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Unformatted text preview: ACM95b/100b Lecture Notes Niles A. Pierce Caltech 2010 Reduction of Order Consider the homogeneous 2nd order linear ODE Ly y 00 + p ( t ) y + q ( t ) y = 0 (1) with p and q continuous on an open interval I . Suppose that one solution y 1 ( t ) is already known. To find a second linearly independent solution we attempt the solution y = v ( x ) y 1 ( x ) . (2) This yields Ly = y 1 v 00 + (2 y 1 + py 1 ) v + ( y 00 1 + py 1 + qy 1  {z } =0 by (1) ) v = 0 (3) or y 1 v 00 + (2 y 1 + py 1 ) v = 0 . (4) This is a first order linear equation that can be solved for v ( x ) and then integrated to obtain v ( x ). Hence, the general solution to (1) is y = c 1 y 1 ( x ) + c 2 v ( x ) y 1 ( x ) . (5) Variation of Parameters Consider the nonhomogeneous 2nd order linear ODE y 00 + p ( t ) y + q ( t ) y = h ( t ) (6) with p,q,h continuous on an open interval I . Suppose we have already obtained the complementary solution y c = c 1 y 1 ( t ) + c 2 y 2 ( t ) (7) which is the general solution to (1). We now attempt to find a particular solution of the formwhich is the general solution to (1)....
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 Winter '09
 NilesA.Pierce

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