# MTH235 - December 7, 2010 1-1 Exam 2 Review This page gives...

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1-1 Exam 2 –Review This page gives a summary of the material required for Exam 2. Subject Matter: 1. Special Second Order Equations: (a) y 00 = f ( t,y 0 ) Set y 0 = v , get v 0 = f ( t,v ) Solve: v = v ( t,C 1 ), Then, integrate to get y = R v ( t,C 1 ) + C 2 (b) y 00 = f ( y,y 0 ) Set y 0 = v , y 00 = dv dt = dv dy dy dt = dv dy v = f ( y,v ) Solve for v = v ( y,C 1 ) Then, y 0 = v ( y,C 1 ) – separable Solve for y = y ( t,C 1 ,C 2 ) — May get implicit solution 2. Reduction of order L ( y ) = y 00 + p ( t ) y 0 + q ( t ) y = 0, Have one non-zero solution y 1 . Get second independent solution by y 2 ( t ) = y 1 ( t ) v ( t ) for some non- constant v ( t ) Plug into equation and get v 0 (2 y 0 1 + py 1 ) + y 1 v 00 = 0 Solve for v 0 , then integrate to get v – Do not need to keep the constants 3. Linear Second Order equations – constant coeﬃcients L ( y ) = ay 00 + by 0 + cy = h ( t ) (a) homogeneous: L ( y ) = 0: Get characteristic polynomial: z ( r ) = ar 2 + br + c , Get roots

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## This note was uploaded on 02/03/2012 for the course MTH 235 taught by Professor Staff during the Fall '11 term at Michigan State University.

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MTH235 - December 7, 2010 1-1 Exam 2 Review This page gives...

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