University:
Princeton University
Course:
PHY 103 | General Physics I
Academic year:

2023
Views:

300

Pages:

1
Author:

Jack Alasco

Solve a differential equation Given: 𝑥 2 𝑦 ′ ′ + 𝑥 𝑦 ′ + ( 𝑥 2 − 𝑛 2 ) 𝑦 = 0 x 2 y ′′ +xy ′ +(x 2 −n 2 )y=0 Method & Answer: 𝑦 = ∑ 𝑛 = 0 ∞ 𝑎 𝑛 𝑥 𝑛 y=∑ n=0 ∞ a n x n 𝑦 ′ = ∑ 𝑛 = 0 ∞ 𝑛 𝑎 𝑛 𝑥 𝑛 − 1 y ′ =∑ n=0 ∞ na n x n−1 𝑦 ′ ′ = ∑ 𝑛 = 0 ∞ 𝑛 ( 𝑛 − 1 ) 𝑎 𝑛 𝑥 𝑛 − 2 y ′′ =∑ n=0 ∞ n(n−1)a n x n−2 Substitute into the equation: 𝑥 2 𝑦 ′ ′ + 𝑥 𝑦 ′ + ( 𝑥 2 − 𝑛 2 ) 𝑦 = 0 x 2 y ′′ +xy ′ +(x 2 −n 2 )y=0 𝑥 2 ∑ 𝑛 = 0 ∞ 𝑛 ( 𝑛 − 1 ) 𝑎 𝑛 𝑥 𝑛 − 2 + 𝑥 ∑ 𝑛 = 0 ∞ 𝑛 𝑎 𝑛 𝑥 𝑛 − 1 + ( 𝑥 2 − 𝑛 2 ) ∑ 𝑛 = 0 ∞ 𝑎 𝑛 𝑥 𝑛 = 0 x 2 ∑ n=0 ∞ n(n−1)a n x n−2 +x∑ n=0 ∞ na n x n−1 +(x 2 −n 2 )∑ n=0 ∞ a n x n =0 Simplify the equation: ∑ 𝑛 = 0 ∞ [ ( 𝑛 ( 𝑛 − 1 ) + 𝑛 + 𝑥 2 𝑎 𝑛 − 𝑛 2 𝑎 𝑛 ] 𝑥 𝑛 = 0 ∑ n=0 ∞ [(n(n−1)+n+x 2 a n −n 2 a n ]x n =0 𝑠 2 + 5 𝑠 + 2 = 0 s 2 +5s+2=0 𝑠 2 + 5 𝑠 + 2 = 0 s 2 +5s+2=0 This implies 𝑠 = 1 s=1. 𝑎 𝑛 = − 𝑎 𝑛 − 2 ( 𝑛 − 1 ) ( 𝑛 − 2 ) a n = (n−1)(n−2) −a n−2 For 𝑛 = 2 n=2: 𝑎 2 = − 𝑎 0 2 ! a 2 = 2! −a 0 For 𝑛 = 4 n=4: 𝑎 4 = − 𝑎 2 4 ! = − 𝑎 0 4 ! a 4 = 4! −a 2 = 4! −a 0 For 𝑛 = 6 n=6: 𝑎 6 = − 𝑎 4 6 ! = − 𝑎 0 6 ! a 6 = 6! −a 4 = 6! −a 0 For 𝑛 = 8 n=8: 𝑎 8 = − 𝑎 6 8 ! = − 𝑎 0 8 ! a 8 = 8! −a 6 = 8! −a 0 Substitute into the y equation: 𝑦 = 𝑎 0 𝑥 𝑛 + 𝑎 1 𝑥 𝑛 − 1 + − 𝑎 0 2 ! 𝑥 2 + − 𝑎 0 4 ! 𝑥 4 + − 𝑎 0 6 ! 𝑥 6 + ⋯ y=a 0 x n +a 1 x n−1 + 2! −a 0 x 2 + 4! −a 0 x 4 + 6! −a 0 x 6 +⋯ Take 𝑎 0 𝑥 2 a 0 x 2 out: 𝑦 = 𝑎 0 𝑥 2 ( 1 − 𝑥 2 2 ! + 𝑥 4 4 ! − 𝑥 6 6 ! + ⋯ ) y=a 0 x 2 (1− 2! x 2 + 4! x 4 − 6! x 6 +⋯) Recognize this as a cosine series: 𝑦 = 𝑎 0 𝑥 2 cos ⁡ 𝑥 y=a 0 x 2 cosx 𝑦 = 𝑎 0 cos ⁡ 𝑥 𝑥 y= x a 0 cosx