This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Here is the idea: substitute y = ∑ n = ∞ c n x n into the differential equation and then equate all coefficients to the righthand side of the equation to determine the coefficients c n . Example 1. Solve Slution. Let y = ∑ n = ∞ c n x n . Then y' = ∑ n = 1 ∞ nc n x n − 1 and y' ' = ∑ n = 2 ∞ n ( n − 1 ) c n x n − 2 . Substituting y ∧ y ' ' into differential equation gives ¿ 2 c 2 + ∑ n = 3 ∞ n ( n − 1 ) c n x n − 2 + ∑ n = ∞ c n x n + 1 Let k = n − 2 ∧ k = n + 1 . Then c k − 1 x k = ¿ 2 c 2 + ∑ k = 1 ∞ ( k + 2 )( k + 1 ) c k + 2 x k + ∑ k = 1 ∞ ¿ This relation generates consecutive coefficients of the assumed solution one at the time as we let k take on the successive integers Example 2. Use power series method to solve given differential equation about ordinary point x = 0, y '' +( cosx ) y = . Solution....
View
Full Document
 Spring '15
 Linear Algebra, Algebra, Power Series, 0 k, 1 k

Click to edit the document details