ࡱ>     wt5@ àbjbj22 =XXnnnn nnn88nD|r4 s{"{{{Y~Y~Y~jllllll$R)`Y~}|Y~Y~Y~nn{{-Y~lnR{{jY~j62"j{s ;jnő0.<ס8j nnnn\Y~Y~Y~Y~Y~Y~Y~ DYb T bChapter 5 Higher order Differential Equations Initial-value and Boundary-value Problems Homogeneous Equations Non-homogeneous Equations Reduction of order Solution of Homogeneous Linear Equations with Constant Coefficients The Method of Undetermined Coefficients The Method of Variation of Parameters Cauchy-Euler Equation Non-linear Differential Equations Exercises In this chapter we discuss the solution of differential equations of order two or more. In the first eight sections underlying theory and certain important methods are presented. Ninth section is devoted to a brief introduction of non-linear higher-order differential equations. The main goal of the present chapter is to find general solutions of linear higher-order differential equations. 5.1 Initial-value and Boundary-Value Problems Initial-value Problem: For a linear differential equation the following problem is called an nth order initial-value problem: Find a solution of the differential equation  EMBED Equation.3  (5.1) subject to y(x0)=y0, y'(x0)=y1, - - - -, y(n-1)(x0)=yn-1 (5.2) Conditions given by (5.2) are called n initial conditions. The following theorem provides existence and uniqueness of solutions of initial-value problems. Theorem 5.1 (Existence and Uniqueness of Solutions). Let an(x), an-1(x), - - - - ,a1 (x), a0(x) and g(x) be continuous on an interval (, and let an (x) (0 for every x in (. If x=x0 is any point in (, then there exists a unique solution y(x) of the initial value problem (5.1)-(5.2) on the interval (. Example 5.1 The initial-value problem  EMBED Equation.3  EMBED Equation.3  y(1)=0, y'(1)=0, y"(1)=0 possesses the trivial solution y=0. Since the third-order equation is linear with constant coefficients, it follows that all conditions of Theorem 5.1 are satisfied. Hence y=0 is the only solution on any interval containing x=1. Example 5.2 Check whether the function y=3e2x+e-2x-3x is a solution of the initial value problem  EMBED Equation.3  EMBED Equation.3  y(0)=4, y'(0)=1, Here a2(x)=1(0, a1(x)= 4(0 for every interval containing x=0, g(x)=12x. a2(x), a1(x) and g(x) are continuous on any interval ( containing x=0. y=3e2x+e-2x-3x is a solution of the initial-value problem on any interval ( containing x=0 by Theorem 5.1. It is also unique solution by the same theorem. Boundary-Value Problem: A boundary value problem consists of solving a linear differential equation of order two or greater in which the dependent variable y or its derivatives are specified at different points. A typical boundary value problem (BVP) is solve the linear differential equation of order 2.  EMBED Equation.3  EMBED Equation.3  (5.3) subject to y(a)=y0, y(b)=y1 (5.4) Conditions in (5.4) are called boundary conditions. Remark 5.1 (i) A solution of the BVP (5.3)-(5.4), is a function ((x) satisfying the differential equation (5.3) on some interval (, containing a and b, whose graph passes through the two points (a,yo) and (b,y1). (ii) For a second-order differential equation, other pairs of boundary conditions could be y'(a)=y0, y(b)=y1 y(a)=y0, y'(b)=y1 y'(a)=y0, y'(b)=y1 Example 5.3 Show that the following boundary-value problem  EMBED Equation.3  y(0)=0, y EMBED Equation.3 =0 has infinitely many solutions. Solution: It can be checked that y=c1cos4x+c2sin 4x is a solution of the equation  EMBED Equation.3  y(0)=0 =c1cos 4.0 + c2 sin 4.0 or c1=0.  EMBED Equation.3  or c2 sin 2 (=0, for any choice c2 Hence the boundary-value problem  EMBED Equation.3  y(0)=0, y ( EMBED Equation.3 )=0 has infinitely many solutions 5.2 Homogeneous Equations A linear nth-order differential equation of the form  EMBED Equation.3  (5.5) where g(x) ( 0 (g(x) is not identically zero) is called a non homogeneous equation. If g(x)=0, that is,  EMBED Equation.3  (5.6) is called homogeneous linear differential equation of nth-order. We shall see in Section 5.3 that in order to solve a non homogeneous differential equation (5.5), we must be able to solve the associated homogeneous equation (5.6). Here we discuss the general solution of (5.6). Throughout the discussion we assume that (i) ai(x), i=0,1,2 - - - - n are continuous, (ii) g(x)=0 as the case for homogeneous equations or continuous, and (iii) an(x) (0 for every x in the interval on which solution is considered. Differential Operator: Let  EMBED Equation.3  The symbol D is said to be a differential operator as it transforms a differentiable function into another function. We can write  EMBED Equation.3 is, D is acting (operating) twice on y. Continuing this process we can write  EMBED Equation.3  We define an nth-order differential operator to be L=an(x)Dn+an-1(x) Dn-1+ - - - - + a1(x)D+ao(x). (5.7) By the two basic properties of differentiation, we have D((f(x))= ( Df(x), where ( is a constant D{f(x) + g(x)} =Df(x)+Dg(x) This means that the differential operator L possesses a linearity property; that is, L operating on a linear combination of two differentiable functions is the same as the linear combination of L operating on the individual functions. In symbol this means that L{(f(x)+(g(x)} = (L(f(x))+( L(g(x)), (5.8) where ( and ( are constants. In view of (5.8) L is called a linear operator. Homogenous equation (5.6) can be expressed in terms of the D notion as L(y)=0. So (5.5) can be written as L(y)=g(x) Superposition Principle: The following theorem tells us that the sum or superposition of two or more solutions of (5.6) is also a solution of (5.6). Theorem 5.2 Let y1, y2, - - - -, yn be solutions of (5.6) on an interval (. Then the linear combination y=(1y1+(2y2+ - - - - +(nyn, where =(i, i=1,2, - - - - n, are arbitrary constants, is also a solution of (5.6) on (. Corollary 5.1 (i) Every constant multiple of a solution of (5.6) is also a solution, that is, (y (x) is a solution of (5.6) whenever y(x) is a solution of (5.6) for arbitrary constant (. (ii) (5.6) always possesses the trivial solution y(x)=0. Linear Dependence and Linear Independence: Definition 5.1 A set of functions f1(x), f2(x), - - - -, fn(x) is said to be linearly independent on an interval ( if the only constants for which (1f1(x)+ (2f2(x)+ - - - - +(nfn(x)=0 for every x in the interval are (1=(2= - - - - (n=0. A set of functions which is not linearly independent is called linearly dependent. Remark 5.2 An equivalent formulation for linearly dependent set: A set of functions f1(x), f2(x), - - - - ,fn(x) are linearly dependent on an interval ( if there exist constants (1,(2, - - - - -(n, not all zero, such that (1f1(x)+ (2f2(x) + - - - - +(nfn(x)=0 for every x in the interval. Let the set consist of two functions only say f1(x) and f2(x). Therefore, assuming (1(0, f1(x)= - EMBED Equation.3 f2(x), that is, f1(x) is a constant multiple of f2(x). Thus if the set of two functions is linearly dependent then one must be constant multiple of the other. Conversely, let f1(x)= (2f2(x) for some constant (2. Then f1(x)+ (2f2(x)=0 for every x in the interval. Hence the set of two functions is linearly dependent because at least one of the constants, namely (1= 1 is not zero. We conclude that a set of two functions is linearly independent when neither function is a constant multiple of the other on the interval. Example 5.4 (i) Let f1(x) = sin 2x, f2(x)=sin x cos x The set of f1( x) and f2(x) is linearly dependent on (-(,() as f1(x) is a constant multiple of f2 (x), as f1(x)=sin 2x=2 sin x cos x on (-(,(). (ii) Let f1(x)=ex, f2(x)=5ex. The set {f1(x), f2(x)} is linearly dependent (iii) Let f1(x)=2+x, f2(x)=2+|x| {f1(x), f2(x)} is linearly independent as f1(x) and f2(x) cannot be multiples of each other. Now we mention results characterizing linearly independent solutions of (5.6) in terms of determinant called Wronskian. Definition 5.2. Let each of the functions f1(x), f2(x), - - - -, fn(x) possesses at least n-1 derivatives. The determinant  EMBED Equation.3  (5.9) where the primes denote derivatives, is called the Wronskian of the functions. Theorem 5.3 Let y1, y2, - - - -, yn be n solutions of (5.6) on an interval (. Then the set of solutions is linearly independent on ( if and only if W(y1, y2 - - - -, yn) (0 for every x in the interval. Definition 5.3 (Fundamental Set of Solutions). Any set y1, y2- - - -, yn of n linearly independent solutions of (5.6) on an interval ( is said to be a fundamental set of solutions on the interval. Theorem 5.4 (Existence of a Fundamental Set) There exists a fundamental set of solutions of (5.6) on an interval (. Theorem 5.5 (General Solution). Let y1, y2- - - -, yn be a fundamental set of