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# Exercícios resolvidos Equações diferenciais elementares 9 ed Boyce e DiPrima

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point. Note that p0 = q0 = 0 . (b) The indicial equation is given by r(r − 1) = 0 , that is, r2 − r = 0 , with roots r1 = 1 and r2 = 0 . (c) In order to find the solution corresponding to r1 = 1 , set y = x ∑∞ n=0 anx n. Upon substitution into the ODE, we have ∞∑ n=0 (n+ 2)(n+ 1)an+1 xn+1 + 2 ∞∑ n=0 (n+ 1)anxn+1 + 6 ex ∞∑ n=0 anx n+1 = 0 . After adjusting the indices in the first two series, and expanding the exponential function, ∞∑ n=1 n(n+ 1)an xn + 2 ∞∑ n=1 nan−1xn + 6 a0x+ (6a0 + 6a1)x2+ + (6a2 + 6a1 + 3a0)x3 + (6a3 + 6a2 + 3a1 + a0)x4 + . . . = 0 . Equating the coefficients, we obtain the system of equations 2a1 + 2a0 + 6a0 = 0 6a2 + 4a1 + 6a0 + 6a1 = 0 12a3 + 6a2 + 6a2 + 6a1 + 3a0 = 0 20a4 + 8a3 + 6a3 + 6a2 + 3a1 + a0 = 0 ... Setting a0 = 1 , solution of the system results in a1 = −4, a2 = 17/3, a3 = −47/12, a4 = 191/120 , . . . . Therefore one solution is y1(x) = x− 4x2 + 173 x 3 − 47 12 x4 + . . . . 5.6 213 The exponents differ by an integer. So for a second solution, set y2(x) = a y1(x) ln x + 1 + c1x+ c2x2 + . . .+ cnxn + . . . . Substituting into the ODE, we obtain aL [y1(x)] · ln x+ 2a y ′1(x) + 2a y1(x)− a y1(x) x + L [ 1 + ∞∑ n=1 cn x n ] = 0 . Since L [y1(x)] = 0 , it follows that L [ 1 + ∞∑ n=1 cn x n ] = −2a y ′1(x)− 2a y1(x) + a y1(x) x . More specifically, ∞∑ n=1 n(n+ 1)cn+1xn + 2 ∞∑ n=1 n cnx n + 6 + (6 + 6c1)x+ + (6c2 + 6c1 + 3)x2 + . . . = −a+ 10ax− 613 ax 2 + 193 12 ax3 + . . . . Equating the coefficients, we obtain the system of equations 6 = −a 2c2 + 8c1 + 6 = 10a 6c3 + 10c2 + 6c1 + 3 = −613 a 12c4 + 12c3 + 6c2 + 3c1 + 1 = 193 12 a ... Solving these equations for the coefficients, a = −6 . In order to solve the remaining equations, set c1 = 0 . Then c2 = −33, c3 = 449/6 , c4 = −1595/24 , . . . . Therefore a second solution is y2(x) = −6 y1(x) ln x+ [ 1− 33x2 + 449 6 x3 − 1595 24 x4 + . . . ] . 15.(a) Note the p(x) = 6x/(x− 1) and q(x) = 3x−1(x− 1)−1 . Furthermore, x p(x) = 6x2/(x− 1) and x2q(x) = 3x/(x− 1) . It follows that p0 = lim x→0 6x2 x− 1 = 0 q0 = lim x→0 3x x− 1 = 0 and therefore x = 0 is a regular singular point. (b) The indicial equation is given by r(r − 1) = 0 , that is, r2 − r = 0 , with roots r1 = 1 and r2 = 0 . 214 Chapter 5. Series Solutions of Second Order Linear Equations (c) In order to find the solution corresponding to r1 = 1 , set y = x ∑∞ n=0 anx n. Upon substitution into the ODE, we have ∞∑ n=1 n(n+ 1)an xn+1 − ∞∑ n=1 n(n+ 1)an xn+ + 6 ∞∑ n=0 (n+ 1)anxn+2 + 3 ∞∑ n=0 anx n+1 = 0 . After adjusting the indices, it follows that ∞∑ n=2 n(n− 1)an−1 xn − ∞∑ n=1 n(n+ 1)an xn+ + 6 ∞∑ n=2 (n− 1)an−2xn + 3 ∞∑ n=1 an−1xn = 0 . That is, −2a1 + 3a0 + ∞∑ n=2 [−n(n+ 1)an + (n2 − n+ 3)an−1 + 6(n− 1)an−2]xn = 0. Setting the coefficients equal to zero, we have a1 = 3a0/2 , and for n ≥ 2 , n(n+ 1)an = (n2 − n+ 3)an−1 + 6(n− 1)an−2 . If we assign a0 = 1 , then we obtain a1 = 3/2 , a2 = 9/4 , a3 = 51/16 , . . . . Hence one solution is y1(x) = x+ 3 2 x2 + 9 4 x3 + 51 16 x4 + 111 40 x5 + . . . . The exponents differ by an integer. So for a second solution, set y2(x) = a y1(x) ln x + 1 + c1x+ c2x2 + . . .+ cnxn + . . . . Substituting into the ODE, we obtain 2ax y ′1(x)− 2a y ′1(x) + 6ax y1(x)− a y1(x) + a y1(x) x + L [ 1 + ∞∑ n=1 cn x n ] = 0, since L [y1(x)] = 0 . It follows that L [ 1 + ∞∑ n=1 cn x n ] = 2a y ′1(x)− 2ax y ′1(x) + a y1(x)− 6ax y1(x)− a y1(x) x . Now L [ 1 + ∞∑ n=1 cn x n ] = 3 + (−2c2 + 3c1)x+ (−6c3 + 5c2 + 6c1)x2+ + (−12c4 + 9c3 + 12c2)x3 + (−20c5 + 15c4 + 18c3)x4 + . . . . Substituting for y1(x) , the right hand side of the ODE is a+ 7 2 ax+ 3 4 ax2 + 33 16 ax3 − 867 80 ax4 − 441 10 ax5 + . . . . 