such as the ratio test. However, even without knowing the formula for a n and y 4

Transcription

1 5.2 Series Solutions near an Ordinary Point, Part I 247 such as the ratio test. However, even without knowing the formula for a n we shall see in Section 5.3 that it is possible to establish that the series in Eq. 20 converge for all x, andfurtherdefinefunctionsy 3 and y 4 that are linearly independent solutions of the Airy equation 12. Thus y = a 0 y 3 x + a 1 y 4 x is the general solution of Airy s equation for < x <. It is worth emphasizing, as we saw in Example 3, that if we look for a solution of Eq. 1 of the form y = a n x x 0 n,thenthecoefficientspx, Qx, and Rx in Eq. 1 must also be expressed in powers of x x 0.Alternatively,wecanmakethe change of variable x x 0 = t,obtaininganewdifferentialequationfory as a function of t, andthenlookforsolutionsofthisnewequationoftheform a n t n.whenwe have finished the calculations, we replace t by x x 0 see Problem 19. In Examples 2 and 3 we have found two sets of solutions of Airy s equation. The functions and defined by the series in Eq. 17 are linearly independent solutions of Eq. 12 for all x, andthisisalsotrueforthefunctionsy 3 and y 4 defined by the series in Eq. 20. According to the general theory of second order linear equations each of the first two functions can be expressed as a linear combination of the latter two functions and vice versa a result that is certainly not obvious from an examination of the series alone. Finally, we emphasize that it is not particularly important if, as in Example 3, we are unable to determine the general coefficient a n in terms of a 0 and a 1.Whatisessential is that we can determine as many coefficients as we want. Thus we can find as many terms in the two series solutions as we want, even if we cannot determine the general term. While the task of calculating several coefficients in a power series solution is not difficult, it can be tedious. A symbolic manipulation package can be very helpful here; some are able to find a specified number of terms in a power series solution in response to a single command. With a suitable graphics package one can also produce plots such as those shown in the figures in this section. PROBLEMS In each of Problems 1 through 14 solve the given differential equation by means of a power series about the given point x 0.Findtherecurrencerelation;alsofindthefirstfourtermsineach of two linearly independent solutions unless the series terminates sooner. If possible, find the general term in each solution. 1. y y, x 0 2. y xy y, x 0 3. y xy y, x 0 = 1 4. y + k 2 x 2 y, x 0, k aconstant 5. 1 xy + y, x x 2 y xy + 4y, x 0 7. y + xy + 2y, x 0 8. xy + y + xy, x 0 = x 2 y 4xy + 6y, x x 2 y + 2y, x x 2 y 3xy y, x xy + xy y, x y + xy + 3y, x y + x + 1y + 3y, x 0 = 2

2 248 Chapter 5. Series Solutions of Second Order Linear Equations In each of Problems 15 through 18: a Find the first five nonzero terms in the solution of the given initial value problem. b Plot the four-term and the five-term approximations to the solution on the same axes. c From the plot in part b estimate the interval in which the four-term approximation is reasonably accurate. 15. y xy y, y0 = 2, y 0 = 1; see Problem x 2 y xy + 4y, y0 = 1, y 0 = 3; see Problem y + xy + 2y, y0 = 4, y 0 = 1; see Problem xy + xy y, y0 = 3, y 0 = 2; see Problem By making the change of variable x 1 = t and assuming that y is a power series in t, find two linearly independent series solutions of y + x 1 2 y + x 2 1y in powers of x 1. Show that you obtain the same result directly by assuming that y is a Taylor series in powers of x 1andalsoexpressingthecoefficientx 2 1inpowersof x Show directly, using the ratio test, that the two series solutions of Airy s equation about x convergeforallx;seeeq.17ofthetext. 21. The Hermite Equation. The equation y 2xy + λy, < x <, where λ is a constant, is known as the Hermite 5 equation. It is an important equation in mathematical physics. a Find the first four terms in each of two linearly independent solutions about x. b Observe that if λ is a nonnegative even integer, then one or the other of the series solutions terminates and becomes a polynomial. Find the polynomial solutions for λ, 2, 4, 6, 8, and 10. Note that each polynomial is determined only up to a multiplicative constant. c The Hermite polynomial H n x is defined as the polynomial solution of the Hermite equation with λ = 2n for which the coefficient of x n is 2 n.findh 0 x,..., H 5 x. 22. Consider the initial value problem y = 1, y0. a Show that y = sin x is the solution of this initial value problem. b Look for a solution of the initial value problem in the form of a power series about x. Find the coefficients up to the term in x 3 in this series. In each of Problems 23 through 28 plot several partial sums in a series solution of the given initial value problem about x, thereby obtaining graphs analogous to those in Figures through y xy y, y0 = 1, y 0 ; see Problem x 2 y xy + 4y, y0 = 1, y 0 ; see Problem y + xy + 2y, y0, y 0 = 1; see Problem x 2 y + 2y, y0, y 0 = 1; see Problem y + x 2 y, y0 = 1, y 0 ; see Problem xy + xy 2y, y0, y 0 = 1 5 Charles Hermite was an influential French analyst and algebraist. He introduced the Hermite functions in 1864, and showed in 1873 that e is a transcendental number that is, e is not a root of any polynomial equation with rational coefficients. His name is also associated with Hermitian matrices see Section 7.3, some of whose properties he discovered.

