# reduction - GE 207K Reduction of Order September 15, 2010...

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GE 207K Reduction of Order September 15, 2010 This page aims to recap our mini-lecture on solving 2nd order ODEs using reduction of order. A 2nd-order linear homogeneous ordinary diﬀerential equation has the general form of y 00 + p ( t ) y 0 + q ( t ) y = 0 . We seek two solutions which we shall call y 1 ( t ) and y 2 ( t ) such that they are independent of one another. In other words, they are not a constant multiple of each other. Then: y 1 ( t ) and y 2 ( t ) are a fundamental set of solutions of the ODE. (You will need the Wronski determinant to verify this). y = c 1 y 1 ( t ) + c 2 y 2 ( t ) is the general solution of ODE (Principle of superposition). If we’re given one solution, say y 1 ( t ) , the we can ﬁnd the second solution using the reduction of order. The procedure is summarized below: We are given a second order homogenous diﬀerential equation along with one solution y 00 + p ( t ) y 0 + q ( t ) y = 0 , y 1 ( t ) , and we seek the second (linearly independent) solution, y 2 . Steps – (Simpliﬁed): 1. Write the diﬀerential equation in the STANDARD FORM y 00 + p ( t ) y 0 + q ( t ) y = 0 . (1) 2. Second solution has the form y 2 = v ( t ) y 1 ( t ) . Function v ( t ) is obtained by solving the diﬀerential equation v 00 + ± 2 y 0 1 y 1 + p ( t ) ² v 0 = 0 . (2) where y 1 is the given ﬁrst solution. 3. Substitute u = v 0 in above equation to turn the 2nd order ODE into a 1st order ODE. 4. Second solution is y 2 ( t ) = v ( t ) y 1 ( t ) . Discard any terms in y 2 ( t ) that appears as a constant multiple of y 1 ( t ) .

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## This note was uploaded on 10/08/2010 for the course GE 207K taught by Professor None during the Fall '10 term at University of Texas at Austin.

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reduction - GE 207K Reduction of Order September 15, 2010...

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