1.11 Some Higher-Order Differential Equations

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1 page 99. Some Higher-Order Differential Equations 99. Some Higher-Order Differential Equations So far we have developed analytical techniques only for solving special types of firstorder differential equations. The methods that we have discussed do not apply directly to higher-order differential equations, and so the solution to such equations usually requires the derivation of new techniques. One approach is to replace a higher-order differential equation by an equivalent system of first-order equations. (This will be developed further in Chapter 7.) For example, any second-order differential equation that can be written in the form d 2 ( y 2 = F x,y, ) (..) where F is a known function, can be replaced by an equivalent pair of first-order differential equations as follows. We let v = /. Then / 2 = /, and so solving Equation (..) is equivalent to solving the following two first-order differential equations = v, (..2) = F(x,y,v). (..3) In general the differential equation (..3) cannot be solved directly, since it involves three variables, x, y, and v. However, for certain forms of the function F, Equation (..3) will involve only two variables, and then we can sometimes solve it for v using one of our previous techniques. Having obtained v, we can then substitute into Equation (..2) to obtain a first-order differential equation for y. We now discuss two forms of F for which this is certainly the case. Case : Second-Order Equations with the Dependent Variable Missing If y does not occur explicitly in the function F, then Equation (..) assumes the form d 2 ( y 2 = F x, ). (..4) Substituting v = / and / = / 2 into this equation allows us to replace it with the two first-order equations = v, (..5) = F(x,v). (..6) Thus, to solve Equation (..4), we first solve Equation (..6) for v in terms of x and then solve Equation (..5) for y as a function of x. Example.. Find the general solution to 2 = ( ) x + x2 cos x, x > 0. (..7)

2 page CHAPTER First-Order Differential Equations Solution: In Equation (..7) the dependent variable is missing, and so we let v = /, which implies that / 2 = /. Substituting into Equation (..7) yields the following equivalent first-order system: = v, (..8) = x (v + x2 cos x). (..9) Equation (..9) is a first-order linear differential equation with standard form x v = x cos x. (..0) An appropriate integrating factor is I(x) = e x = e ln x = x. Multiplying Equation (..0) by x reduces it to d (x v) = cos x, which can be integrated directly to obtain Thus, x v = sin x + c. v = x sin x + cx. (..) Substituting the expression for v from (..) into Equation (..8) gives which we can integrate to obtain = x sin x + cx y(x) = x cos x + sin x + c x 2 + c 2. Case 2: Second-Order Equations with the Independent Variable Missing If x does not occur explicitly in the function F in Equation (..), then we must solve a differential equation of the form d 2 ( y 2 = F y, ). (..2) In this case, we still let v =, as previously, but now we use the chain rule to express / 2 in terms of /. Specifically, we have 2 = = = v.

3 page 0. Some Higher-Order Differential Equations 0 Substituting for / and / 2 into Equation (..2) reduces the second-order equation to the equivalent first-order system = v, (..3) = F(y,v). (..4) In this case, we first solve Equation (..4) for v as a function of y and then solve Equation (..3) for y as a function of x. Example..2 Find the general solution to 2 = 2 y ( ) 2. (..5) Solution: In this differential equation, the independent variable does not occur explicitly. Therefore, we let v = / and use the chain rule to obtain 2 = = = v. Substituting into Equation (..5) results in the equivalent system = v, (..6) v = 2 y v2. (..7) Separating the variables in the differential equation (..7) gives 2 =, (..8) v y which can be integrated to obtain ln v =2ln y +c. Combining the logarithm terms and exponentiating yields v(y) = c ( y) 2, (..9) where we have set c =±e c. Notice that in solving Equation (..7), we implicitly assumed that v 0, since we divided by it to obtain Equation (..8). However, the general form (..9) does include the solution v = 0, provided we allow c to equal zero. Substituting for v into Equation (..6) yields = c ( y) 2. Separating the variables and integrating, we obtain ( y) = c x + d. That is, y = c x + d.

4 page CHAPTER First-Order Differential Equations Solving for y gives y(x) = c x + (d ), (..20) c x + d which can be written in the simpler form y(x) = x + a x + b, (..2) where the constants a and b are defined by a = (d )/c and b = d /c. Notice that the form (..2) does not include the solution y = constant, which is contained in (..20) (set c = 0). This is because in dividing by c, we implicitly assumed that c 0. Thus in specifying the solution in the form (..2), we should also include the statement that any constant function y = k (k a constant) is a solution. Example..3 Determine the displacement at time t of a simple harmonic oscillator that is extended a distance A units from its equilibrium position and released from rest at t = 0. Solution: According to the derivation in Section., the motion of the simple harmonic oscillator is governed by the initial-value problem 2 = ω2 y, (..22) y(0) = A, (0) = 0, (..23) where ω is a positive constant. The differential equation (..22) has the independent variable t missing. We therefore let v = / and use the chain rule to write 2 = v It then follows that Equation (..22) can be replaced by the equivalent first-order system = v, (..24) v = ω2 y. (..25) Separating the variables and integrating Equation (..25) yields which implies that 2 v2 = 2 ω2 y 2 + c, v =± c ω 2 y 2 where c = 2c. Substituting for v into Equation (..24) yields =± c ω 2 y 2. (..26)

