24. x 2 y xy y sec(ln x); 1 e x y 1 cos(ln x), y 2 sin(ln x) 25. y y tan x 26. y 4y sec 2x 28.

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1 16 CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS 11. y 3y y 1 4. x yxyy sec(ln x); 1 e x y 1 cos(ln x), y sin(ln x) ex 1. y y y 1 x 13. y3yy sin e x 14. yyy e t arctan t 15. yyy e t ln t 16. y y y 41x 17. 3y6y6y e x sec x 18. 4y 4y y e x/ 11 x In Problems 5 8 solve the given third-order differential equation by variation of parameters. 5. y ytan x 6. y 4ysec x 7. y y y y e 4x 8. ex y 3y y 1 e x In Problems 19 solve each differential equation by variation of parameters, subject to the initial conditions y(0) 1, y(0) yy xe x/ 0. yyy x 1 1. yy8y e x e x. y4y4y (1x 6x)e x In Problems 3 and 4 the indicated functions are known linearly independent solutions of the associated homogeneous differential equation on (0, ). Find the general solution of the given nonhomogeneous equation. 3. x yxy(x 1 4)y x 3/ ; y 1 x 1/ cos x, y x 1/ sin x Discussion Problems In Problems 9 and 30 discuss how the methods of undetermined coefficients and variation of parameters can be combined to solve the given differential equation. Carry out your ideas. 9. 3y6y30y 15 sin x e x tan 3x 30. yyy 4x 3 x 1 e x 31. What are the intervals of definition of the general solutions in Problems 1, 7, 9, and 18? Discuss why the interval of definition of the general solution in Problem 4 is not (0, ). 3. Find the general solution of x 4 yx 3 y4x y 1 given that y 1 x is a solution of the associated homogeneous equation. 4.7 CAUCHY-EULER EQUATION REVIEW MATERIAL Review the concept of the auxiliary equation in Section 4.3. INTRODUCTION The same relative ease with which we were able to find explicit solutions of higher-order linear differential equations with constant coefficients in the preceding sections does not, in general, carry over to linear equations with variable coefficients. We shall see in Chapter 6 that when a linear DE has variable coefficients, the best that we can usually expect is to find a solution in the form of an infinite series. However, the type of differential equation that we consider in this section is an exception to this rule; it is a linear equation with variable coefficients whose general solution can always be expressed in terms of powers of x, sines, cosines, and logarithmic functions. Moreover, its method of solution is quite similar to that for constant-coefficient equations in that an auxiliary equation must be solved. Cauchy-Euler Equation A linear differential equation of the form a n x n dn y dx n a n1x n1 dn1 y dx n1 a 1x dx a 0y g(x), Copyright 01 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the ebook and/or echapter(s). Editorial review has

2 4.7 CAUCHY-EULER EQUATION 163 where the coefficients a n, a n1,..., a 0 are constants, is known as a Cauchy-Euler equation. The differential equation is named in honor of two of the most prolifi mathematicians of all time. Augustin-Louis Cauchy (French, ) and Leonhard Euler (Swiss, ). The observable characteristic of this type of equation is that the degree k n, n 1,..., 1, 0 of the monomial coefficients x k matches the order k of differentiation d k ydx k : same d a n x n y d n a n1 x n1 y n1.... dx n dx n1 As in Section 4.3, we start the discussion with a detailed examination of the forms of the general solutions of the homogeneous second-order equation ax d y bx cy 0. (1) The solution of higher-order equations follows analogously. Also, we can solve the nonhomogeneous equation ax ybxycy g(x) by variation of parameters, once we have determined the complementary function y c. Note The coefficient ax of y is zero at x 0. Hence to guarantee that the fundamental results of Theorem are applicable to the Cauchy-Euler equation, we focus our attention on finding the general solutions defined on the interval ( ). Method of Solution We try a solution of the form y x m, where m is to be determined. Analogous to what happened when we substituted e mx into a linear equation with constant coefficients, when we substitute x m, each term of a Cauchy-Euler equation becomes a polynomial in m times x m, since a k x k dk y dx k a kx k m(m 1)(m ) (m k 1)x mk a k m(m 1)(m ) (m k 1)x m. same For example, when we substitute y x m, the second-order equation becomes ax d y bx cy am(m 1)xm bmx m cx m (am(m 1) bm c)x m. Thus y x m is a solution of the differential equation whenever m is a solution of the auxiliary equation am(m 1) bm c 0 or am (b a)m c 0. There are three different cases to be considered, depending on whether the roots of this quadratic equation are real and distinct, real and equal, or complex. In the last case the roots appear as a conjugate pair. Case I: Distinct Real Roots Let m 1 and m denote the real roots of (1) such that m 1 m. Then and y x m y 1 x m 1 form a fundamental set of solutions. Hence the general solution is y c 1 x m 1 c x m. () (3) EXAMPLE 1 Distinct Roots Solve x d y x 4y 0. Copyright 01 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the ebook and/or echapter(s). Editorial review has

