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Unformatted text preview: Math 205, Spring 2011 Homework 11: due April 11 Chapter 6, Sections 2 and 3. NOTE: The assignment on 6.5 will be collected on April 18 2 Week 11: 6.2 Constant Coef, Homogeneous DE 6.3 Nonhomogeneous DE (undetermined coef) 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 . 3 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 .  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 ....
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 Spring '08
 zhang
 Math

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