Chapter Second Order Differential Equations

1 Chapter2 Second Order Differential Equations Either mathematics is too big for the human mind or the human mind is more thana machine. - Kurt G del (1906-1978) the last section we sawhow Second Order Differential equationsnaturally appear in the derivations for simple oscillating systems. In thissection we will look at more general Second Order linear Differential Order Differential Equations are typically harder than first most cases students are only exposed to Second Order linear differentialequations. A general form for asecond Order linear Differential equationis givenbya(x)y (x) +b(x)y (x) +c(x)y(x) =f(x).( )One can rewrite this equation using operator terminology. Namely, onefirst defines the Differential operatorL=a(x)D2+b(x)D+c(x), whereD=ddx. Then, Equation ( ) becomesLy=f.( )The solutions of linear Differential Equations are found by making use ofthe linearity ofL. Namely, we consider thevector space1consisting of real-1We assume that the reader has been in-troduced to concepts in linear in the text we will recall the def-inition of a vector space and see that lin-ear algebra is in the background of thestudy of many concepts in the solutionof Differential functions over some domain.

2 Letfandgbe vectors in this alinear operatorif for two vectorsfandgand scalara, we (f+g) =L f+ (a f) =aL typically solves ( ) by finding the general solution of the homoge-neous problem,Lyh=0and a particular solution of the nonhomogeneous problem,Lyp= Differential equationsThen, the general solution of ( ) is simply given asy=yh+yp. This istrue because of the linearity ofL. Namely,Ly=L(yh+yp)=Lyh+Lyp=0+f=f.( )There are methods for finding a particular solution of a nonhomogeneousdifferential equation. These methods range from pure guessing, the Methodof Undetermined Coefficients, the Method of Variation of Parameters, orGreen s functions. We will review these methods later in the solutions to the homogeneous problem,Lyh=0, is not al-ways easy. However, many now famous mathematicians and physicists havestudied a variety of Second Order linear Equations and they have saved usthe trouble of finding solutions to the Differential Equations that often ap-pear in applications.

3 We will encounter many of these in the followingchapters. We will first begin with some simple homogeneous linear differ-ential is also useful in producing the general solution of a homoge-neous linear Differential equation. Ify1(x)andy2(x)are solutions of thehomogeneous equation, then thelinear combination y(x) =c1y1(x) +c2y2(x)is also a solution of the homogeneous equation. This is easily andLy12=0. We considery=c1y1+c2y2. Then, sinceLisa linear operator,Ly=L(c1y1+c2y2)=c1Ly1+c2Ly2=0.( )Therefore,yis a fact, ify1(x)andy2(x)arelinearly independent,theny=c1y1+c2y2is the general solution of the homogeneous problem. A set of functions{yi(x)}ni=1is a linearly independent set if and only ifc1y1(x) +..+cnyn(x) =0impliesci=0, fori=1, .. ,n. Otherwise, they are said to be linearlydependent. Note that forn=2, the general form isc1y1(x) +c2y2(x) = linearly dependent, then the coefficients are not zero andy2(x) = c1c2y1(x)and is a multiple ofy1(x).

4 We see this in the next thaty1(x) =xandy2(x) =4xare linearly setc1y1(x) +c2y2(x) =0 and show that there are nonzero con-stants,c1andc2satisfying this equation. Namely, letc1x+c2(4x) = , forc1= 4c2, this is true for any nonzeroc2. Letc2=1 and wehavec1= Order Differential Equations 33 Next we consider two functions that are not constant multiples of thaty1(x) =xandy2(x) =x2are linearly setc1y1(x) +c2y2(x) =0 and show that it can only be true ifc1=0 andc2=0. Letc1x+c2x2=0,for allx. Differentiating, we have two sets of Equations that must betrue for allx:c1x+c2x2=0,c1+2c2x=0.( )Settingx=0, we getc1=0. Settingx=1, thenc1+c2=0. Thus,c2= approach would be to solve for the constants. Multiplyingthe Second equation byxand subtracting yieldsc2=0. Substitutingthis result into the Second equation, we findc1= Second Order Differential Equations we seek two linearly indepen-dent functions,y1(x)andy2(x). As in the last example, we setc1y1(x) +c2y2(x) =0 and show that it can only be true ifc1=0 andc2=0.

5 Differen-tiating, we havec1y1(x) +c2y2(x) =0,c1y 1(x) +c2y 2(x) =0.( )These must hold for allxin the domain of the we solve for the constants. Multiplying the first equation byy 1(x)and the Second equation byy2(x), we havec1y1(x)y 2(x) +c2y2(x)y 2(x) =0,c1y 1(x)y2(x) +c2y 2(x)y2(x) =0.( )Subtracting gives[y1(x)y 2(x) y 1(x)y2(x)]c1= , eitherc1=0 ory1(x)y 2(x) y 1(x)y2(x) =0. So, if the latter istrue, thenc1=0 and therefore,c2=0. This gives a condition for whichy1(x)andy2(x)are linearly independent:y1(x)y 2(x) y 1(x)y2(x) =0.( )We define this quantity as the Wronskian ofy1(x)andy2(x).Linear independence of the solutions ofa Differential equation can be establishedby looking at the Wronskian of the so-lutions. For a Second Order differentialequation the Wronskian is defined asW(y1,y2) =y1(x)y 2(x) y 1(x)y2(x).The Wronskian can be written as a determinant:W(y1,y2) = y1(x)y2(x)y 1(x)y 2(x) =y1(x)y 2(x) y 1(x)y2(x).

