M257-316Notes_Lecture3

M257-316Notes_Lecture3 - Chapter 3 Lecture 3 - Review of...

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Chapter 3 Lecture 3 - Review of Methods to Solve ODE 3.1 First Order ODE: Separable Equations: dy dx = P ( x ) Q ( y ) (3.1) Z dy Q ( y ) = Z P ( x ) dx + C EG: dy dx = 4 y x ( y 3) µ y 3 y dy = 4 x dx y 3ln | y | =4 l n | x | + C (3.2) y = ln( x 4 y 3 )+ C Ax 4 y 3 =e y Linear First Order Eq. - The Integrating Factor: y 0 ( x P ( x ) y = Q ( x ) (3.3) Can we fnd a Function F ( x ) to multiply (4.3) by in order to turn the leFt hand side into a derivative oF a product: Fy 0 + FPy = FQ (3.4) 17
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Lecture 3 - Review of Methods to Solve ODE ( Fy ) 0 = 0 + F 0 y = FQ (3.5) So let F 0 = FP which is a separable Eq. dF F ( x ) = P ( x ) dx Z dF F = Z P ( x ) dx + C Therefore ln F = Z P ( x ) dx + C (3.6) or F = A e R P ( x ) dx choose A =1 F =e R P ( x ) dx integrating factor Therefore e R P ( x ) dx y 0 +e R P ( x ) dx P ( x ) y R P ( x ) dx Q ( x ) (e R P ( x ) dx y ) 0 R P ( x ) dx Q ( x ) y ( x )=e R P ( x ) dx n R e R x P ( t ) dt Q ( x ) dx + C o (3.7) EG: 1 y 0 +2 y = 0 (3.8) F ( x 2 x e 2 x y 0 2 x 2 y =(e 2 x y ) 0 =0 e 2 x y =? c y ( x )= C e 2 x 3.2 Another Method - Series Solution: Since the unknown solution y ( x ) is deFned implicitly by (3.8) let us look for a series solution: y ( x n =0 a n x n . y 0 = X n =1 a n nx n 1 (3.9) Therefore y 0 y = X n =1 a n nx n 1 + X n =0 2 a n x n In the Frst sum let m = n 1 n m n = m +1 18
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3.3. SECOND ORDER CONSTANT COEFFICIENT LINEAR EQUATIONS: Therefore m =0 a m +1 ( m +1) x m + n =0 2 a n x n =0 n m : m =0 { a m +1 ( m +1)+2 a m } x m (3.10) a m +1 = 2 ( m +1) a m a 1 = 2 a 0 ,a 2 =+ 2 2 2 1 a 0 3 = 2 3 · 2 2 · 2 1 a 0 =( 1) 3 2 3 3!
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This note was uploaded on 05/04/2011 for the course MATH 25 taught by Professor Lo during the Spring '11 term at BC.

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M257-316Notes_Lecture3 - Chapter 3 Lecture 3 - Review of...

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