3 164 CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS SOLUTION Rather than just memorizing equation (), it is preferable to assume y x m as the solution a few times to understand the origin and the difference between this new form of the auxiliary equation and that obtained in Section 4.3. Differentiate twice, dx mxm1, and substitute back into the differential equation: if m 3m 4 0. Now (m 1)(m 4) 0 implies m 1 1, m 4, so y c 1 x 1 c x 4. Case II: Repeated Real Roots If the roots of () are repeated (that is, m 1 m ), then we obtain only one solution namely, y x m 1. When the roots of the quadratic equation am (b a)m c 0 are equal, the discriminant of the coefficients is necessarily zero. It follows from the quadratic formula that the root must be m 1 (b a)a. Now we can construct a second solution y, using (5) of Section 4.. We firs write the Cauchy-Euler equation in the standard form and make the identifications P(x) bax and (b>ax) dx (b>a) ln x. Thus y x m 1 d y dx b ax e(b/a)ln x x m 1 d y dx m(m 1)xm, x d y x 4y x m(m 1)x m x mx m1 4x m x m (m(m 1) m 4) x m (m 3m 4) 0 dx dx c ax y 0 x m 1 x b/a x m 1 dx ; e (b / a)ln x e ln xb / a x b / a x m 1 x b/a x (ba)/a dx ; m 1 (b a)/a x m 1 dx x xm 1 ln x. The general solution is then y c 1 x m 1 c x m 1 ln x. (4) EXAMPLE Repeated Roots Solve SOLUTION 4x d y 8x y 0. The substitution y x m yields 4x d y 8x y xm (4m(m 1) 8m 1) x m (4m 4m 1) 0 when 4m 4m 1 0 or (m 1) 0. Since m 1 1, it follows from (4) that the general solution is y c 1 x 1/ c x 1/ ln x. For higher-order equations, if m 1 is a root of multiplicity k, then it can be shown that x m 1, x m 1 ln x, x m 1 (ln x),..., x m 1 (ln x) k1 Copyright 01 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the ebook and/or echapter(s). Editorial review has

4 4.7 CAUCHY-EULER EQUATION 165 are k linearly independent solutions. Correspondingly, the general solution of the differential equation must then contain a linear combination of these k solutions. Case III: Conjugate Complex Roots If the roots of () are the conjugate pair m 1 a ib, m a ib, where a and b 0 are real, then a solution is But when the roots of the auxiliary equation are complex, as in the case of equations with constant coefficients, we wish to write the solution in terms of real functions only. We note the identity which, by Euler s formula, is the same as y C 1 xi C xi. x i (e ln x ) i e i ln x, x ib cos(b ln x) i sin(b ln x). Similarly, x ib cos(b ln x) i sin(b ln x). Adding and subtracting the last two results yields x ib x ib cos(b ln x) and x ib x ib i sin(b ln x), respectively. From the fact that y C 1 x aib C x aib is a solution for any values of the constants, we see, in turn, for C 1 C 1 and C 1 1, C 1 that y 1 or y 1 x(x i x i) and y x(x i x i) y 1 x cos( ln x) and y ix sin( ln x) are also solutions. Since W(x a cos(b ln x), x a sin(b ln x)) bx a1 0, b 0 on the interval (0, ), we conclude that 0 x y 1 x cos( ln x) and y x sin( ln x) constitute a fundamental set of real solutions of the differential equation. Hence the general solution is y x[c 1 cos( ln x) c sin( ln x)]. (5) _1 (a) solution for 0 x 1 y 10 1 EXAMPLE 3 An Initial-Value Problem Solve 4x y17y 0, y(1) 1, y(1) 1. SOLUTION The y term is missing in the given Cauchy-Euler equation; nevertheless, the substitution y x m yields 5 x 4x y17y x m (4m(m 1) 17) x m (4m 4m 17) 0 when 4m 4m From the quadratic formula we find that the roots are m 1 1 and m 1 i i. With the identifications and b we see from (5) that the general solution of the differential equation is 1 y x 1/ [c 1 cos( ln x) c sin( ln x)] (b) solution for 0 x 100 FIGURE Solution curve of IVP in Example 3 By applying the initial conditions y(1) 1, y(1) 1 to the foregoing solution and using ln 1 0, we then find, in turn, that c 1 1 and c 0. Hence the solution of the initial-value problem is y x 1/ cos( ln x). The graph of this function, obtained with the aid of computer software, is given in Figure The particular solution is seen to be oscillatory and unbounded as x :. Copyright 01 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the ebook and/or echapter(s). Editorial review has