3 5.5 Euler Equations , and 18 to obtain real-valued solutions valid in any interval not containing the origin also see Problems 30 and 31. These results are summarized in the following theorem. Theorem The general solution of the Euler equation 1, PROBLEMS x 2 y + αxy + βy, in any interval not containing the origin is determined by the roots r 1 of the equation If the roots are real and different, then If the roots are real and equal, then If the roots are complex, then where r 1, r 2 = λ ± iµ. Fr = rr 1 + αr + β. y = c 1 x r 1 + c2 x r y = c 1 + c 2 ln x x r y = x λ [c 1 cosµ ln x + c 2 sinµ ln x ], 26 The solutions of an Euler equation of the form x x 0 2 y + αx x 0 y + βy 27 are similar to those given in Theorem If one looks for solutions of the form y = x x 0 r,thenthegeneralsolutionisgivenbyeithereq.24,25,or26with x replaced by x x 0.Alternatively,wecanreduceEq.27totheformofEq.1by making the change of independent variable t = x x 0. The situation for a general second order differential equation with a regular singular point is analogous to that for an Euler equation. We consider that problem in the next section. In each of Problems 1 through 12 determine the general solution of the given differential equation that is valid in any interval not including the singular point. 1. x 2 y + 4xy + 2y 2. x y + 3x + 1y y 3. x 2 y 3xy + 4y 4. x 2 y + 3xy + 5y 5. x 2 y xy + y 6. x 1 2 y + 8x 1y + 12y 7. x 2 y + 6xy y 8. 2x 2 y 4xy + 6y 9. x 2 y 5xy + 9y 10. x 2 2 y + 5x 2y + 8y 11. x 2 y + 2xy + 4y 12. x 2 y 4xy + 4y In each of Problems 13 through 16 find the solution of the given initial value problem. Plot the graph of the solution and describe how the solution behaves as x x 2 y + xy 3y, y1 = 1, y 1 = x 2 y + 8xy + 17y, y1 = 2, y 1 = x 2 y 3xy + 4y, y 1 = 2, y 1 = x 2 y + 3xy + 5y, y1 = 1, y 1 = 1