solutions of (5.6) on an interval (. Then y = (1y1(x) + (2y2(x) +- - - - + (nyn(x) where (i, i=1,2, - - - - n are arbitrary constants, is also a solution of (5.6). It is called the general solution of (5.6) Remark 5.3 Theorem 5.5 states that for any solution y(x) of (5.6) on an interval (, c1, c2, - - - - cn can be found such that y(x) = c1y1(x) + c2y2(x)+ - - - - + cnyn(x) Example 5.5 The set consisting of e-3x and e4x is a fundamental set of solutions of the differential equation y"-y'-12y=0 on (-(,(). y=e-3x is a solution of the given differential equation, that is, y"-y'-12y=9e-3x+3e-3x-12e-3x=0 y=e4x is a solution of the given differential equation, that is, y"-y'-12y = 16e4x-4e4x-12e4x=0 The set of { e-3x, e4x} is linearly independent as  EMBED Equation.3  is a function and not constant. In other words neither is constant multiple of the other and so { e-3x, e4x} is independent.  EMBED Equation.3  Therefore { e-3x, e4x} is a fundamental set of solutions on interval (-(,() 5.3 Non-homogeneous Equations Any function yp, free of arbitrary parameters, that satisfies (5.5) is said to be a particular solution or particular integral of the equation. Theorem 5.6 Let yp be any particular solution of the non homogenous linear nth-order differential equation (5.5) on an interval (, and let y1, y2, - - - -,yn be a fundamental set of solutions of the associated homogenous differential equation (5.6) on (. Then the general solution of the equation on the interval is y=c1y1(x)+ c2y2(x) + - - - - +cnyn(x) +yp, (5.10) where ci, i=1,2, - - - -, n are arbitrary constants. The linear combination yc(x) = c1y1(x)+c2y2(x)+ - - - - +cnyn(x), which is the general solution of (5.6), is called complementary function for non homogeneous differential equation (5.5). Thus, in order to solve (5.5) we first solve associated homogeneous linear differential equation (5.6) and then find a particular solution of (5.5). The general solution of (5.5) is y = complementary function + any particular solution = yc+yp. (5.11) Example 5.6 y=c1e2x+c2e5x+6ex is the general solution of the non homogeneous differential equation y"-7y'+10y=24ex on (-(,(). Verification: We are required to check that yc(x)=c1e2x+c2e5x is the general solution of y"-7y'+10y= 0 and y=6ex is a particular solution of y"-7y'+10y=24ex We have y'c(x) = 2c1e2x+5c2e5x y"c(x)= 4c1e2x+25c2e5x y"-7y'+10y=(4c1e2x+25c2e5x)-7(2c1e2x+5c2e5x) +10c1e2x+10c2e5x =(14c1e2x-14c1e2x)+(35c2e5x-35c2e5x) = 0 Thus, yc(x) is the general solution of y"-7y'+10y=0 We also have y'=6ex y"=6ex, so y"-7y'+10y=6ex-42ex+60ex = 24ex that is y=6ex is a particular solution of y"-7y'+10y=24ex 5.4 Reduction of order Let a2(x) y"+a1(x) y' +ao(x)y=0 (5.12) be linear second-order homogeneous differential equation. The main idea is to discuss procedure to reduce (5.12) to a linear first-order differential equation. Theorem 5.7 If y1 is a nontrivial solution of the second-order homogeneous linear differential equation (5.12) then the substitution y2(x)=y1(x)u(x), followed by the substitution w(x)= u'(x) reduces (5.12) to a first-order linear differential equation. Remark 5.4 (i) First order linear differential equation obtained in Theorem 5.7 can be solved by computing an integrating factor ((x)=e(P(x)dx (see Section 2.3) (ii) This procedure holds also for higher order linear differential equations. Example 5.7 Let y1 be a solution of y"-y=0 on the interval (-(,(). Use reduction of order to find a second solution y2. Verification Let y2(x)=y1(x)u(x)=u(x)ex. Differentiating this product function we get y'2(x)=uex+u' (x)ex y"2(x)=uex+ u' (x)ex+ u' (x)ex+u"(x)ex Therefore, y"2(x)-y2(x) = ex(u"+2u')=0 Since ex ( 0, this equation gives us u"+2u'=0 By substituting u'=w in this equation we get w'+2w=0 This is a linear first-order differential equation Applying integrating factor e(2dx=e2x, we can write  EMBED Equation.3 . By integrating we obtain e2xw=c1 or w = u'=c1e-2x. Integrating again with respect to x we get  EMBED Equation.3  Thus y2(x)=u(x)ex= EMBED Equation.3  By choosing c2 = 0 and c1 = -2 we get y2(x) =e-x Since W(y1,y2)  EMBED Equation.3  = 2 ( 0 for every x((-(,(), the solutions are linearly independent in this interval. 