5.6 215 Equating the coefficients, we obtain the system of equations 3 = a −2c2 + 3c1 = 72a −6c3 + 5c2 + 6c1 = 34a −12c4 + 9c3 + 12c2 = 3316a ... We find that a = 3. In order to solve the second equation, set c1 = 0. Solution of the remaining equations results in c2 = −21/4 , c3 = −19/4 , c4 = −597/64 , . . .. Hence a second solution is y2(x) = 3 y1(x) ln x+ [ 1− 21 4 x2 − 19 4 x3 − 597 64 x4 + . . . ] . 16.(a) After multiplying both sides of the ODE by x , we find that x p(x) = 0 and x2q(x) = x . Both of these functions are analytic at x = 0 , hence x = 0 is a regular singular point. (b) Furthermore, p0 = q0 = 0 . So the indicial equation is r(r − 1) = 0 , with roots r1 = 1 and r2 = 0 . (c) In order to find the solution corresponding to r1 = 1 , set y = x ∑∞ n=0 anx n. Upon substitution into the ODE, we have ∞∑ n=1 n(n+ 1)an xn + ∞∑ n=0 anx n+1 = 0 . That is, ∞∑ n=1 [n(n+ 1)an + an−1] xn = 0 . Setting the coefficients equal to zero, we find that for n ≥ 1 , an = −an−1 n(n+ 1) . It follows that an = −an−1 n(n+ 1) = an−2 (n− 1)n2(n+ 1) = . . . = (−1)na0 (n!)2(n+ 1) . Hence one solution is y1(x) = x− 12x 2 + 1 12 x3 − 1 144 x4 + 1 2880 x5 + . . . . The exponents differ by an integer. So for a second solution, set y2(x) = a y1(x) ln x + 1 + c1x+ c2x2 + . . .+ cnxn + . . . . 216 Chapter 5. Series Solutions of Second Order Linear Equations Substituting into the ODE, we obtain aL [y1(x)] · ln x+ 2a y ′1(x)− a y1(x) x + L [ 1 + ∞∑ n=1 cn x n ] = 0 . Since L [y1(x)] = 0 , it follows that L [ 1 + ∞∑ n=1 cn x n ] = −2a y ′1(x) + a y1(x) x . Now L [ 1 + ∞∑ n=1 cn x n ] = 1 + (2c2 + c1)x+ (6c3 + c2)x2 + (12c4 + c3)x3+ + (20c5 + c4)x4 + (30c6 + c5)x5 + . . . . Substituting for y1(x) , the right hand side of the ODE is −a+ 3 2 ax− 5 12 ax2 + 7 144 ax3 − 1 320 ax4 + . . . . Equating the coefficients, we obtain the system of equations 1 = −a 2c2 + c1 = 3 2 a 6c3 + c2 = − 512a 12c4 + c3 = 7 144 a ... Evidently, a = −1 . In order to solve the second equation, set c1 = 0 . We then find that c2 = −3/4 , c3 = 7/36 , c4 = −35/1728 , . . . . Therefore a second solution is y2(x) = −y1(x) ln x+ [ 1− 3 4 x2 + 7 36 x3 − 35 1728 x4 + . . . ] . 19.(a) After dividing by the leading coefficient, we find that p0 = lim x→0 x p(x) = lim x→0 γ − (1 + α+ β)x 1− x = γ . q0 = lim x→0 x2q(x) = lim x→0 −αβ x 1− x = 0 . Hence x = 0 is a regular singular point. The indicial equation is r(r − 1) + γ r = 0 , with roots r1 = 1− γ and r2 = 0 . (b) For x = 1, p0 = lim x→1 (x− 1)p(x) = lim x→1 −γ + (1 + α+ β)x x = 1− γ + α+ β . 5.6 217 q0 = lim x→1 (x− 1)2q(x) = lim x→1 αβ(x− 1) x = 0 . Hence x = 1 is a regular singular point. The indicial equation is r2 − (γ − α− β) r = 0 , with roots r1 = γ − α− β and r2 = 0 . (c) Given that r1 − r2 is not a positive integer, we can set y = ∑∞ n=0 anx n. Sub- stitution into the ODE results in x(1− x) ∞∑ n=2 n(n− 1)anxn−2 + [γ − (1 + α+ β)x] ∞∑ n=1 nanx n−1 − αβ ∞∑ n=0 anx n = 0. That is, ∞∑ n=1 n(n+ 1)an+1xn − ∞∑ n=2 n(n− 1)anxn + γ ∞∑ n=0 (n+ 1)an+1xn− − (1 + α+ β) ∞∑ n=1 nanx n − αβ ∞∑ n=0 anx n = 0. Combining the series, we obtain γ a1 − αβ a0 + [(2 + 2γ)a2 − (1 + α+ β + αβ)a1]x+ ∞∑ n=2 An x n = 0 , in which An = (n+ 1)(n+ γ)an+1 − [n(n− 1) + (1 + α+ β)n+ αβ] an . Note that n(n− 1) + (1 + α+ β)n+ αβ = (n+ α)(n+ β) . Setting the coefficients equal to zero, we have γ a1 − αβ a0 = 0 , and an+1 = (n+ α)(n+ β) (n+ 1)(n+ γ) an for n ≥ 1 . Hence one solution is y1(x) = 1 + αβ γ ·