4 266 Chapter 5. Series Solutions of Second Order Linear Equations 17. Find all values of α for which all solutions of x 2 y + αxy + 5/2y approachzero as x Find all values of β for which all solutions of x 2 y + βy approach zero as x Find γ so that the solution of the initial value problem x 2 y 2y, y1 = 1, y 1 = γ is bounded as x Find all values of α for which all solutions of x 2 y + αxy + 5/2y approachzero as x. 21. Consider the Euler equation x 2 y + αxy + βy. Find conditions on α and β so that a All solutions approach zero as x 0. b All solutions are bounded as x 0. c All solutions approach zero as x. d All solutions are bounded as x. e All solutions are bounded both as x 0andasx. 22. Using the method of reduction of order, show that if r 1 is a repeated root of rr 1 + αr + β, then x r 1 and x r 1 ln x are solutions of x 2 y + αxy + βy forx > Transformation to a Constant Coefficient Equation. The Euler equation x 2 y + αxy + βy canbereducedtoanequationwithconstantcoefficientsbyachange of the independent variable. Let x = e z,orz = ln x,andconsideronlytheintervalx > 0. a Show that dy dx = 1 dy x dz and b Show that the Euler equation becomes d 2 y dx 2 = 1 d 2 y x 2 dz 2 1 dy x 2 dz. d 2 y dy + α 1 + βy. 2 dz dz Letting r 1 denote the roots of r 2 + α 1r + β, show that c If r 1 are real and different, then d If r 1 are real and equal, then y = c 1 e r 1 z + c 2 e r 2 z = c 1 x r 1 + c 2 x r 2. y = c 1 + c 2 ze r 1 z = c 1 + c 2 ln xx r 1. e If r 1 are complex conjugates, r 1 = λ + iµ, then y = e λz [c 1 cosµz + c 2 sinµz] = x λ [c 1 cosµ ln x + c 2 sinµ ln x]. In each of Problems 24 through 29 use the method of Problem 23 to solve the given equation for x > x 2 y 2y 25. x 2 y 3xy + 4y = ln x 26. x 2 y + 7xy + 5y = x 27. x 2 y 2xy + 2y = 3x 2 + 2lnx 28. x 2 y + xy + 4y = sinln x 29. 3x 2 y + 12xy + 9y 30. Show that if L[y] = x 2 y + αxy + βy, then L[ x r ] = x r Fr for all x < 0, where Fr = rr 1 + αr + β. Henceconcludethatifr 1 r 2 are roots of Fr, then linearly independent solutions of L[y] forx < 0are x r 1 and x r Suppose that x r 1 and x r 2 are solutions of an Euler equation for x > 0, where r 1 r 2,and r 1 is an integer. According to Eq. 24 the general solution in any interval not containing the origin is y = c 1 x r 1 + c 2 x r 2.Showthatthegeneralsolutioncanalsobewrittenas

5 Answers to Problems C H A P T E R 1 Section 1.1, page 8 1. y 3/2ast 2. y diverges from 3/2 ast 3. y diverges from 3/2 ast 4. y 1/2 ast 5. y diverges from 1/2 ast 6. y diverges from 2 ast 7. y = 3 y 8. y = 2 3y 9. y = y y = 3y y andy = 4areequilibriumsolutions;y 4ifinitialvalueispositive;y diverges from 0 if initial value is negative. 12. y andy = 5areequilibriumsolutions;y diverges from 5 if initial value is greater than 5; y 0ifinitialvalueislessthan y isequilibriumsolution;y 0ifinitialvalueisnegative;y diverges from 0 if initial value is positive. 14. y andy = 2areequilibriumsolutions;y diverges from 0 if initial value is negative; y 2ifinitialvalueisbetween0and2;y diverges from 2 if initial value is greater than a dq/dt = q10 6 b q 10 4 g; no 16. dv/dt = kv 2/3 for some k > a dq/dt = q b q 1250 mg 18. a mv = mg kv 2 b v mg/k c k = 2/ y is asymptotic to t 3ast 20. y 0ast 21. y, 0, or depending on the initial value of y 22. y or depending on the initial value of y 23. y or or y oscillates depending on the initial value of y 679

