Linear Differential Equations of Higher Order Basic Theory: Initial-Value Problems d y d y dy

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1 Liear Differetial Equatios of Higher Order Basic Theory: Iitial-Value Problems d y d y dy Solve: a( ) + a ( )... a ( ) a0( ) y g( ) = d d d ( ) Subject to: y( 0) = y0, y ( 0) = y,..., y ( 0) = y Eistece of a Uique Solutio: Let a (), a - (),,a (), a 0 () ad g() be cotiuous o a iterval I, ad let a () 0 for every i this iterval. If = 0 is ay poit i this iterval, the a solutio y() of the iitial value problem eists o the iterval ad is uique. E: Fid the solutio to y + y = 4 + 0si subject to y( π) = 0, y ( π) = give that y = ccos + c si + 4 5cos is a geeral solutio to the D.E. Is that solutio uique? Boudary Value Problems Aother type of problem cosists of solvig a liear DE of order two or greater i which the depedet variable y or its derivatives are specified at differet poits. A problem such as d y dy Solve: a( ) + a ( ) + a0( ) y = g( ) d d Subject to: y( 0) = y0, y( ) = y is called a two poit boudary value problem or BVP. The prescribed values y(0) = y O ad y() = y are called boudary coditios. A solutio of the foregoig problem is a fuctio satisfyig the D.E. o some iterval I, cotaiig 0 ad, whose graph passes through the two poits (0,y 0 ) ad (,y). E: Solve y + 6y = 0 subject to y(0) = 0, y( π/) = 0, give that y = ccos4 + csi4 is a geeral solutio to the D.E. Differece betwee ad IVP ad BVP: I a IVP all values eeded to solve a particular problem are specified at a sigle poit (0) I a BVP all values eeded to solve a particular problem are specified at differet poits (0,, etc ) Liear Depedece ad Liear Idepedece A set of fuctios f (), f (),,f () is said to be liearly depedet o a iterval I if there eists costats c, c,,c, ot all zero, such that cf ( ) + cf( ) c ( ) f = 0 for every i the iterval. If the set of fuctios is ot liearly depedet o the iterval, it is said to be liearly idepedet.

2 I other words a set of fuctios is liearly idepedet o a iterval I if ad oly if the oly costats for which cf ( ) + cf( ) c ( ) f = 0 for every i the iterval are c = c = = c = 0 E: Are the fuctios f( ) = si ad f ( ) = si cos liearly depedet or idepedet? E: Are the fuctios g( ) = ad g ( ) l = liearly depedet or idepedet? E: Are the fuctios h( ) = + 5, h( ) = + 5, h( ) = ad h( ) = liearly 3 4 depedet or idepedet? Also ote that a set of fuctios f, f, f 3,, f is liearly depedet o a iterval if at least oe fuctio ca be epressed as a liear combiatio of the remaiig fuctios. Solutios of a D.E. We are primarily iterested i liearly idepedet solutios of liear D.E. To determie whether a set of solutios of a th order liear D.E. is liearly idepedet ca be doe usig determiats. Wroskia: Suppose each of the fuctios f (), f (),,f () possess at least derivatives. The determiat of f f f f f f f f f f f f ( ) ( ) ( ) is called the Wroskia of the fuctios. = W( f, f,..., f ) Criterio for Liearly Idepedet Fuctios: The set of fuctios f (), f (),,f () is liearly idepedet o I if ad oly if W(f,f,,f ) 0 for at least oe poit i the iterval. The coverse is also true E: Are the followig eamples liearly idepedet? A. f( ) = si, f ( ) = cos B. ( ) m m g = e, g( ) = e, m m α C. α = = D. h( ) e cos β, h( ) e si β, αad β arerealumbers z( ) = e, z ( ) = e, z( ) = e 3

