Exercícios - Pvi e Edos - Cálculo 4 - 20232

Cálculo 4 - Lista 1 Dy/dx + 4y = 1 senda y = \sum_{n=0}^{\infty} c_n x^n \rightarrow dy/dx = \sum_{n=1}^{\infty} c_n n x^{n-1} entao dy/dx + 4y = \sum_{n=1}^{\infty} c_n n x^{n-1} + 4 \sum_{n=0}^{\infty} c_n x^n = 0 \updownarrow \sum_{k=0}^{\infty} c_{k+1} (k+1) x^k + 4 \sum_{k=0}^{\infty} c_k x^k = 0 \sum_{k=0}^{\infty} [c_{k+1} (k+1) + 4 c_k] x^k = 0 logo, c_{k+1} (k+1) + 4c_k = 0 \rightarrow c_{k+1} = \frac{-4}{k+1} c_k k=0 \rightarrow c_1 = - 4c_0 k=1 \rightarrow c_2 = \frac{-1}{2} c_1 = \frac{1}{2} \cdot 4c_0 k=2 \rightarrow c_3 = \frac{-1}{3} c_2 = \frac{-1}{3} \cdot \frac{1}{2} \cdot 4c_0 k=3 \rightarrow c_4 = \frac{-1}{4} c_3 = \frac{4}{4} \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot 4c_0 com sigo c_k = \frac{(-1)^k (4)^k c_0}{k! \cdot k!} logo, y = A + \sum_{k=1}^{\infty} B \frac{(-1)^k (4) x^k}{k!} = A + B \sum_{k=1}^{\infty} \frac{(-1)^k (4x)^k}{k!} y = A + B e^{-4x} \rightarrow y' = -4B e^{-4x} entao y' + 4y = 1 \rightarrow -4B e^{-4x} + 4A + 4B e^{-4x} = 1 \rightarrow 4A = 1 \rightarrow A = \frac{1}{4} Portanto, y(x) = \frac{1}{4} + B e^{-4x} fazendo y(0) = 1,25 = \frac{5}{4} \rightarrow \frac{5}{4} = \frac{1}{4} + B \rightarrow B = 1 \rightarrow y(x) = \frac{1}{4} + e^{-4x} por fim, |y(0,2) = 0,7| (2) y'' + 3x y' + 2y = 0 senda y = \sum_{n=0}^{\infty} c_n x^n \rightarrow y' = \sum_{n=1}^{\infty} c_n n x^{n-1} \rightarrow y'' = \sum_{n=2}^{\infty} c_n n(n-1) x^{n-2} entao, \{ sum } y'' + 3x y' + 2y = \sum_{n=2}^{\infty} c_n n(n-1) x^{n-2} + 3x \sum_{n=1}^{\infty} c_n n x^{n-1} \} \sum_{n=0}^{\infty} c_n x^n = \sum_{n=2}^{\infty} c_n n(n-1) x^{n-2} + 3 \sum_{n=1}^{\infty} c_n n x^{n-1} + 2 \sum_{n=0}^{\infty} c_n x^n = 0 \updownarrow k = n-2 (3) (1 - x^2) y'' - 2 x y' + 3 0 y = 0 Supondo y = \sum_{n=0}^{\infty} c_n x^{n + a} - D y' = \sum_{n=0}^{\infty} c_n(n+a)y^{n+a} -D y'' = \sum_{n=0}^{\infty} c_n(n+a)(n+a-1)x^{n+a-2} (1 - x^2) y'' - 2 x y' + 3 0 y = (1 - x^2)\sum_{n=0}^{\infty} c_n(n+a)(n+a-1)x^{n+a-2} -2 x \sum_{n=0}^{\infty} c_n(n+a)x^{n+a-1} + 3 0 \sum_{n=0}^{\infty} c_n x^{n+a} = 0 x \Bigg[ \sum_{n=0}^{\infty} c_n(n+a)(n+a-1)x^{n-2} - \sum_{n=0}^{\infty} c_n(n+a)(n+a-1)x^n \Bigg] - 2 \sum_{n=0}^{\infty} c_n(n+a)x^{n-1} + 3 0 \sum_{n=0}^{\infty} c_n x^n \Bigg] = 0 x \Bigg[ c_0(n-1)x^{-2} + c_{1}n(n+1)x^{-1} + \sum_{n=2}^{\infty} c_n(n+a)(n+a-1)x^{n-2} \sum_{k=0}^{\infty} c_{k+2}(k+2)(k+1)x^k + 3 \sum_{k=0}^{\infty} c_k k x^k + 2 \sum_{k=0}^{\infty} c_k x^k = 0 2 c_2 + 2c_0 + \sum_{k=2}^{\infty} [c_{k+2}(k+2)(k+1) + (3k+d)c_k]x^k = 0 Logo, 2 c_2 + 2c_0 = 0 \and c_{k+2}(k+2)(k+d)+(3k+d)c_k = 0 c_2 = -c_0 \rightarrow c_{k+2} = -\frac{(3k+1)}{(k+2)(k-1)}c_k c_j = 0 k = 0 \rightarrow c_2 = -\frac{c_0}{2} k = 2 \rightarrow e_1 = -\frac{7}{4.3} e_{2} = \frac{4}{4.3} c_0 k = 4 \rightarrow c_6 = \frac{-13}{6.5} c_4 = -\frac{13.7}{6.5.4.3.2} c_0 k = 6 \rightarrow c_8 = \frac{-19}{8.7} c_6 = \frac{19.13.7}{7.6.5.3.2} c_0 c_0 = 0 k = 1 \rightarrow c_3 = -\frac{4}{3.2} c_1 k = 3 \rightarrow c_5 = -\frac{10}{5.4} c_3 = \frac{10.4}{5.4.3.2} c_1 k = 5 \rightarrow c_7 = -\frac{16 c_5}{7.6} c_5 = \frac{-16}{7.6} c_3 = \frac{-16.4}{7.6.4.3.2} c_1 Então, y = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + c_4 x^4 + c_5 x^5 + ... y = c_0 + c_1 x -\frac{c_0}{2} x^2 - \frac{4}{12} c_1 x^3 + ... + y' = c_3 - c_0 x - c_3 x^3 y(0) = 1 \rightarrow c_0 = 1 y(0) = 1 \rightarrow c_3 = 1 Logo, a solução é \boxed{y = 1 + x - \frac{x^2}{2} - \frac{x^3}{3} + ... } = 0 logo, c_0 \nu (\nu - 1)x^{-2} + c_1 \nu (\nu - 1)x^{-2} + \sum_{k=0}^{\infty} \left[ c_{k+2}(k+2+\nu)(k+\nu+1)x^k \right] - \sum_{k=0}^{\infty} \left[ c_k (k+\nu)(k+\nu - 1)x^k \right] - 2 \sum_{k=0}^{\infty} c_k (k+\nu)x^k + 30 \sum_{k=0}^{\infty} c_k x^k \right] = 0 c_0 \nu (\nu - 1)x^{-2} + c_1 nu (\nu - 1)x^{-2} + \sum_{k=0}^{\infty} \left[ c_{k+2}(k+2+\nu)(k+\nu+1)x^k \right] - c_k (k+\nu)(k+\nu)(k+\nu - 1)x^k - 2 c_k (k+\nu)x^k + 30 c_k x^k = 0 fazendo, \nu(\nu-1) = 0 \rightarrow \nu = 0 \mbox{ ou } \nu = 1 c_{k+2}(k+2+\nu) - c_k(k+\nu)[k+\nu-1] + 30 c_k = 0 c_{k+2}(k+2+\nu) - c_k(k+\nu)(k+\nu-1) + 30 c_k = 0 c_{k+2} = \frac{(k+\nu)(k+\nu-1)-30}{k+2+\nu} c_k com \nu = 0, c_{k+2} = \frac{k(k-1)-20}{k+2} c_k k=0 \rightarrow c_2 = \frac{-30}{2} c_0 = -15 c_0 k=1 \rightarrow c_3 = \frac{-30}{3} c_3 = -10 c_1 com \nu = 1, c_{k+2} = \frac{k(k+1)-30}{k+3} c_k k=0 \rightarrow c_2 = \frac{-30}{3} = -10 c_0 k=1 \rightarrow c_3 = \frac{2-30}{4} c_3 = -7 c_3 (x - x) y' - xy = (x - 2) \sum_{n=1}^{\infty} c_n n