5.5 Homogeneous Linear Equations with Constant Coefficients We consider in this section equations of the type  EMBED Equation.3  (5.13) where the coefficients an, an-1, - - - - a2, a1, a0 are real constants and an ( 0. We focus mainly our attention to the case n=2, similar discussion is possible for other higher numbers. It is interesting to note that all solutions of (5.13) for any n in general and n=2 in particular are exponential functions or are constructed out of exponential function. Let us consider the special case n=2 of (5.13) of the form ay"+by'+cy=0 (5.14) If we try a solution of the form y=emx, then after substituting y'=memx and y"=m2emx equation (5.14) gives us a m2 emx+bm emx+cemx = 0 or emx(am2+bm+c)=0 Since emx(0 for all x, am2+bm+c=0 (5.15) (5.15) is called the auxiliary equation. Equation (5.14) can be satisfied by the roots of (5.15) Roots of (5.15) are  EMBED Equation.3   EMBED Equation.3  We know that (i) m1 and m2 are real and distinct if b2-4ac>0 (ii) m1 and m2 are real and equal if b2-4ac=0 (iii) m1 and m2 are conjugate complex numbers if b2-4ac<0 Case (i) Distinct Real Roots Let m1 and m2 be two distinct real roots of (5.15). We find two solutions y1= em1x and y2=em2x We can check that y1 and y2 are linearly independent on (-(,() and form a fundamental set y=c1em1x+c2em2x (5.16) is the general solution of (5.14) Case (ii) Repeated Roots If m1=m2 we obtain only one exponential solution, y1=em1x. A second solution y2= em1x  EMBED Equation.3  =em1x(dx=x em 1x In this equation we have used b/a=2m1 The general solution in this case is y=c1em1x+c2xem1x (5.17) Case (iii) Conjugate Complex Roots If m1 and m2 are complex, then m1=(+i( and m2 =(-i( , where ( and ( are real and >0, i2=-1. y1=c1e((+i()x and y2=c2e((-i()x are two linearly independent solutions. Thus y=y1+y2=c1e((+i()x+c2e((-i()x is the general solution of (5.14) y is in complex form. By applying Euler's formula ei(=cos(+i sin (, where ( is any real number, we write the general solution in real form. From this formula it follows that ei(x = cos (x+i sin (x  EMBED Equation.3   EMBED Equation.3  Since y=c1e((+i()x+c2e((-i()x is a solution of (5.14) for every choice of c1 and c2, choices c1 = c2 = 1 and c1=1 and c2 = -1 give in turn two solutions y3 = e((+i()x +e((-i()x and y4 = e((+i()x -e((-i()x But y3 = e(x(ei(x+e-i(x) = 2 e(xcos(x and y4 = e(x(ei(x-e-i(x) = 2 i e(x sin (x In view of Corollary 5.1 e(x cos(x and e(x sin (x are real solutions of (5.14). Moreover, these solutions form a fundamental set on (-(,(). Consequently the general solution is y =c1 e(xcos(x + c2 e(x sin (x = e(x(c1cos(x + c2 sin (x) (5.18) Example 5.8 Solve the following differential equations: 2y"-5y-3y=0 y"+5y'-6y=0 y"+8y'+16y=0 y"+4y'+7y=0 Solution of (i) The auxiliary equation is 2m2-5m-3=0 which can be written as (2m+1)(m-3)=0 Therefore two roots are m1=- EMBED Equation.3 , m2=3 The solution is of the form (5.16), that is, y=c1 EMBED Equation.3 +c2e3x (ii) The auxiliary equation is m2+5m-6=0. This can be written in the form (m-1)(m+6)=0. Roots are m1=1, m2= - 6. Then the solution is of the form (5.16), that is, y=c1ex+c2e-6x The auxiliary equation is m2+8m+16=0 or (m+4)2 = 0. Roots are m1=m2= -4. The solution is of the form (5.17), that is, y=c1em1x+c2xem1x=c1e-4x+c2xe-4x The auxiliary equation is m2+4m+7=0. Roots m1 and m2 are given by  EMBED Equation.3  The solution is of the form (5.18), that is,  EMBED Equation.3 x) Example 5.9 Solve the initial-value problem y"+3y'+2y=0 y(0) = 1, y'(0)=2 Solution: The auxiliary equation is m2+3m+2=0 Roots m1 and m2 are m1 = - 1 and m2 = -2 Therefore the solution is of the form (5.17), that is, y=c1e-x+c2e-2x To find c1 and c2 we use initial conditions y(0)=1 and y'(0)=2 y(0)=1=c1e-0+c2e-0=c1+c2 or c1+c2=1 y'=-c1e-x-2c2e-2x y'(0)= -c1e-0-2c2e-0 = -c1-2c2=2 or c1+2c2= -2 Thus c1+c2=1 c1+2c2=-2 This gives c2= - 3 and c1=4 Therefore, y = 4e-x-3e-2x Remark 5.5 : In general, to solve an nth-order differential equation (5.13) we must solve an nth-degree polynomial equation: anmn+an-1mn-1+- - - - +a2m2+a1m+a0=0n-1+ . . . a2m2+a1m+a0=0 If all roots of this equation are real and distinct, then the general solution of (512) is If all roots (say m1,m2,- - - - mn) of this equation are real and distinct, then the general solution of (5.13) is y=c1em1x+c2em 2x+ - - - -+ cnemnx. It is difficult to summarize other two cases because the roots of any auxiliary equation of degree greater than 2 can occur in many