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Unformatted text preview: GE 207K Reduction of Order September 16, 2011 Summary : Given a linear second order differential equation y 00 + p ( t ) y + 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 (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 + q ( t ) y = 0 , y 1 ( t ) , and we seek the second (linearly independent) solution, y 2 . The general procedure is summarized on the next page. The simplified procedure skips all the algebra work involved. Steps – (Simplified): 1. Write the differential equation in the STANDARD FORM y 00 + p ( t ) y + 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 1 y 1 + p ( t ) v = 0 . (2) where y 1 is the given first solution. 3. Substitute u = v 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 02/03/2012 for the course M 427K taught by Professor Fonken during the Fall '08 term at University of Texas at Austin.
 Fall '08
 Fonken
 Differential Equations, Equations

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