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# Exercícios resolvidos Equações diferenciais elementares 9 ed Boyce e DiPrima

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2n(2n+ 1)! (x− 1)n) y2(x) = 1 + ∞∑ n=1 (−1)n α(−α)(1− α 2)(4− α2) . . . ((n− 1)2 − α2) n! · 3 · 5 . . . (2n− 1) (x− 1) n . 13.(a) Here xp(x) = 1− x and x2q(x) = λx, which are both analytic at x = 0. In fact, p0 = lim x→0 x p(x) = 1 and q0 = lim x→0 x2q(x) = 0 . (b) The indicial equation is r(r − 1) + r = 0 , with roots r1,2 = 0 . (c) Set y = a0 + a1x+ a2x2 + . . .+ anxn + . . . . Substitution into the ODE results in ∞∑ n=2 n(n− 1)an xn−1 + ∞∑ n=1 nan x n−1− − ∞∑ n=0 nan x n + λ ∞∑ n=0 anx n = 0 . 5.6 207 That is, ∞∑ n=1 n(n+ 1)an+1 xn + ∞∑ n=0 (n+ 1)an+1 xn− − ∞∑ n=1 nan x n + λ ∞∑ n=0 anx n = 0 . It follows that a1 + λa0 + ∞∑ n=1 [ (n+ 1)2an+1 − (n− λ)an ] xn = 0 . Setting the coefficients equal to zero, we find that a1 = −λa0 , and an = (n− 1− λ) n2 an−1 , n = 2, 3, . . . . That is, for n ≥ 2 , an = (n− 1− λ) n2 an−1 = . . . = (−λ)(1− λ) . . . (n− 1− λ) (n!)2 a0 . Therefore one solution of the Laguerre equation is y1(x) = 1 + ∞∑ n=1 (−λ)(1− λ) . . . (n− 1− λ) (n!)2 xn. Note that if λ = m , a positive integer, then an = 0 for n ≥ m+ 1 . In that case, the solution is a polynomial y1(x) = 1 + m∑ n=1 (−λ)(1− λ) . . . (n− 1− λ) (n!)2 xn . 5.6 2.(a) P (x) = 0 only for x = 0. Furthermore, xp(x) = −2− x and x2q(x) = 2 + x2. It follows that p0 = lim x→0 (−2− x) = −2 q0 = lim x→0 (2 + x2) = 2 and therefore x = 0 is a regular singular point. (b) The indicial equation is given by r(r − 1)− 2r + 2 = 0 , that is, r2 − 3r + 2 = 0 , with roots r1 = 2 and r2 = 1 . 4. The coefficients P (x) , Q(x) , and R(x) are analytic for all x ∈ R . Hence there are no singular points. 208 Chapter 5. Series Solutions of Second Order Linear Equations 5.(a) P (x) = 0 only for x = 0. Furthermore, xp(x) = 3 sin xx and x 2q(x) = −2. It follows that p0 = lim x→0 3 sin x x = 3 q0 = lim x→0 −2 = −2 and therefore x = 0 is a regular singular point. (b) The indicial equation is given by r(r − 1) + 3r − 2 = 0 , that is, r2 + 2r − 2 = 0 , with roots r1 = −1 + √ 3 and r2 = −1− √ 3 . 6.(a) P (x) = 0 for x = 0 and x = −2. We note that p(x) = x−1(x+ 2)−1/2, and q(x) = −(x+ 2)−1/2. For the singularity at x = 0, p0 = lim x→0 1 2(x+ 2) = 1 4 q0 = lim x→0 −x2 2(x+ 2) = 0 and therefore x = 0 is a regular singular point. For the singularity at x = −2, p0 = lim x→−2 (x+ 2)p(x) = lim x→−2 1 2x = −1 4 q0 = lim x→−2 (x+ 2)2q(x) = lim x→−2 −(x+ 2) 2 = 0 and therefore x = −2 is a regular singular point. (b) For x = 0: the indicial equation is given by r(r − 1) + 1 4 r = 0 , that is, r2 − 34r = 0 , with roots r1 = 34 and r2 = 0 . For x = −2: the indicial equation is given by r(r − 1)− 1 4 r = 0 , that is, r2 − 54r = 0 , with roots r1 = 54 and r2 = 0 . 7.(a) P (x) = 0 only for x = 0. Furthermore, xp(x) = 12 + sin x 2x and x 2q(x) = 1. It follows that p0 = lim x→0 xp(x) = 1 q0 = lim x→0 x2q(x) = 1 and therefore x = 0 is a regular singular point. 5.6 209 (b) The indicial equation is given by r(r − 1) + r + 1 = 0 , that is, r2 + 1 = 0 , with complex conjugate roots r = ± i . 8.(a) Note that P (x) = 0 only for x = −1. We find that p(x) = 3(x− 1)/(x+ 1), and q(x) = 3/(x+ 1)2. It follows that p0 = lim x→−1 (x+ 1)p(x) = lim x→−1 3(x− 1) = −6 q0 = lim x→−1 (x+ 1)2q(x) = lim x→−1 3 = 3 and therefore x = −1 is a regular singular point. (b) The indicial equation is given by r(r − 1)− 6r + 3 = 0, that is, r2 − 7r + 3 = 0 , with roots r1 = (7 + √ 37 )/2 and r2 = (7− √ 37 )/2 . 10.