3 Solutios of liear Differetial Equatios A liear th order DE of the form d y d y dy a( ) + a ( )... a ( ) a0( ) y = d d d is said to be homogeeous, whereas a equatio d y d y dy a( ) + a ( )... a ( ) a0( ) y g( ) = d d d with g() ot idetically zero, is said to be ohomogeeous (The word homogeeous here does ot refer to the coefficiets that are homogeeous fuctios) E: y + 3y + 5y = is a ohomogeeous secod order liear differetial equatio ad y + 3y + 5y = 0is the associated homogeeous equatio. For ow o we will make the followig assumptios whe statig defiitios ad theorems about liear equatios o some iterval I,. The coefficiet fuctios a i (), i = 0,,,, are cotiuous. g() is cotiuous 3. a () 0 for every i the iterval. Superpositio Priciple - Homogeeous Equatios: Let y, y,...,y k be liearly idepedet solutios of the homogeeous th order differetial equatio o a iterval I, the the liear combiatio y = c y () + c y () c k y k (), where the c i, i = 0,,,k are arbitrary costats, is also solutio o the iterval, we call this the geeral solutio of the homogeeous D.E. Ay set y,y,..., y of liearly idepedet solutios of the homogeeous liear th order differetial equatio o a iterval I is said to be a fudametal set of solutios o the iterval. Corollaries to the Superpositio Theorem: A costat multiple y = c y () of a solutio y () of a homogeeous liear DE is also a solutio. A homogeeous liear DE always possesses the trivial solutio y = 0. E: The fuctios y = ad y = l are both solutios of the homogeeous liear 3 equatio y y + 4y = 0 o the iterval (0, ). By the superpositio priciple the liear combiatio y = c + c is also a solutio of the equatio o the iterval. l

4 Nohomogeeous Liear Differetial Equatios Ay fuctio y p, free of arbitrary parameters, that satisfies a ohomogeeous liear D.E. is said to be a particular solutio or particular itegral of the equatio. A easy eample would be y p= 3 is a particular solutio to y + 9y = 7. The particle solutio is t ecessary restricted to costats. Let y p be a give (or particular) solutio of the ohomogeeous liear th order differetial equatio o the iterval I, ad let yc = c y () + c y () c k y k () deote the geeral solutio of the associated homogeeous equatio o the iterval, the the geeral solutio of the ohomogeeous equatio o the iterval is defied to be y = y c + y p = c y ( ) + c y ( ) c y ( ) + y p ( ) E: The ohomogeeous liear differetial equatio y 6y + y 6y = 3 has a particular solutio yp = ad its associated homogeeous equatios has a geeral 3 solutioy = ce + ce + ce. Fid the geeral solutio to the ohomogeeous D.E. c 3 Reductio of Order If you have a kow solutio to a secod order liear differetial equatio oe iterestig thig that occurs with these types of equatios is that you ca use that solutio to costruct a secod solutio. Suppose y ( ) is a kow solutio to a( ) y + a( ) y + a0( ) y = 0. We assume just like previously that a() 0 for every i some iterval I. The process we use to a fid a secod solutio y ( ) cosists of reducig the order of the equatio to a first order liear D.E., which we already kow how to solve, through substitutio. Suppose that y() is a otrivial solutio of the previous D.E. ad that y() is defied o I. We seek a secod solutio, y(), so that y(), y(), are a liearly idepedet o I. If y() ad y() are liearly idepedet the the quotiet y()/ y(), is o-costat that is y()/ y()= u() or y()= u() y(). The fuctio u() ca be foud by substitutig y () = u()y () ito the give differetial equatio. The Geeral Case Suppose we divide a( ) y + a( ) y + a0( ) y = 0 by a () i order to put the equatio i the stadard form y + P( ) y + Q( ) y = 0 where P() ad Q() are cotiuous o some iterval I. Let us suppose further that y () is a kow solutio of the stadard form o I ad that y () 0 for every i the iterval. If we defie y()= u()y () it follows that y = uy + yu ad y = uy + yu + yu Substitutig ito the stadard form gives

