y 0 is y cos, y sin We seek prticulr solution of the form Y( ) C ( )cos C ( )sin We sustitute the epression for Y( ) nd its derivtives into the inhomogeneous eqution We hve system of liner equtions for C ( ) nd C ( ) C cos C sin 0 cos sin Here is the Wronskin W( ) C sin C cos sin cos cos We solve this system using Krmer's rule nd get sin C ( ) nd C ( ) cos cos Using techniques of integrtion, we get C ( ) nd cos C ( ) tg The prticulr solution is sin Y( ) cos tgsin cos cos cos The generl solution of the inhomogeneous eqution is sin y ( ) y ( ) Y ( ) CcosCsin cos cos Method of Undetermined Coefficients For some simple differentil equtions, (primrily constnt coefficient equtions), nd some simple inhomogeneities we re le to guess the form of prticulr solution Y( ) This form will contin some unknown prmeters We sustitute this form into the differentil eqution to determine the prmeters nd thus determine prticulr solution Consider n n th order inhomogeneous eqution with constnt coefficients - -

36 Differentil Equtions ( n) ( n) y ny y 0y f( ) We cn guess the form of prticulr solution, if f ( ) is one of few simple forms f ( ) Pn ( ) e cos( ) or f ( ) Pn ( ) e sin( ), where Pn ( ) is polynomil function with degree n Then prticulr solution Y( ) is given y ( ) k ( ) cos( ) ( ) Y Q e R e sin( ) where n n n n Qn( ) A0 A An, nd Rn( ) B0 B Bn, where the constnts A i nd B i hve to e determined The power k is equl to 0 if i is not root of the chrcteristic eqution If i is simple root, then k nd k r if i is root of mutiplicity r Remrk: If the nonhomogeneous term f() stisfies the following N f ( ) f( ) f( ) fn( ) fi( ), i where fi ( ) re of the forms cited ove, then we split the originl eqution into N equtions ( n) ( n) y ny y 0y fi( ), i,,, N Then find prticulr solution Yi ( ) A prticulr solution to the originl eqution is given y N Y( ) Y( ) Y( ) YN( ) Yi( ) i Emple: Find generl solution to the eqution y y 4y e sin 8e Solution: We split the eqution into the following three equtions: () 4 y y y e, () y y 4y sin, () y y 4y 8e The roots of the chrcteristic eqution 4 0 re nd 4 Prticulr solution to Eqution (): Since, nd 0, then i, which is not one of the roots Then k 0 The prticulr solution is given s Y Ae - -

37 If we plug it into the eqution (), we get Differentil Equtions 4Ae 6Ae 4Ae e, which implies A, tht is, Y e Prticulr solution to Eqution (): Since 0, nd, then i i, which is not one of the roots Then k 0 The prticulr solution is given s Y Acos Bsin If we plug it into the eqution (), we get ( Acos Bsin ) ( Asin Bcos ) 4( Acos Bsin ) sin, which implies 5AB0, Esy clcultions give A5B A nd 7 5 B Tht is 7 5 Y cos sin 7 7 Prticulr solution to Eqution (): Since, nd 0, then i, which is one of the roots Then k The prticulr solution is given s Y Ae If we plug it into the eqution (), we get A ( ) e A( ) e 4Ae 8e, 8 8 which implies A, tht is Y e 5 5 A prticulr solution to the originl eqution is 5 8 Y( ) e cos sin e A generl solution to the originl eqution is y( ) Ce Ce e cos sin e

38 LITERATURE Literture [] BUBENÍK, F: Mthemtics for Engineers Vydvtelství ČVUT, Prh997 ISBN [] DEMLOVÁ, M, HAMHALTER, J: Clculus I Vydvtelství ČVUT, Prh 998 ISBN [] HARSHBARGER, RJ-REYNOLDS, JJ: Clculus with Applictions DCHeth nd Compny, Leington990 ISBN [4] JAMES, G: Modern Engineering Mthemtics Addison-Wesley, 99 ISBN [5] JAMES, G: Advnced Modern Engineering Mthemtics Addison-Wesley, 99 ISBN [6] PAVELKA, P-PINKA, P: Integrální počet funkcí jedné proměnné Mtemtik III, VŠB-TU Ostrv 999 ISBN X [7] VLČEK, J-VRBICKÝ, J: Diferenciální rovnice Mtemtik IV, VŠB-TU Ostrv 997 ISBN