x^{n-1} - x \sum_{n=0}^{\infty} c_n x^n = 0 \rightarrow \sum_{n=1}^{\infty} c_n n x^n - 2 \sum_{\substack{n=1 \\ k=n}}^{\infty} c_n n x^{n-1} - \sum_{\substack{n=0 \\ k=n+1}}^{\infty} c_n x^{n+1} = 0 \sum_{\substack{k=1}}^{\infty} c_k x^k - 2 \sum_{\substack{k=0}}^{\infty} c_{k+1} (k+1) x^k - \sum_{\substack{k=0}}^{\infty} c_{k-1} x^k = 0 \sum_{k=1}^{\infty} c_k x^k - 2 c_1 - 2 \sum_{\substack{k=0}}^{\infty} c_{k+3} (k+3) x^k - \sum_{\substack{k=0}}^{\infty} c_{k-3} x^k = 0 - 2 c_1 + \sum_{k=0}^{\infty} [c_k k - 2 c_{k+1} (k+1) - c_{k-1}] x^k = 0 \text{Logo:} -2e_j = 0 \rightarrow c_j = 0 c_k k - 2 c_{k+1} (k+1) - c_{k-1} = 0 c_{k+1} = \frac{c_{k-1} - kc_k}{k+1} k=1 \rightarrow c_2 = \frac{c_0 - c_1}{2} = \frac{c_0}{2} k=2 \rightarrow c_3 = \frac{c_2 - c_3}{3} = \frac{-c_2}{3} = \frac{-c_0}{3 \cdot 2} k=3 \rightarrow c_4 = \frac{c_3 - c_3}{4} = \frac{1}{4} \left( \frac{c_0}{2} + \frac{c_0}{3 \cdot 2} \right) = \frac{1}{4} \left( \frac{4c_0}{3 \cdot 2} \right) k=4 \rightarrow c_5 = \frac{c_3 - c_4}{5} = \frac{1}{5} \left(-\frac{c_0}{3 \cdot 2} - \frac{4c_0}{4 \cdot 3 \cdot 2} \right) = -\frac{1}{5 \cdot 4 \cdot 3 \cdot 2} \left( 8c_0 \right) \text{Com solução} \gamma = c_0 + c_2 x + c_2 x^2 + c_3 x^3 + c_4 x^4 + 1 ... \gamma = c_0 + \frac{c_0}{2} x - \frac{c_0}{3 \cdot 2} x^2 + \frac{4c_0}{4 \cdot 3 \cdot 2} x^3 - \frac{8c_0}{5 \cdot 4 \cdot 3 \cdot 2} x^4 + ... \gamma = (x-2) \sum_{k=0}^{\infty} \frac{c_0 x^k}{k!} \text{Com} \gamma(0) =1 \rightarrow c_0 \cdot 1 \rightarrow c_0 = 1 \rightarrow \boxed{\gamma = (x-2)^2 e^x} \quad \text{e} \quad \boxed{\gamma(2) = 0} Cálculo 4 - Lista 2 (1) y^{\prime\prime} - y^{\prime} + xy = 0 \gamma = \sum_{n=0}^{\infty} c_n x^n \rightarrow \gamma^{\prime} = \sum_{n=1}^{\infty} c_n n x^{n-1} \rightarrow \gamma^{\prime\prime} = \sum_{n=2}^{\infty} c_n n(n-1)x^{n-2} \gamma^{\prime\prime} - \gamma^{\prime} + xy = \sum_{n=2}^{\infty} c_n n(n-1) x^{n-2} - \sum_{n=1}^{\infty} c_n n x^{n-2} + x \sum_{n=0}^{\infty} c_n x^n = 0 \sum_{n=2}^{\infty} c_n n(n-1)x^{n-2} - \sum_{n=1}^{\infty} c_n n x^{n-1} + \sum_{n=0}^{\infty} c_n x^{n+1} = 0 \quad x = 0 \sum_{k=0}^{\infty} c_{k+2}(k+2)(k+1) x^k - \sum_{k=0}^{\infty} c_{k+1}(k+1) x^k + \sum_{k=0}^{\infty} c_k x^k = 0 \sum_{k=0}^{\infty} c_{k-2}(k+2)(k+1)x^k - \sum_{k=0}^{\infty} c_{k+1}(k+1) x^k + \sum_{k=0}^{\infty} c_k x^k = 0 2c_2 - c_j + \sum_{k=1}^{\infty} \left[ c_{k+2} (k+1) (k+2) - c_{k+1} (k+1) + c_{k-2} \right] x^k = 0 \\ logo, \\ 2c_2 - c_j = 0 \ \Rightarrow \ c_{k+2} (k+1)(k+2) - c_{k+1} (k+1) + c_{k-1} = 0 \\ \\\ c_j = 2c_2 \ \Rightarrow \ c_2 = \frac{c_j}{2} \\ \\\ c_{k+2} = \frac{c_{k+1}}{k+2} - \frac{c_{k-1}}{(k+2)(k+1)} \\ \\\ k_0, k_1 = 0 \\ k = 1 \ \Rightarrow c_3 = \frac{c_2}{3} - \frac{c_0}{3 \cdot 2} = - \frac{c_0}{3 \cdot 2} \\ k = 2 \ \Rightarrow c_4 = \frac{c_3}{4} - \frac{c_1}{4 \cdot 3} = \frac{c_3}{4} = - \frac{c_0}{4 \cdot 3 \cdot 2} \\ k = 3 \ \Rightarrow c_5 = \frac{c_4}{5} - \frac{c_2}{5 \cdot 4} = \frac{c_4}{5} = - \frac{c_0}{5 \cdot 4 \cdot 3 \cdot 2} \\ \\ thus, \\ c_x = - \frac{c_0}{k!} c_0 = 0 \\ k = 1 \ \Rightarrow c_3 = \frac{c_2}{3} = \frac{c_1}{3 \cdot 2} \\ k = 2 \ \Rightarrow c_4 = \frac{c_3}{4} - \frac{c_1}{4 \cdot 3} = \frac{c_1}{4 \cdot 3 \cdot 2} = - \frac{c_1}{4 \cdot 3} \\ k = 3 \ \Rightarrow c_5 = \frac{c_4}{5} - \frac{c_2}{5 \cdot 4} = - \frac{c_1}{5 \cdot 4 \cdot 3 \cdot 2} = \frac{c_1}{5 \cdot 4 \cdot 3 \cdot 2} \\ c_5 = - \frac{4c_1}{5 \cdot 4 \cdot 3 \cdot 2} \\ \\ Portanto, \\ \psi = c_0 \left( 1 - \frac{x^3}{6} - \frac{x^4}{24} - \frac{x^5}{120} - \ldots \right) \\ + c_3 \left( x + \frac{x^2}{2} + \frac{x^3}{6} - \frac{x^4}{24} - \frac{4x^5}{120} - \ldots \right) \\ \\ 2) \ \y'' - y' + x^2 y = 0 \\ Da questao, temos que. \sum_{n=2}^{\infty} c_n n(n-1)x^{n-2} - \sum_{n=1}^{\infty} c_n n x^{n-1} + \sum_{n=0}^{\infty} c_n x^{n+2} = 0 \\ n = n-2 \\ k = n-2 \\ k = n-1 \\ n = n+2 \\ \sum_{k=0}^{\infty} c_k (k+2)(k+1)x^k - \sum_{k=0}^{\infty} c_k (k+1)x^k \\ + \sum_{k=0}^{\infty} c_k x^k = 0 \\ 2c_2 + 6c_3 - c_j - 2c_2 + \sum_{k=2}^{\infty} [c_{k+2} (k+2)(k+1) \\ - c_{k+1} (k+1) + c_{k-2}] x^k = 0 \\ logo, \\ 6c_3 - c_j = 0 \ \Rightarrow \ c_{k+2} (k+2)(k+1)-c_{k+1} (k+1)+c_{k-2}=0 \\ \\ c_3=\frac{c_j}{6} \\ e, \\ c_{k+2}=\frac{c_{k+1}}{k+2} - \frac{c_{k-2}}{(k+2)(k+1)} C_1 = 0 k = 2 \rightarrow C_1 = \frac{C_3}{4} - \frac{c_0}{4 \cdot 3} = -\frac{c_0}{12} k = 3 \rightarrow C_5 = \frac{C_4}{5} - \frac{c_1}{5 \cdot 4} = \frac{C_4}{5} = -\frac{c_0}{60} c_0 = 0 k = 2 \rightarrow C_1 = \frac{C_3}{4} - \frac{c_0}{4 \cdot 3} = \frac{C_3}{4} = \frac{C_1}{24} k = 3 \rightarrow C_5 = \frac{C_4}{5} - \frac{c_1}{5 \cdot 4} = \frac{C_1}{24} \frac{C_1}{5 \cdot 4} = -\frac{C_1}{29} \therefore y = c_0 \left(1 - \frac{x^4}{12} - \frac{x^5}{60} - \ldots \right) + C_1 \left(x + \frac{x^3}{6} + \frac{x^4}{24} - \frac{x^5}{24} - \right) 1 - x^2 y'' - 2xy' + 2xy = 0 Sendo \text{ } y = \sum_{n=0}^{\infty} C_n x^n \to y'' = \sum_{n=1}^{\infty} C_n n(n-1)x^{n-2} {n = 0} y'' = \sum_{n=2}^{\infty} n(n-1)x^{n-2} y'' - x^2 y'' - 2 xy' + 2 y = 0 \sum_{n=2}^{\infty} C_n n(n-1)x^{n-2} - x^2 \sum_{n=2}^{\infty} C_n n(n-1)x^{n-2} -2x \sum C_n n x^{n-1} + 2 \sum_{n=0}^{\infty} C_n x^n = 0 \sum_{n=2}^{\infty} C_n n(n-1)x^{n-2} - \sum_{n=2}^{\infty} C_n n(n-1) \underbrace{{n-2}_{k = n}} x^{n-2} - \sum_{n=2}^{\infty} 2 C_n n x^n + \sum_{n=0}^{\infty} 2 C_n x^n = 0 \sum_{k=0}^{\infty} C_{k+2} (k+2)(k+1)x^{k} - \sum_{k=2}^{\infty} C_k k(k-1) x^k - \sum_{k=0}^{\infty} 2C_k x^k (k+1) x^k + \sum_{k=0}^{\infty} 2C_k k x^k = 0 2c_2 + 2c_0 + x(6c_3 - 2c_1 + 2c_0) + \sum_{k=2}^{\infty} [c_{k + 2} (k+2)(k+1) -c_x x(k-1)-2 c_x k + 2 c_x] x^k = 0 Logo, 2C_2 + 2C_0 = 0 \rightarrow C_2 = -C_0 6c_3 - 2c_1 + 2c_0 = 0 \rightarrow C_3 = 0 c_{k+2} (k+2)(k+1) - c_k k(k-1) - 2c_k k + 2C_k = 0 c_{k+2} = \frac{(k^2 + k - 2)c_k}{(k+2)(k+1)} = \frac{(k+2)(k-1)}{(k+2)(k-1)} c_k c_{k+2} = \frac{k - 1}{k + 1} c_k k = 2 \rightarrow c_4 = \frac{2 - 1}{2 + 1} c_2 = -\frac{c_0}{3} k = 3 \rightarrow c_5 = \frac{2}{4} k = 4 \rightarrow c_6 = \frac{3}{5} c_4 = \frac{3}{5}\left(-\frac{c_0}{3}\right) = -\frac{c_0}{5} k = 5 \rightarrow c_7 = \frac{5}{8} c_5 = 0 k = 6 \rightarrow c_8 = \frac{c_0}{7} c_8 = -\frac{c_0}{4} Portanto, \[ y = c_0 \left( 1 - x^2 - \frac{x^3}{3} - \frac{x^6}{5} - \frac{x^8}{7} \ldots \right) + c_1 x \] 1) y'' + (1 + x^2)y = 0 Subs \[ y = \sum_{n=0}^{\infty} c_n x^{n} - y' = \sum_{n=1}^{\infty} c_n n x^{n-1} \rightarrow y'' = \sum_{n=2}^{\infty} c_n n(n-1)x^{n-2} \] \[ y'' + y + x^2 y = \sum_{n=2}^{\infty} c_n n(n-1)x^{n-2} + \sum_{n=0}^{\infty} c_n x^{n} + \sum_{n=0}^{\infty} c_n x^{n+2} = 0 \] \[ \sum_{n=2}^{\infty} c_n n(n-1)x^{n-2} + \sum_{n=0}^{\infty} c_n x^{n} + \sum_{n=0}^{\infty} c_n x^{n+2} = 0 \] \[ \sum_{n=0}^{\infty} c_{n+2}(n+1)(k+2)x^k + \sum_{n=0}^{\infty} c_k x^k + \sum_{k=-2}^{\infty} c_k x^{k} = 0 \] + c_2 + 6 c_3 x + c_0 + c_1 x + \sum_{k=-2}^{\infty} [c_{k+2}(k+1)(k+2)x^k = 0] Coefs. 