5 This implies that we must have ( ) ( ) y + Py + Qy = u y + Py + Qy + yu + y + Py u = 0. = zero ( ) yu + y + Py u = 0 Through the substitutio w = u we ca tur the previous equatio ito the followig homogeeous liear D.E. yw + y + Py w = 0 Notice this equatio is also separable. If we separate we get Now itegrate ( ) dw y + d + Pd = w y 0 lw + ly = Pd + c lwy wy = ce ce w = u = y If we itegrate agai we ca fid u() = Pd + c Pd Pd Pd ce d y Lastly substitutig ito the origial form of y which was y () = u()y () gives P( ) d ce ( ) ( ) y = y ( y( ) ) E: Give that y() = e is a solutio of y y = 0 o the iterval (-, ). Use the reductio of order to fid the secod solutio. E: Give that y() = 3 is a solutio of y 6y = 0, use the reductio of order to fid a secod solutio o the iterval (-, ). E: Give that y() = is a solutio of y 3y + 4y = 0, use the reductio of order to fid a secod solutio o the iterval (0, ). si E: Give that y ( ) = is a solutio of y + y + ( ) y = 0, use the reductio of 4 order to fid a secod solutio o the iterval (0, ). d

6 Homogeeous Liear Equatios with Costat Coefficiets For a liear st order D.E. y + ay = 0we ca see that y = ce a is a solutio. The we ca also seek to determie whether epoetial solutios eist for higher order equatios that have costat coefficiets. Auiliary Equatio Cosider the special case of secod order equatio ay + by + cy = 0 where a, b, ad c are costats. If we try to fid a solutio of the form y = e m the after substitutig m m y = me ad y = me the origial equatio becomes m m m m am e + bme + ce = 0 or e ( am + bm + c) = 0 sice e m 0 for all, it is apparet the oly way y = e m ca satisfy the D.E. is if m is chose as a root of the quadratic equatio am + bm + c = 0. This is called the auiliary equatio of the differetial equatio. Give that there are always two roots m ad m to the quadratic there will be three correspodig cases. m ad m are real ad distict (discrimiat > 0). m ad m are real ad equal (discrimiat = 0) 3. m ad m are comple cojugate umbers (discrimiat < 0) Case : Distict Real Roots Uder the assumptio that the auiliary equatio has two uequal real roots m ad m, we fid two solutios, y = e m ad y = e m. We have see that these fuctios are liearly idepedet ad hece form a fudametal set. Therefore the geeral solutio is m m y = ce + ce Case : Repeated Real Roots Whe m = m, we obtai oly oe epoetial solutio, y = e m from the quadratic, that is m = - b/a it follows from the reductio of order formula that the secod solutio is Therefor the geeral solutio is e e y e d e d e d e ( ba / ) m m m m m ( ) = = = = m m ( e ) e. y = ce + ce. m m Case 3: Cojugate Comple Roots If m ad m are comple, the we ca write m = α + iβ ad m = α iβwhere α ad β are real ad i =. Formally there is o differece betwee this ad case ecept that we are dealig with comple umbers. So the geeral solutio is ( α + βi) ( α βi) y = Ce + Ce Usually i practice we prefer to work with real fuctios so we use Euler s formula: iθ e = cosθ + isiθ where ϴ is a real umber. It follows from this formula that

7 iβ iβ e = cos β + isi β ad e = cos β isi β From this we ca see that iβ iβ e + e = cos β ad i β iβ e e = isi β Lookig back at the origial geeral solutio, if we let C=C= ad C= ad C=- we obtai the two solutios ( i ) ( i ) y e α + β e α = + β ( i ) ( i ) ad y e α + β e α = β. Usig Euler s formula these become α iβ iβ α α iβ iβ α y = e ( e + e ) = e cos β ad y = e ( e e ) = ie si β Therefore the geeral solutio y = cy + cy ca be writte as y = ce α cos β + ce α si β = e α ccos β + c si β E: Solve the followig DE's a. 5 y 4 y y = 0 b. 4y + 4 y + y = 0 c. y + y + y = 0 d. y + k y = 0 where k is a real costat ( ) Higher Order Equatios I geeral to solve the th order D.E. where the coefficiets are real costats, we must solve the th degree polyomial auiliary equatio am + a m am + am + a = 0 0 Higher order polyomials have the three types of roots as quadratics: distict real, repeated real, ad/or comple cojugates. There are just more combiatios of how these roots ca come up. If all roots are distict reals such that m m... m the the geeral solutios is m m m y = ce + ce +... ce If there are repeated real roots, say m has multiplicity k, the the geeral solutio is m m m k m y = ce + ce + ce + + c e 3... Comple cojugates would also occur the same as before. If a comple cojugate pair is repeated the you would for the previous repeated roots eample ad multiply each pair by a ascedig power of util you ehausted the multiplicity. Due to the umber of roots of a polyomial we ca have ay combiatios of the previous. For eample a fifth degree equatio ca have 3 real distict ad comple, distict real a repeated real ad a comple cojugate pair, etc. E: Solve y + 3y + y + 6y = 0 E: Solve 3 y + 5 y + 0 y 4 y = 0 E: Solve d 4 y d y y = 0 4 d d k