5 166 CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS The next example illustrates the solution of a third-order Cauchy-Euler equation. EXAMPLE 4 Third-Order Equation Solve x d 3 y 3 dx d y 3 5x 7x 8y 0. SOLUTION dx mxm1, The first three derivatives of y x m are d y dx m(m 1)xm, so the given differential equation becomes d 3 y dx 3 m(m 1)(m )xm3, x d 3 y 3 dx d y 3 5x 7x 8y x3 m(m 1)(m )x m3 5x m(m 1)x m 7xmx m1 8x m x m (m(m 1)(m ) 5m(m 1) 7m 8) x m (m 3 m 4m 8) x m (m )(m 4) 0. In this case we see that y x m will be a solution of the differential equation for m 1, m i, and m 3 i. Hence the general solution is y c 1 x c cos( ln x) c 3 sin( ln x). Nonhomogeneous Equations The method of undetermined coefficient described in Sections 4.5 and 4.6 does not carry over, in general, to nonhomogeneous linear differential equations with variable coefficients. Consequently, in our next example the method of variation of parameters is employed. EXAMPLE 5 Variation of Parameters Solve x y3xy3y x 4 e x. SOLUTION Since the equation is nonhomogeneous, we first solve the associated homogeneous equation. From the auxiliary equation (m 1)(m 3) 0 we fin y c c 1 x c x 3. Now before using variation of parameters to find a particular solution y p u 1 y 1 u y, recall that the formulas u 1 W 1 > W and u W > W, where W 1, W, and W are the determinants defined on page 158, were derived under the assumption that the differential equation has been put into the standard form yp(x)y Q(x)y f(x). Therefore we divide the given equation by x, and from y 3 x y 3 x y x e x we make the identification f(x) x e x. Now with y 1 x, y x 3, and W x x 3 0 x 1 3x x 3, W 1 3 x e x 3x x 5 e x, W x 0 1 x e x 3 e x, x we fin u 1 x5 e x x e x and u x 3 x3 e x e x. x 3 The integral of the last function is immediate, but in the case of u 1 we integrate by parts twice. The results are u 1 x e x xe x e x and u e x. Hence y p u 1 y 1 u y is y p (x e x xe x e x )x e x x 3 x e x xe x. Finally, y y c y p c 1 x c x 3 x e x xe x. Copyright 01 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the ebook and/or echapter(s). Editorial review has

