Higher Order Linear Differential Equations

1 Higher OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsHigher Order Linear Differential EquationsMath 240 Calculus IIIS ummer 2015, Session IITuesday, July 28, 2015 Higher OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsAgenda1. Linear Differential Equations of ordernLinear Differential operatorsFamiliar stuffAn example2. Homogeneous constant-coefficient Linear differentialequationsHigher OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsIntroduc tionWe now turn our attention to solvinglinear differentialequations of general form of such an equation isa0(x)y(n)+a1(x)y(n 1)+ +an 1(x)y +an(x)y=F(x),wherea0, a1, .. , an,andFare functions defined on general strategy is to reformulate the above equation asLy=F,whereLis an appropriate Linear fact,Lwillbe alinear Differential OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsLinear Differential operatorsRecall that the mappingD:Ck(I) Ck 1(I)defined byD(f) =f is a Linear transformation.

2 ThisDis called thederivative Order derivative operatorsDk:Ck(I) C0(I)are defined by composition:Dk=D Dk 1,so thatDk(f) = Differential operator of ordernis a linearcombination of derivative operators of Order up ton,L=Dn+a1Dn 1+ +an 1D+an,defined byLy=y(n)+a1y(n 1)+ +an 1y +any,where theaiare continous functions then a lineartransformationL:Cn(I) C0(I). (Why?) Higher OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsExamples ExampleIfL=D2+ 4xD 3x, thenLy=y + 4xy haveL(sinx)= sinx+ 4xcosx 3xsinx,L(x2)= 2 + 8x2 e3xD, (2x 3e2x)= 12e2x 2e3x+ (3 sin2x)= 3e3xsin 2x 6 cos 2xHigher OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsHomogene ous and nonhomogeneous equationsConsider the generaln-th Order Linear Differential equationa0(x)y(n)+a1(x)y(n 1)+ +an 1(x)y +an(x)y=F(x),wherea06= 0anda0, a1.

3 , an,andFare functions on (x)is nonzero onI, then we may divide by it and relabel,obtainingy(n)+a1(x)y(n 1)+ +an 1(x)y +an(x)y=F(x),which we rewrite asLy=F(x),whereL=Dn+a1Dn 1+ +an 1D+ (x)is identically zero onI, then the equation ishomogeneous, otherwise it OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsThe general solutionIf we have a homogeneous Linear Differential equationLy= 0,its solution set will coincide withKer(L).In particular, thekernel of a Linear transformation is a subspace of its set of solutions to a Linear Differential equation of ordernis a subspace ofCn(I).It is called thesolution of the solutions space a vector space, the solution space has a basis{y1(x), y2(x), .. , yn(x)}consisting of the vector space can be written as a linearcombination of basis vectorsy(x) =c1y1(x) +c2y2(x) + +cnyn(x).This expression is called thegeneral OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsThe WronskianWe can use the WronskianW[y1, y2.]

4 , yn](x) = y1(x)y2(x) yn(x)y 1(x)y 2(x) y n(x)..y(n 1)1(x)y(n 1)2(x) y(n 1)n(x) to determine whether a set of solutions is linearly , y2, .. , ynbe solutions to then-th Order differentialequationLy= 0whose coefficients are continuous onI. IfW[y1, y2, .. , yn](x) = 0at any single pointx I, then{y1, y2, .. , yn}is linearly summarize, the vanishing or nonvanishing of the Wronskianon an intervalcompletely characterizesthe Linear dependenceor independence of a set of solutions toLy= OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsThe WronskianExampleVerify thaty1(x) = cos 2xandy2(x) = 3 6 sin2xaresolutions to the Differential equationy + 4y= 0on( , ).Determine whether they are linearly independent on [y1, y2](x)= cos 2x3 6 sin2x 2 sin 2x 12 sinxcosx = 6 sin 2xcos 2x+ 6 sin 2xcos 2x= 0 They are linearly dependent. In fact,3y1 y2= OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsNonhomog eneous equationsConsider the nonhomogeneous Linear Differential equationLy= homogeneous equationisLy= {y1, y2.}

5 , yn}arenlinearly independent solutions tothen-th Order equationLy= 0on an intervalI, andy=ypisany particular solution toLy=FonI. Then every solution toLy=FonIis of the formy= c1y1+c2y2+ +cnyn+yp,=yc+ypfor appropriate constantsc1, c2, .. , expression is thegeneral solutiontoLy= of the general solution areIthecomplementary function,yc, which is the generalsolution to the associated homogeneous equation,Itheparticular solution, OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsSomethin g slightly newTheoremIfy=upandy=vpare particular solutions toLy=f(x)andLy=g(x), respectively, theny=up+vpis a solution toLy=f(x) +g(x). haveL(up+vp) =L(up) +L(vp)=f(x) +g(x). OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsAn exampleExampleDetermine all solutions to the Differential equationy +y 6y= 0of the formy(x) =erx, whereris a (x) =erxinto the equation yieldserx(r2+r 6) =r2erx+rerx 6erx= 0, we just need(r+ 3)(r 2) = , the twosolutions of this form arey1(x) =e2xandy2(x) =e this be a basis for the solution space?