8 Differetial Operators I calculus, differetiatio ca be deoted by the capital letter D that is, dy/d = Dy. The symbol D is called the differetial operator because it trasforms a differetiable fuctio ito aother fuctio. E: D(cos4) = -4si4. Higher order derivatives ca be epressed i terms of D as well: d dy d y = = D ( Dy) = D y d d d Where y represets a sufficietly differetiable fuctio. Polyomial epressios ivolvig D, such as D + 3, D + 3D - 4 are also differetial operators. For eample ( D + 3)(5 + ) = ( D + 3)(5 ) + ( D + 3)( ) = D(5 ) D( ) + 3 = I geeral, we defie a th order differetial operator or polyomial operator to be L = a ( ) D + a ( ) D a( ) D + a( ) 0 As a cosequece of differetiatio two basic properties eist for L:. L(cf())= c(l(f()), c is a costat. L{f() + g()} = L(f()) + L(g()) the differetial operator L possess a liearity property; that is, L operatig o a liear combiatio of two differetiable fuctios is the same as the liear combiatio L operatig o the idividual fuctios. We say L is a liear operator. Ay liear DE ca be epressed i terms of D. For eample y + 5 y + 6 y = 5 3ca be epressed as D y + 5Dy + 6 y = ( D + 5 D + 6 ) y = 5 3. We ca write a liear th order differetial equatios as L(y) = 0 or L(y) = g() The liear differetial polyomial operators ca also be factored uder the same rules as polyomial fuctios. If r is a root of L the (D r) is a factor or L. The previous eample could also be writte as ( D + 5 D + 6 ) y = ( D + )( D + 3) y = 5 3. Aihilator Operator If L is a liear differetial operator with costat coefficiets ad y = f () is a sufficietly differetiable fuctio such that L(y) = 0 the L is said to be a aihilator of the fuctio. For eample the costat fuctio y = k is aihilated by D sice Dk = 0. The fuctio y = is aihilated by the differetial operator D sice D(D())=D()=0. The differetial operators D aihilates power fuctios up to y = -. As a immediate cosequece of this ad the fact that differetiatio ca be doe term by term, a polyomial

9 c + c + c c o ca be aihilated by D. We ofte wat to fid the differetial operator of lowest order that will aihilate a fuctio. While D 5 aihilates, D 3 is the smallest oe. 3 4 E: Fid the differetial operator that aihilates The differetial operator ( D α) aihilates each of the fuctios α α α α e, e, e,..., e This is due to the fact from the previous lesso where if α is a root of the auiliary equatio ( m α) is a factor ad the geeral solutio to the homogeous D.E. is α α α α y = ce + ce + c e + + c e 3... E: Fid the differetial operator that aihilates the give fuctio 3 a. e b. 4 e 0e The differetial operator [ D αd + ( α + β )] aihilates each of the fuctios α α α α e cos β, e cos β, e cos β,..., e cos β α α si, α e e si, e si,..., α β β β e si β This ca be see by lookig at the equatio [ m αm + ( α + β )] = 0 whe α ad π are real umbers. It has comple roots α ± iβ both of multiplicity foud by the quadratic formula. E: Fid the differetial operator that aihilates 5e cos 9e si Whe α = 0 ad = a special case is cos β + = 0 si β ( D β ) If we wat to aihilate the sum of two or more fuctios the differetial operators L ad L of y ad y respectively their product L L will aihilate c y + c y E: Fid the differetial operator that aihilates 7 + 6si 4 E: Fid the differetial operator that aihilates 3 a. 5 + e b. e + e 4 5 c. + si3 e