combinations. The Method of Undetermined Coefficients In order to solve non homogeneous linear differential equations with constant coefficients  EMBED Equation.3  (5.19) One must find complementary function yc, that is, the general solution of (5.13) (See Theorem 5.6) and a particular solution of (5.19). A process of finding a particular solution yp of (5.19) is known as the method of Undetermined Coefficients. The underlying idea in this method is to guess about the form of yp that is motivated by the form of g(x) in (5.19). The method is limited to those equations of the type (5.19) where g(x) is of the following forms: g(x) is constant g(x) is polynomial function (function of the form g(x)= a0+a1x+a2x2+- - - -+anxn g(x) =e(x, exponential function. g(x) = sin (x or cos (x or finite sum and products of these functions. It may be observed that this method is not applicable in cases where g(x) =lnx, g(x)= EMBED Equation.3 , g(x)=tan x, g(x)=sin-1x etc. The method is illustrated through the following examples. Example 5.10 Find a general form of a particular solution yp for the following equations 3y" + 2y = 5e2x + 2x3 3y" + 2y = x2e 3x 3y + 2y = 20 sin 2x Solution : (a) The particular solution yp will be of the form yp = Ae2x+B+Cx+Dx2+Ex3 (b) The general form of yp will be of the form yp = Ae-3x +Bxe-3x+ Cx2e-3x (c) yp=A sin 2x + B cos 2x Example 5.11 Find a particular solution yp of differential equation y-y'+y=2 sin 3x Solution: A natural first guess for a particular solution would be A sin 3x. But since successive differentiations of sin 3x produce sin 3x and cos 3x, we are prompted instead to assume a particular solution that includes both of these terms: yp=A cos 3x + B sin 3x. Differentiating yp and substituting the results into the differential equation gives: yp'= - 3A sin 3x + 3B cos 3x yp" = -9A cos 3x 9B sin 3x yp"-yp'+yp= -9A cos 3x - 9B sin 3x +3A sin 3x - 3B cos 3x+A cos 3x +B sin 3x = 2 sin 3x or y"p-yp'=yp=(-8A-3B) cos 3x+(3A-8B) sin 3x = 2 sin 3x + 0 cos 3x Comparing the coefficients of cos 3x and sin 3x we get -8A-3B=0 3A-8B=2 Solving for A and B we get A =  EMBED Equation.3  B = EMBED Equation.3  Thus a particular solution yp is given by yp= EMBED Equation.3 sin 3x Example 5.12 Find a particular solution of y" +3y'+2y=5x2. We guess that yp is of the form yp=A+Bx+Cx2 yp=B+2Cx yp=2C yp"+3yp'+2yp = 2C+3B+6Cx+2A+2Bx+2C x2=5x2 or (2C+3B+2A)+(6C+2B)x+2Cx2=0+0x+5x2 This implies that 2A+3B+2C=0 2B+6C=0 2C=5. Thus C=  EMBED Equation.3  Therefore yp= EMBED Equation.3  Example 5.13 Solve the differential equation y"-10y'+25y=30x+3 by undetermined coefficients. Solution: Step 1. Find the complementary function of y"-10y+25y=0 Step 2. Find yp. Step 1. The auxiliary equation is m2-10m +25=0 (m-5)2=0 m1=5, m2=5 Solution is of the form (5.17), that is, y=c1 e5x+c2xe5x Step 2. Let yp = Ax+B yp'=A yp"=0 0-10A+25(Ax+B)=30x+3 (-10A+25B)+25Ax=30x+3 This implies -10A+25B=3, 25A = 30 Thus A= EMBED Equation.3   EMBED Equation.3 3 gives B= EMBED Equation.3  yp= EMBED Equation.3  The general solution is y=c1e5x+c2xe5x+ EMBED Equation.3  Example 5.14 Solve the differential equation y"+4y=3 sin 2x by undetermined coefficients. Solution: Step 1 Find complementary function Auxiliary equation is m2+4=0 m=( 2i Solution is of the form (5.19), that is, yc(x)=e0.x(c1cos 2x + c2sin 2x) Step 2 Finding a particular solution yp. yp=Ax sin2x+ Bxcos2x yp'=A sin 2x+2Ax cos 2x+B cos 2x-2Bx sin 2x yp"=2A cos 2x+2A cos 2x-4Ax sin 2x-2B sin 2x 2B sin 2x-4Bx cos 2x = 4A cos 2x-4Ax sin 2x-4B sin 2x-4 Bx cos 2x yp" + 4yp= (4A cos 2x 4Ax sin 2x 4 B sin 2x 4Bx cos 2x) + (4Ax sin 2x + 4Bx cos 2x)=3 sin 2x or 4A cos 2x-4B sin 2x = 3 sin 2x -4B=3 or B= - EMBED Equation.3  A= 0 yp= -  EMBED Equation.3 x cos 2x y= yc+yp = c1 cos 2x+c2 sin 2x -  EMBED Equation.3 x cos 2x Undetermined Coefficients-Annihilator Approach: Differential equation (5.19) can be written in terms of operators D, D2, D3, - - - - Dn as Ly=g(x) (5.20) where L=anDn+an-1Dn-1+ - - - - +a1D+a0 (5.21) L is said to be an annihilator operator of a function f if L (f(x)) = 0, where f(x), is sufficiently differentiable. The differential operator Dn annihilates each of the functions 1,x,x2,- - -, xn-1 (5.22) The differential operator (D-()n annihilates each of the functions e(x, xe(x, x2e(x, - - - -, xn-1e(x (5.23) Example 5.15 Find a