“main”2007/2/16page 99iiiiiiii1.11 Some Higher-Order Differential Equations 991.11 Some Higher-Order Differential EquationsSo far we have developed analytical techniques only for solving special types of first-order differential equations. The methods that we have discussed do not apply directly tohigher-order differential equations, and so the solution to such equations usually requiresthe derivation of new techniques. One approach is to replace a higher-order differentialequation by an equivalent system of first-order equations. (This will be developed furtherin Chapter 7.) For example, any second-order differential equation that can be written inthe formd2ydx2= Fx,y,dydx(1.11.1)where F is a known function, can be replaced by an equivalent pair of first-order differ-ential equations as follows. We let v = dy/dx. Then d2y/dx2= dv/dx, and so solv-ing Equation (1.11.1) is equivalent to solving the following two first-order differentialequationsdydx= v, (1.11.2)dvdx= F(x,y, v). (1.11.3)In general the differential equation (1.11.3) cannot be solved directly, since it involvesthree variables, x, y, and v. However, for certain forms of the function F , Equa-tion (1.11.3) will involve only two variables, and then we can sometimes solve it forv using one of our previous techniques. Having obtained v, we can then substitute intoEquation (1.11.2) to obtain a first-order differential equation for y. We now discuss twoforms of F for which this is certainly the case.Case 1: Second-Order Equations with the Dependent Variable MissingIf y does not occur explicitly in the function F , then Equation (1.11.1) assumes theformd2ydx2= Fx,dydx. (1.11.4)Substituting v = dy/dx and dv/dx = d2y/dx2into this equation allows us to replaceit with the two first-order equationsdydx= v, (1.11.5)dvdx= F(x,v). (1.11.6)Thus, to solve Equation (1.11.4), we first solve Equation (1.11.6) for v in terms ofx and then solve Equation (1.11.5) for y as a function of x.Example 1.11.1 Find the general solution tod2ydx2=1xdydx+ x2cos x,x>0. (1.11.7)“main”2007/2/16page 100iiiiiiii100 CHAPTER 1 First-Order Differential EquationsSolution: In Equation (1.11.7) the dependent variable is missing, and so we let v =dy/dx, which implies that d2y/dx2= dv/dx. Substituting into Equation (1.11.7) yieldsthe following equivalent first-order system:dydx= v, (1.11.8)dvdx=1x(v + x2cos x). (1.11.9)Equation (1.11.9) is a first-order linear differential equation with standard formdvdx− x−1v = x cos x. (1.11.10)An appropriate integrating factor isI(x) = e−x−1dx= e−ln x= x−1.Multiplying Equation (1.11.10) by x−1reduces it toddx(x−1v) = cos x,which can be integrated directly to obtainx−1v = sin x + c.Thus,v = x sin x + cx. (1.11.11)Substituting the expression for v from (1.11.11) into Equation (1.11.8) givesdydx= x sin x + cxwhich we can integrate to obtainy(x) =−x cos x + sin x + c1x2+ c2. Case 2: Second-Order Equations with the Independent Variable MissingIf x does not occur explicitly in the function F in Equation (1.11.1), then we mustsolve a differential equation of the formd2ydx2= Fy,dydx. (1.11.12)In this case, we still letv =dydx,as previously, but now we use the chain rule to express d2y/dx2in terms of dv/dy.Specifically, we haved2ydx2=dvdx=dvdydydx= vdvdy.“main”2007/2/16page 101iiiiiiii1.11 Some Higher-Order Differential Equations 101Substituting for dy/dx and d2y/dx2into Equation (1.11.12) reduces the second-orderequation to the equivalent first-order systemdydx= v, (1.11.13)dvdy= F(y,v). (1.11.14)In this case, we first solve Equation (1.11.14) for v as a function of y and then solveEquation (1.11.13) for y as a function of x.Example 1.11.2 Find the general solution tod2ydx2=−21 − ydydx2. (1.11.15)Solution: In this differential equation, the independent variable does not occur ex-plicitly. Therefore, we let v = dy/dx and use the chain rule to obtaind2ydx2=dvdx=dvdydydx= vdvdy.Substituting into Equation (1.11.15) results in the equivalent systemdydx= v, (1.11.16)vdvdy=−21 − yv2. (1.11.17)Separating the variables in the differential equation (1.11.17) gives1vdv =−21 − ydy, (1.11.18)which can be integrated to obtainln |v|=2ln|1 − y|+c.Combining the logarithm terms and exponentiating yieldsv(y) = c1(1 − y)2, (1.11.19)where we have set c1=±ec. Notice that in solving Equation (1.11.17), we implicitlyassumed that v = 0, since we divided by it to obtain Equation (1.11.18). However, thegeneral form (1.11.19) does include the solution v = 0, provided we allow c1to equalzero. Substituting for v into Equation (1.11.16) yieldsdydx= c1(1 − y)2.Separating the variables and integrating, we obtain(1 − y)−1= c1x + d1.That is,1 − y =1c1x + d1.“main”2007/2/16page 102iiiiiiii102 CHAPTER 1 First-Order Differential EquationsSolving for y givesy(x) =c1x + (d1− 1)c1x + d1, (1.11.20)which can be written in the simpler formy(x) =x + ax + b, (1.11.21)where the constants a and b are defined by a = (d1− 1)/c1and b = d1/c1. Noticethat the form (1.11.21) does not include the solution y = constant, which is containedin (1.11.20) (set c1= 0). This is because in dividing by c1, we implicitly assumed thatc1= 0. Thus in specifying the solution in the form (1.11.21), we should also include thestatement that any constant function y = k (k a constant) is a solution. Example 1.11.3 Determine the displacement at time t of a simple harmonic oscillator that is extended adistance A units from its equilibrium position and released from rest at t = 0.Solution: According to the derivation in Section 1.1, the motion of the simple har-monic oscillator is governed by the initial-value problemd2ydt2=−ω2y, (1.11.22)y(0) = A,dydt(0) = 0, (1.11.23)where ω is a positive constant. The differential equation (1.11.22) has the independentvariable t missing. We therefore let v = dy/dt and use the chain rule to writed2ydt2= vdvdyIt then follows that Equation (1.11.22) can be replaced by the equivalent first-ordersystemdydt= v, (1.11.24)vdvdy=−ω2y. (1.11.25)Separating the variables and integrating Equation (1.11.25) yields12v2=−12ω2y2+ c,which implies thatv =± c1− ω2y2where c1= 2c. Substituting for v into Equation (1.11.24) yieldsdydt=± c1− ω2y2. (1.11.26)“main”2007/2/16page 103iiiiiiii1.11 Some Higher-Order Differential Equations 103Setting t = 0 in this equation and using the initial conditions (1.11.23), we find thatc1= ω2A2. Equation (1.11.26)