10 Udetermied Coefficiets: Aihilator Approach Suppose that L(y) = g() is a liear DE with costat coefficiets ad that g() cosists of fiite sums ad products of the fuctios that we have aihilators. I other words g() is a liear combiatio of fuctios of the form k (costat), m, m e α, m e α cosβ, ad m e α siβ where m is a oegative iteger ad α ad β are real umbers. We ow kow that such a fuctio g() ca be aihilated by a differetial operator L of lowest order cosistig of a product of the operatorsd,( D α), ad [ D αd + ( α + β )]. Applyig L to both sides of the equatio L(y) = g() yields L L(y) = L g() = 0. By solvig the homogeeous higher order equatio L L(y) = 0, we ca discover the form of a particular solutio yp for the origial o-homogeeous equatio L(y) = g(). We the substitute this assumed form ito L(y) = g() to fid the eplicit particular solutio yp. This procedure for determiig yp, is called the method of udetermied coefficiets. I previous sectios it was stated that the geeral solutio of a o-homogeeous liear DE L(y) = g() is y = yc + yp, where yc is the geeral solutio of the associated homogeeous equatio L(y) = 0 ad yp is the particular solutio of the o-homogeeous equatio. Sice we ow kow how to fid both of these whe the coeeficiets are costats we ca fid the geeral solutio to a o-homogeeous liear D.E. Steps to Solve Udetermied Coeffieiets: Aihilator Approach If the D.E. L(y) = g() has costat coefficiets, ad the fuctio g() has a differetial aihilator the: i. Fid the complimetary (geeral) solutio yc for the associated homogeeous equatio L(y) = 0. ii. Apply the differetial operator L that aihilates the fuctio g() o both sides of the homogeeous equatio L(y) = g(). iii. Fid the geeral solutio of the associated higher-order homogeeous D.E. L L(y) = 0 iv. Delete from the solutio i step(iii) all those terms that are duplicated i the complimetary solutio y c foud i step (i). Form a liear combiatio y p of the terms that remai. This is the form of a particular solutio of L(y) = g(). v. Substitute y p foud i the step (iv) ito L(y) = g(). Match coefficiets of the various fuctios o each side of the equality, ad solve the resultig system of equatios for the ukow coefficiets i yp. vi. With the particular solutio foud i step (v), form the geeral solutio y = y c + y p of the give D.E. E: Solve y + 3 y + y = 4 3 E: Solve y 3 y = 8e + 4si E: Solve y + 8y = 5 + e

11 E: Solve y + y = cos cos E: Determie the form of a particular solutio for E: Determie the form of a particular solutio for y y + y = e 0 cos y 4 y + 4 y = e + 3e 5 The method of udetermied coefficiets is ot applicable to liear D.E. with variable coefficiets or is it applicable to liear equatios with costat coefficiets whe g() is a fuctio that does ot have a aihilator such as g() = l, g() = /, g() = ta, g() = si - Variatio of Parameters: It ca be see that we ca fid a particular solutio of a liear first-order D.E. of the form y p = u ()y () o a iterval. Where y () is a geeral solutio to the associated homogeeous D.E. To adapt this method of Variatio of Parameters to a liear secodorder D.E. we begi by puttig the equatio i stadard form. y + P( ) y + Q( ) y = f( ) For the liear secod-order differetial equatio we seek a particular solutio y p = u ()y () +u ()y () where y ad y form a fudametal set of solutios o I of the associated homogeeous D.E. therefore y + P( ) y + Q( ) y = 0 ad y + P( ) y + Q( ) y = 0 We also impose the assumptio yu + yu = 0 i order to simplify the first derivative ad thereby the secod derivate of y p. We differetiate y p twice givig us y p = uy + yu + uy + yu = uy + uy ad y p = uy + yu + uy + u y the substitute these ito the origial D.E. ad group terms u y + Py + Qy + u y + Py + Qy + yu + yu = f which gives you [ ] [ ] ( ) ( ) yu + yu = f Now sice we are goig to seek two ukows, u ad u, we eed two equatios. These are yu yu yu + yu = f + = ad ( ) 0 From liear algebra Cramer's rule is a way of obtaiig the solutio of the system i terms of determiats. y y Let W =, W 0 y y y = f ( ) y, ad y 0 W = the y f ( ) W yf( ) W yf( ) u = = ad u = = W W W W Now to fid u ad u we itegrate. To repeat this process for each problem will be too time cosumig so it s best to just kow the formulas for u ad u. Oce you have u ad u, create your particular solutio y = u y + u y p ad the form your geeral solutio