5 page 03. Some Higher-Order Differential Equations 03 Setting t = 0 in this equation and using the initial conditions (..23), we find that c = ω 2 A 2. Equation (..26) therefore gives =±ω A 2 y 2. By separating the variables and integrating, we obtain where b is an integration constant. Thus, arcsin(y/a) = ±ωt + b, y(t) = A sin(b ± ωt). The initial condition y(0) = A implies that sin b =, and so we can choose b = π/2. We therefore have That is, y(t) = A sin(π/2 ± ωt) y(t) = A cos ωt. Consequently the predicted motion is that the mass oscillates between ±A for all t. This solution makes sense physically, since the simple harmonic oscillator does not include dissipative forces that would slow the motion. Remark In Chapter 6 we will see how to solve the initial-value problem (..22), (..23) in just a few lines of work without requiring any integration! Exercises for. Skills Be familiar with the strategy of solving a higher-order differential equation by replacing it with an equivalent system of first-order differential equations, and be able to carry out this strategy in particular instances. Problems For Problems 3, solve the given differential equation.. y = 2x y + 4x (x )(x 2)y = y. 3. y + 2y )(y ) 2 = y. 4. y = (y ) 2 tan y. 5. y + y tan x = (y ) 2. d 2 ( ) x = y 2x y = 6x t d2 x = 2(t + 2 ). 9. y α(y ) 2 βy = 0, where α and β are nonzero constants. 0. y 2x y = 8x 4.. ( + x 2 )y = 2xy. 2. y + y (y ) 2 = ye y (y ) y y tan x =, 0 x<π/2. In Problems 4 5, solve the given initial-value problem. 4. yy = 2(y ) 2 + y 2, y(0) =, y (0) = y = ω 2 y, y(0) = a, y (0) = 0, where ω,a are positive constants.

6 page CHAPTER First-Order Differential Equations 6. The following initial-value problem arises in the analysis of a cable suspended between two fixed points y = a + (y ) 2, y(0) = a, y (0) = 0, where a is a nonzero constant. Solve this initial-value problem for y(x). The corresponding solution curve is called a catenary. 7. Consider the general second-order linear differential equation with dependent variable missing: y + p(x)y = q(x). Replace this differential equation with an equivalent pair of first-order equations and express the solution in terms of integrals. 8. Consider the general third-order differential equation of the form y = F(x,y ). (..27) (a) Show that Equation (..27) can be replaced by the equivalent first-order system du = u 2, du 2 = u 3, du 3 = F(x,u 3), where the variables u,u 2,u 3 are defined by u = y, u 2 = y, u 3 = y. (b) Solve y = x (y ). 9. A simple pendulum consists of a particle of mass m supported by a piece of string of length L. Assuming that the pendulum is displaced through an angle θ 0 radians from the vertical and then released from rest, the resulting motion is described by the initial-value problem d 2 θ 2 + g L sin θ = 0, θ(0) = θ 0, dθ (0) = 0. (..28) (a) For small oscillations, θ <<, we can use the approximation sin θ θ in Equation (..28) to obtain the linear equation d 2 θ 2 + g L θ = 0, θ(0) = θ 0, dθ (0) = 0. Solve this initial-value problem for θ as a function of t. Is the predicted motion reasonable? (b) Obtain the following first integral of (..28): dθ 2g =± L (cos θ cos θ 0). (..29) (c) Show from Equation (..29) that the time T (equal to one-fourth of the period of motion) required for θ to go from 0 to θ 0 is given by the elliptic integral of the first kind L T = 2g θ0 0 cos θ cos θ0 dθ. (..30) (d) Show that (..30) can be written as L π/2 T = du, g 0 k 2 sin 2 u where k = sin(θ 0 /2). [Hint: First express cos θ and cos θ 0 in terms of sin 2 (θ/2) and sin 2 (θ 0 /2).].2 Chapter Review Basic Theory of Differential Equations This chapter has provided an introduction to the theory of differential equations. A differential equation involves one or more derivatives of an unknown function, and the highest-order derivative is the order of the differential equation. For an nth-order differential equation, the general solution contains n arbitrary constants, and all solutions can be obtained by assigning appropriate values to the constants. This chapter is concerned mainly with first-order differential equations, which may be written in the form

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