6 696 Answers to Problems Section 5.2, page a n+2 = a n /n + 2n + 1 x = 1 + x 2 2! + x 4 4! + x 6 6! + = x 2n 2n! = cosh x x = x + x 3 3! + x 5 5! + x 7 7! + = x 2n+1 2n + 1! = sinh x 2. a n+2 = a n /n + 2 x = 1 + x x x = x 2n 2 n n! x = x + x x x = 2 n n!x 2n+1 2n + 1! 3. n + 2a n+2 a n+1 a n x = x x x 14 + x = x x x x a n+4 = k 2 a n /n + 4n + 3; a 2 = a 3 x = 1 k2 x k4 x k 6 x m+1 k 2 x 4 m+1 = m + 34m + 4 m=0 x = x k2 x k4 x k 6 x [ ] 1 m+1 k 2 x 4 m+1 = x m + 44m + 5 m=0 Hint: Let n = 4m in the recurrence relation, m = 1, 2, 3, n + 2n + 1a n+2 nn + 1a n+1 + a n, n 1; a 2 = 1 2 a 0 x = x x x 4 +, x = x 1 6 x x x a n+2 = n 2 2n + 4a n /[2n + 1n + 2], n 2; a 2 = a 0, a 3 = 1 4 a 1 x = 1 x x x 6 +, x = x 1 4 x x x a n+2 = a n /n + 1, n, 1, 2,... x = 1 x x x =1 + 1 n x 2n n 1 x = x x 3 n=1 2 + x x =x + n=1 1 n x 2n n 8. a n+2 = [n a n+1 + a n + a n 1 ]/n + 1n + 2, n = 1, 2,... a 2 = a 0 + a 1 /2 x = x x x 14 + x = x x x x n + 2n + 1a n+2 + n 2n 3a n x = 1 3x 2, x = x x 3 / n + 2a n+2 n 2a n x = 1 x 2 4, x = x x 3 12 x x x 2n+1 4 n 2n 12n + 1

7 Answers to Problems n + 2a n+2 n + 1a n x = 1 + x x x n n 2 4 2n x 2n + x = x x x x n n 3 5 2n + 1 x 2n n + 2n + 1a n+2 n + 1na n+1 + n 1a n x = 1 + x x x x n n! +, x = x 13. 2n + 2n + 1a n+2 + n + 3a n x = x x x 6 n 3 5 2n n x 2n + 2n! x = x x x 5 20 x n n n x 2n+1 + 2n + 1! 14. 2n + 2n + 1a n+2 + 3n + 1a n+1 + n + 3a n x = x x x 24 + x = x x x x a y = 2 + x + x x x 4 + c about x < a y = 1 + 3x + x x x 4 + c about x < a y = 4 x 4x x x 4 + c about x < a y = 3 + 2x 3 2 x x x 4 + c about x < x = x x x 16 + x = x x x x a x = 1 λ 2! x 2 λλ 4 + x 4 λλ 4λ 8 x 6 + 4! 6! x = x λ 2 x 3 λ 2λ 6 + x 5 λ 2λ 6λ 10 x 7 + 3! 5! 7! b 1, x, 1 2x 2, x 2 3 x 3, 1 4x x 4, x 4 3 x x 5 c 1, 2x, 4x 2 2, 8x 3 12x, 16x 4 48x , 32x 5 160x x 22. b y = x x 3 /6 + Section 5.3, page φ 0 = 1, φ 0, φ iv 0 = 3 2. φ 0, φ 0 = 2, φ iv 0 3. φ 1, φ 1 = 6, φ iv 1 = φ 0, φ 0 = a 0, φ iv 0 = 4a 1 5. ρ =, ρ = 6. ρ = 1, ρ = 3, ρ = 1 7. ρ = 1, ρ = 3 8. ρ = 1 9. a ρ = b ρ = c ρ = d ρ = e ρ = 1 f ρ = 2 gρ = h ρ = 1 iρ = 1 jρ = 2 k ρ = 3 l ρ = 1 m ρ = n ρ =