12 y = y c + y p E: Solve y 4 y + 4y = ( + ) e E: Solve 4 y + 36 y = csc 3 E: Solve y y = Higher-Order Equatios The same process ca be used for liear th order o-homogeeous D.E. equatios put i stadard form ( ) ( ) y + P ( ) y +... P( ) y + P( ) y = f( ) 0 If y = cy + cy c y is the complimetary solutio to the associated homogeeous c D.E., the a particular solutio isyp = u( ) y( ) + u( ) y( ) u( ) y( ) where the u, k =,,..., 3 k are determied by the equatios yu + yu yu = 0 yu + yu yu = 0... ( ) ( ) ( ) y u + y u y u = f( ) Cramer's Rule gives Wk u k =, k =,,..., W where W is the Wroskia of y, y,..., y ad W k is the determiat obtaied by replacig the kth colum of the Wrokskia by (0,0,0,...,f()) Cauchy-Euler Equatio: A liear D.E. of the form d y d y dy 0 a + a a + a y = g( ) d d d where the coefficiets a, a-,, a0 are costats, is kow as the Cauchy-Euler equatio. The disguisig characteristic is that the degree of matches the order of the derivative. We first look at the geeral solutio to the secod order homogeeous equatio a y + by + cy = 0 the the solutio to higher order equatios will follow. We try a solutio of the form y = m, where m is to be determied. Similar to what happeed whe we substituted y = e m, whe we substitute y = m, each term of a Cauchy-Euler equatio becomes a polyomial i m times m.

13 If m y =, m =, ad y = mm ( ) m the m m m y m a y + by + cy = amm ( ) + bm + c = am( m ) + bm + c m m m = ( am( m ) + bm + c) = + + = m ( am ( b a) m c) 0 Lettig m = 0 achieves othig, so the roots of am + ( b a) m + c = 0, which will be the auillary equatio will give us the solutios of the D.E. There are three differet cases to cosider Case : Distict Real Roots m Let m ad m deote the real roots such that m m. The y = ad y = for a m m fudametal set of solutios. Therefore the geeral solutio to the D.E. is m m y = c + c Case : Repeated Real Roots If the roots are repeated, m = m, we obtai oe solutio y =. With the m ( b a) discrimiat beig zero the root is m =. We ca use the reductio of order a formula to costruct a secod solutio by first puttig the secod order Cauchy-Euler equatio i stadard form b b c b b y + y a + y = 0 where P( ) = ad P( ) d d l a a a = = a b l a b b b a m e m a m m a a m d m So = = = = = m y d d d l For higher order equatios if m has multiplicity k the it ca be show that m m m m 3 m k, l, (l ), (l ),... (l ) are liearly idepedet solutios. Case 3: Cojugate Comple Roots If the roots are cojugate pairs m = α + iβ ad m = α iβ the the solutio is α + iβ α iβ y = C + C But we wat to write the solutio i terms of real fuctios oly, so we use Euler's formula ad get α y = ( ccos βl + c si βl ) ( ) ( ) E: Solve E: Solve E: Solve 4 = 0 y y y 4 y + 8y + y = 0 + =, y () =, y () = 4 y 7y 0

14 E: Solve E: Solve = 0 3 y y y y y 3y + 3y = e 4