P∞
P∞ P∞ n−2 + n=1 cn nxn+1 + n=0 cn xn+1 =P 0 nx n=2 (n − 1)nc P P ∞ ∞ ∞ 2c2 x0 + 6c3 x n=4 (n − 1)ncn xn−2 + n=1 cn nxn+1 + c0 x1 n=1 cn xn+1
=0
Hacemos k = n−2 para la primera serie, yk = n+1para la segunda y tercera series. P P ∞
∞
0 k+2−2 2cP + k=2 ck−1 (k − 1)xk−1+1 + 2 x + 6c3 x k=2 (k + 2 − 1)(k + 2)ck+2 x ∞ k−1+1 c0 x =0 k=2 ck−1 x P∞ 2c2 + 6c3 x + c0 x Pk=2 (k + 1)(k + 2)ck+2 xk + ck−1 (k − 1)xk + ck−1 xk = 0 ∞ 2c2 + 6c3 x + c0 x k=2 [(k + 1)(k + 2)ck+2 + ck−1 (k − 1) + ck−1 ]xk = 0 (k + 1)(k + 2)ck+2 + ck−1 (k − 1) + ck−1 1
Entonces tenemos 2c2 = 0 ; c2 = 0 6c3 + c0 = 0 c3 = − 61 c0
kck−1 (k + 1)(k + 2) Sustituyendo k = 2, 3, 4, ... en la formula se obtiene 2c1 c4 = 3(4) = 16 c1 3c2 c5 = 4(5) = 0 ← c2 = 0 4c3 2 1 c6 = 5(6) = 15 (− 16 c0 ) = − 45 c0 5c4 5 1 5 c7 = 6(7) = 42 ( 6 c1 ) = 136 c1 6c5 6 c8 = 7(8) = 56 (0) = 0 7c6 7 1 7 (− 45 )c0 = − 72(45) c0 c9 = 8(9) = 72 8c7 4 5 5 c10 = 9(10) = 45 ( 136 c1 ) = 45(34) c1 9c8 9 = 110 (0) = 0 c11 = 10(11) 10c9 5 7 7 c12 = 11(12) = 66 (− 72(45) c0 ) = − 66(72)(9) c0 ck+2 =
[(k−1)+1]ck−1 (k+1)(k+2)
=
Por tanto,
y = c0 + c1 x + c2 x2 + c3 x3 + c4 x4 + c5 x5 + c6 x6 + c7 x7 + c8 x8 + c9 x9 + ... 5 5 1 6 7 7 y = c1 [ 61 x4 + 136 x7 + 9(34) x10 ] − c0 [ 45 x + 72(45) x9 + 66(72)(9) x12 ]
3.
y00 − 2xy0 + y = 0
Sutituyendo: P ∞
y = Pn=0 cn xn ∞ y0 = Pn=1 cn nxn−1 ∞ y00 = n=2 (n − 1)ncn xn−2
En P la ecuación original
P∞ P∞ ∞ n−2 − 2xP n=1 cn nxn−1 + n=0 cn xn = 0 n=2 (n − 1)ncn x P∞ P ∞ ∞ n−2 n (n − 1)ncn x − 2 n=1 cn nx + n=0 P cn xn = 0 n=2 P∞ P∞ ∞ n−2 n 2c2 n=3 (n − 1)ncn x − 2 n=1 cn nx + c0 n=1 cn xn = 0 Hacemos y k =P n para las dos y P tres. P∞ k = n − 2 para la serie uno ∞ ∞ k k + c 2c2 k=1P (k + 2 − 1)(k + 2)ck+2 xk+2−2P− 2 k=1 ck kx 0 k=1 ck x = 0 P∞ ∞ ∞ k k k 2c2 + c0 Pk=1 (k + 1)(k + 2)ck+2 x + k=1 ck kx + k=1 ck x = 0 ∞ 2c2 + c0 Pk=1 (k + 1)(k + 2)ck+2 xk − 2ck kxk + ck xk = 0 ∞ 2c2 + c0 k=1 (k + 1)(k + 2)ck+2 xk − 2ck kxk + ck xk = 0
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