differential operator that annihilates the function 4e2x-10xe2x Solution: n=2, (=2, (D-2)2 is a differential operator which annihilates 4e2x-10xe2x, that is, (D-2)2 (4e2x-10xe2x)=0. The differential operator [D2-2(D+((2+(2)]n annihilates each of the functions. e(xcos (x, xe(xcos (x, x2e(xcos (x, - - - - xn-1e(xcos (x, e(sin (x, xe(x sin (x, x2e(x sin (x, - - - - xn-1e(xsin (x, (5.24) Remark 5.6 (i) If L annihilates y1 and y2 then it also annihilates their linear combination, that is (y1, +(y2, where ( and ( are real numbers. (ii) Let L1 and L2 be annihilator operator for y1 and y2 respectively. However L1 (y2) ( 0 and L2(y1)( 0 . Then L1L2 annihilates ( y1+(y2. Steps for solution: (i) Find the complementary solution yc of L(y)=0 (ii) Operate on both sides of L(y)=g(x) with a differential operator L1 that annihilates the function g(x). (iii) Find the general solution of the higher-order homogeneous differential equation L1L(y)=0. (iv) Delete from the solution in step (iii) all those terms that are duplicate in the complementary solution yc found in step (i). From a linear combination yp of the terms that remain. This is the form of a particular solution of L(y)=g(x). (v) Substitute yp found in step (iv) into L(y)=g(x). Match coefficients of the various functions on each side of the equality, and solve the resulting system of equations for the unknown coefficients in yp. (vi) With the particular solution found in step (v), form the general solution y=yc+ yp of the given differential equation. Example 5.16: Solve y"+3y'+2y=4x2 using undetermined coefficients. Step: 1. Solve the homogeneous equation y"+3y'+2y=0 The auxiliary equation is m2+3m+2=0 Roots of this equation are m1= -1 and m2= -2, and so complementary function is of the form (5.16), that is, yc=c1e-x+c2e-2x Step: 2. Now, since 4x2 is annihilated by the differential operator D3, we find that D3(D2+3D+2)y=4D3x 2 is the same as D3(D2+3D+2)y=0 (5.25) The auxiliary equation of the fifth order in (5.25), m3(m2+3m+2)=0 or m3(m+1)(m+2)=0, has roots m1=m2 = m3=0, m4=-1, and m5=-2.  Thus its general solution must be y=c1+c2x+c3x2+ c4e-x+c5e-2x (5.26) The terms in the box in (5.26) constitute the complementary function of the given equation. We can very well argue that a particular solution yp yp of the given equation should also satisfy (5.25). This means that the terms remaining in (5.26) must be the basic form of yp : yp yp yp yp yp=A+Bx+Cx2 (5.27) where, c1, c2, c3 are replaced by A,B and C respectively. For (5.27) to be a particular solution of the given equation, it is necessary to find specific coefficients A, B and C. Differentiating (5.27), we obtain y'p=B+2Cx, y"p=2C; Substituting these values into the given equation y+3y'+2y=4x2, we get yp"+3y'p+2yp=2C+3B+6Cx+2A+2Bx+2Cx2=4x2 or (2C+3B+2A)+(6C+2B)x+2Cx2=(constant terms)0 + 0x+4x2 Comparing constant terms, coefficients of x and x2, we get 2C+3B+2A=0, 6C+2B=0, and 2C=4 This implies C=2, B= -6, and A=7. Thus yp =7-6x+2x2. Step 3. The general solution of the given equation is y=yc+ yp or y=c1e-x+c2e-2x+7-6x+2x2 Example: 5.17 Solve the differential equation y"-9y=54 by undetermined coefficient approach. Solution: Applying D to the differential equation we obtain D(D2-9)y=0 The auxiliary equation is m(m2-9)=0. Roots are m1=0, m2=3, m3= -3 Then general solution is y=c1e3x+c2e-3x+c3. yc and a particular solution is yp = A. Putting values of yp,yp', yp" in the given differential equation we get 0-9A=54 or A = - 6 Thus, the general solution is y=c1 e3x+c2e-3x-6 Example 5.18 Solve y"-2y'+5y=exsin x using Undetermined Coefficients - Annihilator approach. Solution: Applying D2-2D+2 to the differential equation we obtain (D2-2D+2)(D2-2D+5)y=0 Then the general solution is y=ex(c1cos 2x+c2sin 2x) + ex(c3cos x +c4sin x), yc and yp=Aexcos x+Bex sin x. yp'=Aex cos x - Aex sin x + Bex sin x + Bex cos x yp" = (Aex cos x Aex sin x)- (Aex sin x + Aex cos x) +(Bex sin x + Bex cos x) +( Bex cos x Bex sin x) Substituting yp, yp,' yp" in the given equation we get: (Aex cos x Aex sin x) (Aex sin x +Aex cos x) +(Bexsin x + Bex cosx) + (Bex cos x Bex sin x) -2A excos x -2Aex sin x + 2Bex sin x + 2Bex cos x + 5(Aexcos x+Bex sin x) = ex sin x +0.ex cos x + 0. constant term or 3Aex cos x + 3Bex sin x = ex sin x. Equating coefficients gives A=0 and B=1/3. The general solution is y=ex (c1 cos 2 x + c2 sin 2x) +  EMBED Equation.3 ex sin x 5.7 The Method of Variation of Parameters The method of variation of parameters described below is applied to solve a linear second order non-homogeneous differential equation of the form a2(x)y"+a1(x)y' +ao(x)y = g(x) which can be written in the standard form y"+P(x)y' + Q (x)y=f (x) (5.28) In (5.28) we assume that P(x), Q(x) and f (x) are continuous on some interval (. As we have seen earlier in Section 5.5 there is no difficulty in finding the complementary function yc of (5.28) when P(x) and Q(x) are constant functions. Step 1. Find complementary function yc of (5.28) of the form yc=c1y1+c2y2 Step 2. Find Wronksian W of y1 and y2, that is,  EMBED Equation.3  Step 3. Write W1  EMBED Equation.3 , W2 =  EMBED Equation.3  and find u1 and u2 by integrating u1' =  EMBED Equation.3  Step 4. Find a particular solution which is of the form yp=u1y1+u2y2 Step 5. The general solution of the equation is y=yc+yp Example 5.19 Solve the differential equation y"-y=xex by applying the method of variation of parameters. Solution: Corresponding homogeneous equation is y"-y = 0 The auxiliary equation is m2-1=0. Roots are m1=1, m2= -1 The complementary function is y=c1ex+c2e-x  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  Integrating u'1 and u'2 we get  EMBED Equation.3  u2 = -(xe2x/4) + (e2x/8) The general solution is y=yc + yp where yc = c1 ex+c2e-x yp = u1y1+u2y2 = EMBED Equation.3 ex (xe2x/4)e-x+(e2x/8)e-x Thus y=c1ex+c2e-x+ EMBED Equation.3 x4ex- EMBED Equation.3 xex+ EMBED Equation.3 ex Example 5.20 Apply the method of variation of parameters to solve the differential equation y"-y=coshx Solution The auxiliary equation is m2-1=0, so m1=1 and m2=-1 and complementary function yc = c1ex+c2e-x=c1y1+c2y2  EMBED Equation.3  f(x)=coshx = EMBED Equation.3  (e-x+ex)  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  The general solution is y=yc+yp 5.8 Cauchy-Euler Equation Second-order equations of the form  EMBED Equation.3  (5.29) where a, b and c are constants, a (0, and g(x) is continuous on a given interval are called Cauchy-Euler equations. By putting y=xm, y' = mxm-1, y" = m(m-1)xm-2 in (5.29) we get  EMBED Equation.3 am(m-1)xm+bmxm+cxm = (am(m-1)+bm+c)xm Thus y=xm is a solution of  EMBED Equation.3  (5.30) whenever m is a solution of the auxiliary equation am (m-1) + bm+c=0 or am2 + (b-a) m +c=0 (5.31) There are three different cases to be considered: Case 1: District Real Roots Let m1 and m2 be real roots of (5.31) such that m1(m2. Then y1 = xm1 and y2=xm2 form a fundamental set of solutions. Hence the general solution is y=c1xm1+c2xm2 (5.32) Case 2: Repeated Roots If the roots of (5.31) are repeated, that is, m1=m2 then the general solution is of the form y=c1xm1+c2xm1lnx (5.33) Case 3. Conjugate Complex Roots If the roots of (5.31) are the conjugate pair m1= ( +i(, m2= ( -i(, where ( and ( >o are real , then a solution is y=c1x ( +i( + c2x( -i( This solution can be written in the real form as y=x( [c1cos (( ln x) +c2 sin (( ln x)] (5.34) Verification: xi(=(elnx)i(=ei(lnx which, by Euler's formula, is the same as xi(=cos (( lnx)+i sin ((lnx) Similarly, x -i(=cos ((lnx) - i sin ((lnx) By adding and subtracting, the last two results yield xi(+x-i(= 2 cos ((lnx) and xi(-x-i(= 2 i sin ((lnx), respectively. By the fact y=c1x(+i(+c2x(-i( is a solution for any values of the constants, we see, in turn, for c1=c2=1 and c1=1,c2= -1 that y1= x((xi(+x-i() and y2= x((xi(-x-i() or y1=2x(cos ((lnx) and y2=2ix(sin ((lnx) are also solutions. Since Wronskian for x(cos ((lnx) and x(sin ((lnx) in (x2(-1( 0, (>0, on the interval (0, (), we conclude that y1=x(cos ((lnx) and y2=x(sin ((lnx) constitute a fundamental set of real solutions of the differential equation. Hence we get the general solution in the real form y=x([c1os ((lnx) + c2 sin ((lnx)]. Remark 5.7 The method described above holds true for similar equations of order n. Example: 5.21 solve the differential equations x2y"-2y=0 x2y"-3xy'-2y=0 x2y"+xy'+y=0 subject to initial conditions y(1)=1,y'(1)=2 Solution (a) The auxiliary equation is m2-m-2=0 or (m+1) (m-2) =0 so m1= -1, m2=2 The general solution is y=c1x-1+c2x2 (b) The auxiliary equation is m2-4m-2=0  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  The general solution is y=c1x2+ EMBED Equation.3 + c2x2- EMBED Equation.3  (c) The auxiliary equation is m2+1=0 so that the general solution is given by y=c1cos (lnx) +c2 sin (lnx). y'= -c1 EMBED Equation.3 sin (lnx)+c2  EMBED Equation.3 cos (lnx) The initial conditions imply c1= 1 and c2=2. Thus y=cos (lnx)+2 sin (lnx). 