6 4.7 CAUCHY-EULER EQUATION 167 Reduction to Constant Coefficient The similarities between the forms of solutions of Cauchy-Euler equations and solutions of linear equations with constant coefficients are not just a coincidence. For example, when the roots of the auxiliary equations for aybycy 0 and ax ybxycy 0 are distinct and real, the respective general solutions are y c 1 e m 1x c e m x and y c 1 x m 1 c x m, x 0. (5) In view of the identity e ln x x, x 0, the second solution given in (5) can be expressed in the same form as the first solution y c 1 e m 1 ln x c e m ln x c 1 e m 1t c e m t, where t ln x. This last result illustrates the fact that any Cauchy-Euler equation can always be rewritten as a linear differential equation with constant coefficients by means of the substitution x e t. The idea is to solve the new differential equation in terms of the variable t, using the methods of the previous sections, and, once the general solution is obtained, resubstitute t ln x. This method, illustrated in the last example, requires the use of the Chain Rule of differentiation. EXAMPLE 6 Changing to Constant Coefficient Solve x yxyy ln x. SOLUTION With the substitution x e t or t ln x, it follows that dx dt dt dx 1 x dt d y dx 1 d x dx dt dt 1 x ; Chain Rule 1 x d y 1 dt x dt 1 x 1 x d y dt Substituting in the given differential equation and simplifying yields d y y t. dt dt Since this last equation has constant coefficients, its auxiliary equation is m m 1 0, or (m 1) 0. Thus we obtain y c c 1 e t c te t. By undetermined coefficients we try a particular solution of the form y p A Bt. This assumption leads to B A Bt t, so A and B 1. Using y y c y p, we get y c 1 e t c te t t. ; Product Rule and Chain Rule dt. By resubstituting e t x and t ln x we see that the general solution of the original differential equation on the interval (0, ) is y c 1 x c x ln x ln x. Solutions For x < 0 In the preceding discussion we have solved Cauchy-Euler equations for x 0. One way of solving a Cauchy-Euler equation for x 0 is to change the independent variable by means of the substitution t x(which implies t 0) and using the Chain Rule: dt dx dt dx dt and d y dx d dt dt dt dx d y dt. Copyright 01 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the ebook and/or echapter(s). Editorial review has

7 168 CHAPTER 4 HIGHER-ORDER DIFFERENTIAL EQUATIONS See Problems 37 and 38 in Exercises 4.7. A Different Form A second-order equation of the form a(x x 0 ) d y dx b(x x 0) cy 0 dx is also a Cauchy-Euler equation. Observe that (6) reduces to (1) when x 0 0. We can solve (6) as we did (1), namely, seeking solutions of y (x x 0 ) m and using dx m(x x 0) m1 and d y dx m(m 1)(x x 0) m. Alternatively, we can reduce (6) to the familiar form (1) by means of the change of independent variable t x x 0, solving the reduced equation, and resubstituting. See Problems 39 4 in Exercises 4.7. (6) EXERCISES 4.7 In Problems 1 18 solve the given differential equation. 1. x yy 0. 4x yy 0 3. xyy0 4. xy3y0 5. x yxy4y 0 6. x y5xy3y 0 7. x y3xyy 0 8. x y3xy4y x y5xyy x y4xyy x y5xy4y 0 1. x y8xy6y x y6xyy x y7xy41y x 3 y6y x 3 yxyy xy (4) 6y x 4 y (4) 6x 3 y9x y3xyy 0 In Problems 19 4 solve the given differential equation by variation of parameters. 19. xy4yx 4 0. x y5xyy x x 1. x yxyy x. x yxyyx 4 e x 3. x yxyy ln x 4. x yxyy 1 x 1 In Problems 5 30 solve the given initial-value problem. Use a graphing utility to graph the solution curve. 5. x y3xy0, y(1) 0, y(1) 4 6. x y5xy8y 0, y() 3, y() 0 Answers to selected odd-numbered problems begin on page ANS x yxyy 0, y(1) 1, y(1) 8. x y3xy4y 0, y(1) 5, y(1) xyy x, y(1) 1, y(1) 1 x y5xy8y 8x 6, y 1 0, y1 0 In Problems use the substitution x e t to transform the given Cauchy-Euler equation to a differential equation with constant coefficients. Solve the original equation by solving the new equation using the procedures in Sections x y9xy0y 0 3. x y9xy5y x y10xy8y x 34. x y4xy6y ln x 35. x y3xy13y 4 3x 36. x 3 y3x y6xy6y 3 ln x 3 In Problems 37 and 38 use the substitution t xto solve the given initial-value problem on the interval (, 0) x yy 0, y(1), y(1) x y4xy6y 0, y() 8, y() 0 In Problems 39 and 40 use y (x x 0 ) m to solve the given differential equation. 39. (x 3) y8(x 1)y 14y (x 1) y(x 1)y 5y 0 Copyright 01 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the ebook and/or echapter(s). Editorial review has