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 DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Week 7: 0. 4.1 2nd Order Linear Equations 1. 4.2 Constant Coef, Homogeneous DE 2. 4.3 Undetermined Coef 3. 4.5 Springs (Homog case) 2 General Solution: (1) to solve the 2nd order linear DE y + a 1 y + a 2 y = 0 find two linearly independent solutions y 1 and y 2 , then the general solution is y H = c 1 y 1 + c 2 y 2 . (2) to solve the nonhomogeneous 2nd order linear DE y + a 1 y + a 2 y = F, find a particular solution y p , then the general solution is y = y H + y p , where y H is the solution of (1). Problem 1: Determine all values r so y = e rx is a solution to y 4 y + 3 y = 0 . Find the general solution. Solution: For y = e rx , we have y = re rx and y = r 2 e rx . Substitution in the DE gives 0 = y 4 y + 3 y = r 2 e rx 4 re rx + 3 e rx = ( r 2 4 r + 3) e rx . Now r 2 4 r + 3 = ( r 3)( r 1) = 0 has roots r 1 = 1 and r 2 = 3 , so we get solutions y 1 = e x and y 2 = e 3 x , so the general solution is y = c 1 e x + c 2 e 3 x . 4 Thursday: Constant Coef, Homogeneous DE When the DE y + a 1 y + a 2 y = 0 has coefficients a 1 and a 2 that are constant, the two linearly independent solutions y 1 and y 2 in the general solution y H = c 1 y 1 + c 2 y 2 may be determined using the roots of the characteristic polynomial P ( r ) = r 2 + a 1 r + a 2 . Three cases: (1) For distinct real roots r 1 and r 2 , y = c 1 e r 1 x + c 2 e r 2 x . (2) For a repeated real root r 1 the 2nd independent solution is y 2 = xe r 1 x , and y = c 1 e r 1 x + c 2 xe r 1 x . (3) For a pair of complex roots r = a + bi and r = a bi with a and b real, b > 0, y = c 1 e ax cos( bx ) + c 2 e ax sin( bx ) . 6 Tues: Recitation on 4.2. Problem 2: Solve y 6 y + 34 y = 0 . Solution: The characteristic polynomial is r 2 6 r + 34 , so the quadratic formula gives roots 6 ± √ 6 2 4 · 34 2 = 3 ± 5 i. (why? ) So y = e 3 x ( c 1 cos 5 x + c 2 sin 5 x ) is the general solution. *Postscript: We may solve the IVP y + 4 y + 4 y = 0 , y (0) = 1 , y (0) = 4 ....
View
Full
Document
This note was uploaded on 02/29/2008 for the course MATH 205 taught by Professor Zhang during the Spring '08 term at Lehigh University .
 Spring '08
 zhang
 Linear Algebra, Algebra, Linear Equations, Equations

Click to edit the document details