TESTETESTEINDEXAÇÃ0 1726751313012 Lista de exercÃ_cios 5 pdf

Lista de exercicios 5 — Teoria de Frobenius (do livro de Boyce e Di Prima) Problems In each of Problems | through 8: 6. x2(1—x)y”-(1+x)y’ +2xy =0 a. Find all the regular singular points of the given differential 7. (x —2)*(x +2) y" + 2xy’ + 3(x —2)y =0 equation. 8. (4—x?)y” + Ixy’ +3y =0 b. Determine the indicial equation and the exponents at the Jy each of Problems 9 through 12: singularity for each regular singular point. a. Show that x = 0 is a regular singular point of the given 1. xy” + 2xy' + 6e*y =0 differential equation. 2. x?y"”—x(24+x)y +(24+x2)y =0 b. Find the exponents at the singular point x = 0. 3. y"+4xy’ +6y =0 c. Find the first three nonzero terms in each of two solutions 4, 2x(x +2)y"+y’—xy =0 (not multiples of each other) about x = 0. 1 " ye = 5. x?y"4+ <(x4+sinx)y’ +y =0 9. xy"+y'—y=0 2 10. xy’ +2xy'’+6e*y =0 (see Problem 1) ll. xy’+y=0 15. Consider the differential equation 12. x*y” +(sinx) y’ — (cosx)y =0 xy" baxy’ + By =0, 13. a. Show that where qa and @ are real constants anda 4 0. (nx) y” + ly +y=0 a. Show that x = 0 is an irregular singular point. 2 oo b. By attempting to determine a solution of the form S* a,x"*", has a regular singular point at x = 1. n=0 b. Determine the roots of the indicial equation at x = 1. show that the indicial equation for r is linear and _ that, c. Determine the first three nonzero terms in the series consequently, there is only one formal solution of the assumed Co 1 a,(x — 1)'*" corresponding to the larger root. form. . . na c. Show that if 3/a = —1,0,1,2, ..., then the formal series You can assume x — 1 > 0. solution terminates and therefore is an actual solution. For other d. What would you expect the radius of convergence of the values of /a, show that the formal series solution has a zero series to be? radius of convergence and so does not represent an actual solution 14. In several problems in mathematical physics, it is necessary to in any interval. study the differential equation 16. Consider the differential equation _ " _ / _ _ ” Qa , x1 —ax)y"+(y-U+a+8)x)y'—asy=0, (25) y +8 y+ Fy =o, (26) where a, 3, and y are constants. This equation is known as the hypergeometric equation. where a 4 O and @ # O are real numbers, and s and f¢ are positive a. Show that x = 0 is a regular singular point and that the roots _integers that for the moment are arbitrary. of the indicial equation are 0 and 1 — y. a. Show thatifs > lorft > 2, then the point x = 0 is an b. Show that x = 1 is a regular singular point and that the roots irregular singular point. of the indicial equation are 0 and y — a — 8. b. Try to find a solution of equation (26) of the form c. Assuming that 1 — + is not a positive integer, show that, in 00 the neighborhood of x = 0, one solution of equation (25) is y= S> anx'*", x >0. (27) a a(a+l1 B41 n=0 yay = 1 yp SEDO oy ’ youu yy + D2! Show that if s = 2 andt = 2, then there is only one possible What would you expect the radius of convergence of this series value of r for which there is a formal solution of equation (26) of to be? the form (27). d. Assuming that 1 — 7 is not an integer or zero, show that a c. Show that if s = 1 and +t = 3, then there are no solutions of second solution for 0 < x < lis equation (26) of the form (27). d. Show that the maximum values of s and ¢ for which the yo(x) =x! (: +4 (a-7y+tDG-7+D x+ indicial equation is quadratic in r [and hence we can hope to find (2—y)1! two solutions of the form (27)] are s = 1 and t = 2. These are (a-y+)D(a-7+2)(6-y+D(6-7y4+2) » precisely the conditions that distinguish a “weak singularity,” or (2—7)(3—7)2! wore a regular singular point, from an irregular singular point, as we defined them in Section 5.4. €. Show that the point at infinity is a regular singular point and As a note of caution, we point out that although it is sometimes that the roots of the indicial equation are a and 3. See Problem : : : : : possible to obtain a formal series solution of the form (27) at an 32 of Section 5.4. : : : : ws : irregular singular point, the series may not have a positive radius of convergence. See Problem 15 for an example.