2c_2 + c_0 = 0 \rightarrow c_2 = -\frac{c_0}{2} 6c_3 + c_1 = 0 \rightarrow c_3 = -\frac{c_1}{6} c_{k+2} (k+1)(k+2) + c_k + c_{k-2} = 0 c_{k+2} = \frac{-c_k - c_{k-2}}{(k+1)(k+2)} k = 2 \rightarrow c_4 = \frac{-c_2 - c_0}{4,3} = \frac{-1}{12} \left( \frac{c_0}{2} - c_0 \right) = \frac{c_0}{24} k = 3 \rightarrow c_5 = \frac{-c_3 - c_1}{5,4} = \frac{-1}{5,1} \left( \frac{c_1}{6} - c_0 \right) = \frac{5c_1}{120} k = 4 \rightarrow c_6 = \frac{-c_4 - c_0}{6,5} = \frac{-1}{6,5} \left( \frac{c_0}{24} + \frac{c_0}{2} \right) = \frac{-19c_0}{720} Portanto, y = c_0 \left( 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{19}{720}x^6 + \ldots \right) + c_1 \left( x - \frac{x^3}{6} + \frac{5}{120}x^5 + \ldots \right) 5) y'' - 4xy' + (4x^2 - 2)y = 0 y'' - 4xy' + 4x^2 y - 8x^2 y = \sum_{n=2}^{\infty} c_n n(n-1)x^{n-2} -4x\sum_{n=1}^{\infty} c_n n x^{n-1} + 4x^2 \sum_{n=0}^{\infty} c_n x^{n} - \sum_{n=0}^{\infty} c_n x^{n} = 0 \sum_{n=2}^{\infty} c_n n(n-1)x^{n-2} - 4\sum_{n=1}^{\infty} c_n n x^{n-1} + 4\sum_{n=0}^{\infty} c_n x^{n+2} - \sum_{n=0}^{\infty} c_n x^{n} = 0 -2\sum_{k=0}^{\infty} c_n x^k = 0 -2\sum_{k=0}^{\infty} c_k x^k + \sum_{k=0}^{\infty} c_{k+2}(k+2)(k+1)x^k + 4\sum_{k=1}^{\infty} c_{k+2}x^k = 0 -c_0 - 2c_3 x + 2c_2 + 6c_3 x - 4c_1 - 5c_0 x + \sum_{k=2}^{\infty} \left[ c_{k+2}(k+2)(k+1)-4c_{k+1}(k+1)-2c_k+4c_{k+2} \right] x^k = 0 \begin{cases} -2c_0 + 2c_2 - 9c_1 = 0 \rightarrow c_2 = 2c_0 + c_0 -lc_2 + 6c_3 - 5c_2 = 0 \rightarrow c_3 = \frac{c_1}{3} + \frac{c_2}{4} \end{cases} c_{k+2}(k+2)(k+1)-4c_{k+1}(k+1)-2c_k+4c_{k+2}=0 \rightarrow c_{k+2} \left[(k+2)(k+1)+1\right] = 4c_{k+1}(k+1)+2c_k c_{k+2} = \frac{4c_{k+1}(k+1)+2c_k}{(k+2)(k+1)+1} + \frac{2c_k}{c_{k+2}(k+1)+1} c_1=0 k=0 \rightarrow c_2 = \frac{4c_1+2c_0}{6} + \frac{2c_0}{6} = \frac{2}{3}(c_1+c_0) = \frac{2c_0}{3} k=1 \rightarrow c_3 = \frac{8c_2+2c_1}{10} + \frac{2c_1}{10} = \frac{1}{5}c_2 + \frac{1}{5}c_1 c_3 = \frac{8}{15}c_0 c_0=0 c_2=\frac{2}{3}c_1, c_3=\frac{1}{5}c_1 Portanto, y = c_1\left(1+\frac{2}{3}x^2+\frac{8}{15}x^3+\ldots \right) + c_1\left(x+\frac{2}{3}x^2+\frac{4}{5}x^3+\ldots \right)