6 Thus, the definition of a Wronskian can be generalized to a set ofnfunctions{yi(x)}ni=1using ann Differential if the set of functions{1,x,x2}are compute the (y1,y2,y3) = y1(x)y2(x)y3(x)y 1(x)y 2(x)y 3(x)y 1(x)y 2(x)y 3(x) = 1xx2012x002 =2.( )Since,W(1,x,x2) =26=0, then the set{1,x,x2}is linearly Coefficient EquationsThe simplest Second Order Differential equationsare those withconstant coefficients. The general form for a homogeneous constant coeffi-cient Second Order linear Differential equation is given asay (x) +by (x) +cy(x) =0,( )wherea,b, andcare to ( ) are obtained by making a guess ofy(x) =erx. Insertingthis guess into ( ) leads to the characteristic equationar2+br+c=0.( )Namely, we compute the derivatives ofy(x) =erx, to gety(x) =rerx, andThe characteristic equation foray +by +cy=0 isar2+br+c= of this quadratic equation leadto solutions of the Differential (x) =r2erx. Inserting into ( ), we have0=ay (x) +by (x) +cy(x) = (ar2+br+c) the exponential is never zero, we find thatar2+br+c= real, distinct roots,r1andr2, givesolutions of the formy(x) =c1er1x+ roots of this equation,r1,r2, in turn lead to three types of solutionsdepending upon the nature of the roots.

7 In general, we have two linearly in-dependent solutions,y1(x) =er1xandy2(x) =er2x, and the general solutionis given by a linear combination of these solutions,y(x) =c1er1x+ two real distinct roots, we are done. However, when the roots are real,but equal, or complex conjugate roots, we need to do a little more work toobtain usable y 6y=0y(0) =2,y (0) = characteristic equation for this problem isr2 r 6=0. Theroots of this equation are found asr= 2, 3. Therefore, the generalsolution can be quickly written down:y(x) =c1e 2x+ Order Differential Equations 35 Note that there are two arbitrary constants in the general , one needs two pieces of information to find a particularsolution. Of course, we have the needed information in the form ofthe initial also needs to evaluate the first derivativey (x) = 2c1e 2x+3c2e3xin Order to attempt to satisfy the initial conditions. Evaluatingyandy atx=0 yields2=c1+c20= 2c1+3c2( )These two Equations in two unknowns can readily be solved to givec1=6/5 andc2=4/5.

8 Therefore, the solution of the initial valueproblem is obtained asy(x) =65e 2x+ roots,r1=r2=r, give solu-tions of the formy(x) = (c1+c2x) the case when there is a repeated real root, one has only one solution,y1(x) =erx. The question is how does one obtain the Second linearly in-dependent solution? Since the solutions should be independent, we musthave that the ratioy2(x)/y1(x)is not a constant. So, we guess the formy2(x) =v(x)y1(x) =v(x)erx. (This process is called the Method of Reduc-tion of Order . See )For constant coefficient Second Order Equations , we can write the equa-tion as(D r)2y=0,whereD=ddx. We now inserty2(x) =v(x)erxinto this equation. First weFor more on the Method of Reduction ofOrder, see (D r)verx=v ,0= (D r)2verx= (D r)v erx=v , ify2(x)is to be a solution to the Differential equation, thenv (x)erx=0for allx. So,v (x) =0, which implies thatv(x) =ax+ ,y2(x) = (ax+b) loss of generality, we can takeb=0 anda=1 to obtain the secondlinearly independent solution,y2(x) =xerx.

9 The general solution is theny(x) =c1erx+ +6y +9y= this example we haver2+6r+9=0. There is only one root,r= 3. From the above discussion, we easily find the solutiony(x) =(c1+c2x)e Differential equationsWhen one has complex roots in the solution of constant coefficient equa-tions, one needs to look at the solutionsy1,2(x) =e( i ) make use of Euler s formula2, which is treated in Section sFormulaisfoundusingMaclaurin series expansionex=1+x+12x2+13!x3+ .Letx=i and findei =1+i +12(i )2+13!(i )3+ .=1 12 2+14! 4+ i[ 13! 3+15! 5+ ].=cos +isin .ei x=cos x+isin x.( )Then, the linear combination ofy1(x)andy2(x)becomesAe( +i )x+Be( i )x=e x[Aei x+Be i x]=e x[(A+B)cos x+i(A B)sin x] e x(c1cos x+c2sin x).( )Thus, we see that we have a linear combination of two real, linearly inde-pendent solutions,e xcos xande xsin roots,r= i , give solutionsof the formy(x) =e x(c1cos x+c2sin x). +4y= characteristic equation in this case isr2+4=0.

10 The rootsare pure imaginary roots,r= 2i, and the general solution consistspurely of sinusoidal functions,y(x) =c1cos(2x) +c2sin(2x), since =0 and = +2y +4y= characteristic equation in this case isr2+2r+4=0. The rootsare complex,r= 1 3iand the general solution can be written asy(x) =[c1cos( 3x) +c2sin( 3x)]e +4y= is an example of a nonhomogeneous problem. The homoge-neous problem was actually solved in According to thetheory, we need only seek a particular solution to the nonhomoge-neous problem and add it to the solution of the last example to get thegeneral particular solution can be obtained by purely guessing, makingan educated guess, or using the Method of Variation of will not review all of these techniques at this time. Due to thesimple form of the driving term, we will make an intelligent guessofyp(x) =Asinxand determine whatAneeds to be. Inserting thisguess into the Differential equation gives( A+4A)sinx=sinx.