(a) P (x) = 0 for x = 2 and x = −2. We note that p(x) = 2x(x− 2)−2(x+ 2)−1, and q(x) = 3(x− 2)−1(x+ 2)−1. For the singularity at x = 2, lim x→2 (x− 2)p(x) = lim x→2 2x x2 − 4 , which is undefined. Therefore x = 0 is an irregular singular point. For the singu- larity at x = −2, p0 = lim x→−2 (x+ 2)p(x) = lim x→−2 2x (x− 2)2 = − 1 4 q0 = lim x→−2 (x+ 2)2q(x) = lim x→−2 3(x+ 2) x− 2 = 0 and therefore x = −2 is a regular singular point. (b) The indicial equation is given by r(r − 1)− 1 4 r = 0 , that is, r2 − 54r = 0 , with roots r1 = 54 and r2 = 0 . 11.(a) P (x) = 0 for x = 2 and x = −2. We note that p(x) = 2x/(4− x2), and q(x) = 3/(4− x2). For the singularity at x = 2, p0 = lim x→2 (x− 2)p(x) = lim x→2 −2x x+ 2 = −1 q0 = lim x→2 (x− 2)2q(x) = lim x→2 3(2− x) x+ 2 = 0 and therefore x = 2 is a regular singular point. 210 Chapter 5. Series Solutions of Second Order Linear Equations For the singularity at x = −2 , p0 = lim x→−2 (x+ 2)p(x) = lim x→−2 2x 2− x = −1 q0 = lim x→−2 (x+ 2)2q(x) = lim x→−2 3(x+ 2) 2− x = 0 and therefore x = −2 is a regular singular point. (b) For x = 2: the indicial equation is given by r(r − 1)− r = 0 , that is, r2 − 2r = 0 , with roots r1 = 2 and r2 = 0 . For x = −2: the indicial equation is given by r(r − 1)− r = 0 , that is, r2 − 2r = 0 , with roots r1 = 2 and r2 = 0 . 12.(a) P (x) = 0 for x = 0 and x = −3. We note that p(x) = −2x−1(x+ 3)−1, and q(x) = −1/(x+ 3)2. For the singularity at x = 0, p0 = lim x→0 x p(x) = lim x→0 −2 x+ 3 = −2 3 q0 = lim x→0 x2q(x) = lim x→0 −x2 (x+ 3)2 = 0 and therefore x = 0 is a regular singular point. For the singularity at x = −3, p0 = lim x→−3 (x+ 3)p(x) = lim x→−3 −2 x = 2 3 q0 = lim x→−3 (x+ 3)2q(x) = lim x→−3 (−1) = −1 and therefore x = −3 is a regular singular point. (b) For x = 0: the indicial equation is given by r(r − 1)− 2 3 r = 0 , that is, r2 − 53r = 0 , with roots r1 = 53 and r2 = 0 . For x = −3: the indicial equation is given by r(r − 1) + 2 3 r − 1 = 0 , that is, r2 − 13r − 1 = 0 , with roots r1 = (1 + √ 37 )/6 and r2 = (1− √ 37 )/6 . 13.(a) Note that p(x) = 1/x and q(x) = −1/x. Furthermore, xp(x) = 1 and x2q(x) = −x. It follows that p0 = lim x→0 (1) = 1 q0 = lim x→0 (−x) = 0 5.6 211 and therefore x = 0 is a regular singular point. (b) The indicial equation is given by r(r − 1) + r = 0 , that is, r2 = 0 , with roots r1 = r2 = 0 . (c) Let y = a0 + a1x+ a2x2 + . . .+ anxn + . . . . Substitution into the ODE results in ∞∑ n=0 (n+ 2)(n+ 1)an+2 xn+1 + ∞∑ n=0 (n+ 1)an+1xn − ∞∑ n=0 anx n = 0 . After adjusting the indices in the first series, we obtain a1 − a0 + ∞∑ n=1 [n(n+ 1)an+1 + (n+ 1)an+1 − an]xn = 0 . Setting the coefficients equal to zero, it follows that for n ≥ 0 , an+1 = an (n+ 1)2 . So for n ≥ 1 , an = an−1 n2 = an−2 n2(n− 1)2 = . . . = 1 (n!)2 a0 . With a0 = 1 , one solution is y1(x) = 1 + x+ 1 4 x2 + 1 36 x3 + . . .+ 1 (n!)2 xn + . . . . For a second solution, set y2(x) = y1(x) ln x + b1x+ b2x2 + . . .+ bnxn + . . . . Sub- stituting into the ODE, we obtain L [y1(x)] · ln x+ 2 y ′1(x) + L [ ∞∑ n=1 bn x n ] = 0 . Since L [y1(x)] = 0 , it follows that L [ ∞∑ n=1 bn x n ] = −2 y ′1(x) . More specifically, b1 + ∞∑ n=1 [n(n+ 1)bn+1 + (n+ 1)bn+1 − bn]xn = = −2− x− 1 6 x2 − 1 72 x3 − 1 1440 x4 − . . . . 212 Chapter 5. Series Solutions of Second Order Linear Equations Equating the coefficients, we obtain the system of equations b1 = −2 4b2 − b1 = −1 9b3 − b2 = −1/6 16b4 − b3 = −1/72 ... Solving these equations for the coefficients, b1 = −2, b2 = −3/4, b3 = −11/108, b4 = −25/3456 , . . . . Therefore a second solution is y2(x) = y1(x) ln x+ [ −2x− 3 4 x2 − 11 108 x3 − 25 3456 x4 − . . . ] . 14.(a) Here x p(x) = 2x and x2q(x) = 6xex . Both of these functions are analytic at x = 0 , therefore x = 0 is a regular singular