6 Check ! The general solution isy(x) =c1e2x+c2e OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsAn exampleExampleDetermine the general solution to the Differential equationy +y 6y= know the complementary function,yc(x) =c1e2x+c2e the particular solution, we might guess something of theformyp(x) =ce5x. What shouldcbe? We want8e5x=y p+y p 6yp= (25c+ 5c 6c) then solve8 = 24cto findc= general solution isy(x) =c1e2x+c2e 3x+ OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsIntroduc tionWe just found solutions to the Linear Differential equationy +y 6y= 0of the formy(x) =erx. In fact, we found all technique will often work. Ify(x) =erxtheny (x) =rerx,y (x) =r2erx,.. , y(n)(x) = ifrn+a1rn 1+ +an 1r+an= 0theny(x) =erxis asolution to the Linear Differential equationy(n)+a1y(n 1)+ +an 1y +any= s develop this approach more OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsThe auxiliary polynomialConsider the homogeneous Linear Differential equationy(n)+a1y(n 1)+ +an 1y +any= 0withconstant as a Linear differentialoperator, the equation isP(D)y= 0, whereP(D) =Dn+a1Dn 1+ +an 1D+ Linear Differential operator with constant coefficients, such asP(D), is called apolynomial Differential (r) =rn+a1rn 1+ +an 1r+anis called theauxiliary polynomial, and the equationP(r) = 0theauxiliary OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsThe auxiliary polynomialExampleThe equationy +y 6y= 0has auxiliary polynomialP(r)

7 =r2+r the auxiliary polynomials for the following + 2y 3y= 02.(D2 7D+ 24)y= 2y 4y + 8y= 0r2+ 2r 3r2 7r+ 24r3 2r2 4r+ 8 The roots of the auxiliary polynomial will determine thesolutions to the Differential OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsPolynomi al Differential operators commuteThe key fact that will allow us to solve constant-coefficientlinear Differential Equations is that polynomial differentialoperators (D)andQ(D)are polynomial Differential operators, thenP(D)Q(D) =Q(D)P(D). our purposes, it will suffice to consider the case wherePandQare polynomial Differential operators will allow us toturn a root of the auxiliary polynomial into a solution to thecorresponding Differential OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsLinear polynomial Differential operatorsIn our example,y +y 6y= 0,with auxiliary polynomialP(r) =r2+r 6,the roots ofP(r)arer= 2andr= equivalentstatement is thatr 2andr+ 3are Linear factors ofP(r).

8 The functionsy1(x) =e2xandy2(x) =e 3xare solutions toy 1 2y1= 0andy 2+ 3y2= 0, general solution to the Linear Differential equationy ay= 0isy(x) = OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsTheoremS upposeP(D)andQ(D)are polynomial Differential operatorsP(D)y1= 0 =Q(D) (D)Q(D), thenLy1= 0 = (D)Q(D)y2=P(D)(Q(D)y2)=P(D)0 = 0P(D)Q(D)y1=Q(D)P(D)y1=Q(D)(P(D)y1)=Q(D) 0 = theorem implies that, since(D 2)y1= 0and(D+ 3)y2= 0,the functionsy1(x) =e2xandy2(x) =e 3xare solutions toy +y 6y= (D2+D 6)y= (D 2)(D+ 3)y= OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsLinear polynomial Differential operatorsFurthermore, solutions produced from different roots of theauxiliary polynomial are (x) =e2xandy2(x) =e 3x, thenW[y1, y2](x)= e2xe 3x2e2x 3e 3x =e x 1 12 3 = 5e x6= OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsDistinct Linear factorsIf we can factor the auxiliary polynomial into distinct linearfactors, then the solutions from each Linear factor will combineto form a fundamental set of the general solution toy y 2y= auxiliary polynomial isP(r) =r2 r 2= (r 2)(r+ 1).

9 Its roots arer1= 2andr2= functionsy1(x) =e2xandy2(x) =e xsatisfy(D 2)y1= 0 = (D+ 1) ,y1andy2are solutions to the original we have 2 solutions to a2nddegree equation, theyconstitute a fundamental set of solutions;the general solution isy(x) =c1e2x+c2e OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsMultiple rootsWhat can go wrong with this process?The auxiliarypolynomial could have a multiple this case, we wouldget one solution from that root, but not enough to form thegeneral , there are Differential equation(D r)my= 0has the followingmlinearly independent solutions:erx,xerx,x2erx,.. , xm OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsMultiple rootsExampleDetermine the general solution toy + 4y + 4y= The auxiliary polynomial isr2+ 4r+ It has the multiple rootr= Therefore, two linearly independent solutions arey1(x) =e 2xandy2(x) =xe The general solution isy(x) =e 2x(c1+c2x).

10 Higher OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsComplex rootsWhat happens if the auxiliary polynomial has complex roots?Can we recover real solutions?Yes!TheoremIfP(D)y= 0is a Linear Differential equation with real constantcoefficients and(D r)mis a factor ofP(D)withr=a+biandb6= 0, (D)must also have the factor(D r)m, factor contributes the complex solutionse(a bi)x, xe(a bi)x, .. , xm 1e(a bi)x, real and imaginary parts of the complex solutions arelinearly independent real solutionsxkeaxcosbxandxkeaxsinbxfork= 0,1, .. , m OrderLinearDifferentialEquationsMath 240 Linear DELineardifferentialoperatorsFamiliar stuffExampleHomogeneousequationsComplex rootsExampleDetermine the general solution toy + 6y + 25y= The auxiliary polynomial isr2+ 6r+ Its has rootsr= 3 Two independent real-valued solutions arey1(x) =e 3xcos 4xandy2(x) =e 3xsin The general solution isy(x) =e 3x(c1cos 4x+c2sin 4x).