5.9 Non linear Differential Equations The main objective of this book, in general and this chapter in particular, is to study linear differential equations. However, we present here few important general features along with solution of two classes of non linear differential equations, one in which dependent variable is missing and the other where independent variable is missing. General features of non linear equations There are several significant differences between linear and non linear differential equations. We have seen in Theorem 5.5 that a linear combination of solutions of homogeneous linear differential equations is also a solution. Non linear equations do not possess this property called superposability. As we have seen in Sections (5.5)-(5.7) linear differential equations with constant coefficients can be solved. This does not hold for nonlinear differential equations. Even when we can solve a non linear first-order differential equation in the form of one parameter family, this family does not, as a rule, represent a general solution. In other words non linear first-order differential equations can possess singular solutions where as linear equations cannot. There is a major difference in the realm of solvability of two classes of differential equations. Given a linear differential equation there is a strong possibility that we can find some form of a solution that one can look at an explicit solution or a series solution (to be discussed in Chapter 6). On the other hand, non linear differential equations of higher order are not amenable to solution by analytic methods. We have to rely only on numerical and qualitative analysis of non linear differential equations. It may be pointed out that non linear differential equations represent significant real world problems but their discussion is beyond the preview of this book. Special kind of non linear differential equations Case 1: Dependent Variable y missing Second order differential equations of the form F(x, y',y")=0, where dependent variable y is missing can sometimes be solved using first-order methods. This can be reduced to the first order by substitution u=y'. Example 5.22 Solve  EMBED Equation.3  Solution: Let u= EMBED Equation.3  Then the given differential equation can be written as  EMBED Equation.3 = 2xu2 or  EMBED Equation.3  = 2xdx or ( u-2 du = ( 2x dx -u-1=x2 + c12 The constant of integration is written as c12 for convenience. Since u-1= EMBED Equation.3 , it follows that  EMBED Equation.3 = -  EMBED Equation.3  or ( dy = -(  EMBED Equation.3  or y = -  EMBED Equation.3  Case 2: Independent variable x Missing We consider equation of the form F(y,y',y")=0 Let u=y'= EMBED Equation.3 . By chain rule y"= EMBED Equation.3  The given differential equation can be written as F(y,u, u EMBED Equation.3 )=0 Example: 5.23 Solve EMBED Equation.3  Solution: F(y,u, u EMBED Equation.3 )= EMBED Equation.3  or  EMBED Equation.3  ln|u| = ln |y|+c1 which in turn gives u=c2y, where the constant ( ec1 has been designated as c2 Substitute u= EMBED Equation.3 , separate variables once again, integrate, and relabel constant, then we have  EMBED Equation.3  or ln|y| = c2x+c3 or y=c4ec2x 5.10 Exercises Initial-value and Boundary-value Problems Give an example to show that the condition an(x)(0 is essential for the validity of Theorem 5.1 Show that y=c1ex+c2e-x is the general solution of the differential equation y"-y=0 on the interval (-(,(). Find a member of the family that is a solution of the initial value problem: y"-y=0, y(0)=0,y'(0)=1 Show that y"-2y'+xy=sin x y(()=0, y'(()=3 has a unique solution on (0 x2y"-xy'-3y=x2 ln x x2y"-xy'-2y=0 x2y"+xy'+9y=0 3x2y"+6xy+y=0 4x2y"+y=0 x2y"+5xy'+4y=0 x2y"-5xy'+8y=8x6, subject to initial conditions  EMBED Equation.3  Nonlinear Differential Equations Solve the following nonlinear differential equations y"+(y')2+1=0 x2y"+(y')2=0 y'y"=4x, subject to initial conditions y(1) = 5, y'(1)=2 2y"=3y2, subject to initial conditions y(0) = 1, y'(0)=1. 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