8 698 Answers to Problems 10. a x = 1 α2 2! x 2 22 α 2 α 2 x 4 42 α α 2 α 2 x 6 4! 6! [2m 22 α 2 ] 2 2 α 2 α 2 x 2m 2m! x = x + 1 α2 3! x α 2 1 α 2 x 5 + 5! + [2m 12 α 2 ] 1 α 2 x 2m+1 + 2m + 1! b x or x terminates with x n as α = n is even or odd. c n, y = 1; n = 1, y = x; n = 2, y = 1 2x 2 ; n = 3, y = x 4 3 x x = x x x 6 +, ρ = x = x 1 12 x x x 7 +, 12. x = x x x 5 +, ρ = x = x 1 12 x x x 6 +, 13. x = 1 + x x x 6 +, ρ = π/2 x = x x x x 7 +, 14. x = x x x 6 +, ρ = 1 x = x 1 6 x x x 5 +, 15. Cannot specify arbitrary initial conditions at x ; hence x isasingularpoint. 16. y = 1 + x + x 2 2! + + x n n! + =ex 17. y = 1 + x x x 6 2n x n n! y = 1 + x x x y = 1 + x + x 2 + +x n + = 1 1 x 20. y = a x + x 2 2! + + x n n! + x ! + x 4 4! + + x n n! + = a 0 e x + 2 e x 1 x x 2 2 = ce x 2 2x x y = a x x ! x ! + + 1n x 2n 2 n n! x + x 2 2 x 3 = a 0 e x2 /2 + 3 x x x + x 2 2 x 3 3 x x , 1 3x 2, 1 10x x 4 ; x, x 5 3 x 3, x 14 3 x x a 1, x, 3x 2 1/2, 5x 3 3x/2, 35x 4 30x 2 + 3/8, 63x 5 70x x/8 c P 1, 0; P 2, ± ; P 3, 0, ± ; P 4, ± , ± ; P 5, 0, ± , ±

9 Answers to Problems 699 Section 5.4, page x, regular 2. x, regular; x = 1, irregular 3. x, irregular; x = 1, regular 4. x, irregular; x =±1, regular 5. x = 1, regular; x = 1, irregular 6. x, regular 7. x = 3, regular 8. x, 1, regular; x = 1, irregular 9. x = 1, regular; x = 2, irregular 10. x, 3, regular 11. x = 1, 2, regular 12. x, regular 13. x, irregular 14. x, regular 15. x, regular 16. x, ±nπ, regular 17. x, ±nπ, regular 18. x, irregular; x =±nπ, regular 19. y = a 0 1 x x Irregular singular point 23. Regular singular point 24. Regular singular point 25. Irregular singular point 26. Irregular singular point 27. Irregular singular point Section 5.5, page y = c 1 x 1 + c 2 x 2 2. y = c 1 x + 1 1/2 + c 2 x + 1 3/2 3. y = c 1 x 2 + c 2 x 2 ln x 4. y = c 1 x 1 cos2ln x + c 2 x 1 sin2ln x 5. y = c 1 x + c 2 x ln x 6. y = c 1 x c 2 x y = c 1 x 5+ 29/2 + c 2 x 5 29/2 8. y = c 1 x 3/2 cos 1 2 3ln x + c2 x 3/2 sin 1 2 3ln x 9. y = c 1 x 3 + c 2 x 3 ln x y = c 1 x 2 2 cos2ln x 2 + c 2 x 2 2 sin2ln x 2 y = c 1 x 1/2 cos ln x + c2 x 1/2 sin ln x 12. y = c 1 x + c 2 x y = 2x 3/2 x y = 2x 1/2 cos2lnx x 1/2 sin2lnx 15. y = 2x 2 7x 2 ln x 16. y = x 1 cos2lnx 17. α<1 18. β>0 19. γ = α>1 21. a α<1andβ>0 b α<1andβ 0, or α = 1andβ>0 c α>1andβ>0 d α>1andβ 0, or α = 1andβ>0 e α = 1andβ>0 24. y = c 1 x 1 + c 2 x y = c 1 x 2 + c 2 x 2 ln x ln x y = c 1 x 1 + c 2 x x 27. y = c 1 x + c 2 x 2 + 3x 2 ln x + ln x y = c 1 cos2lnx + c 2 sin2lnx + 1 sinln x y = c 1 x 3/2 cos 3 2 ln x + c 2 x 3/2 sin 3 ln x x > 0: c 1 = k 1, c 2 = k 2 ; x < 0: c 1 1 r 1 = k 1, c 2 = k 2