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Unformatted text preview: SOLODE The general Second Order ODE is y 00 ( t ) = f ( t,y ( t ) ,y ( t ) ) . We will only deal with the linear ones: y 00 = p ( t ) y ( t ) q ( t ) y ( t ) + g ( t ) or y 00 + p ( t ) y + q ( t ) y = g ( t ) . Example: Lets start simple: ay 00 + by + cy = 0 . ( CCH ) This equation has constant coefficients a,b,c, and is homogeneous [as g = 0 ]. What to do? SOLODE The general Second Order ODE is y 00 ( t ) = f ( t,y ( t ) ,y ( t ) ) . We will only deal with the linear ones: y 00 = p ( t ) y ( t ) q ( t ) y ( t ) + g ( t ) or y 00 + p ( t ) y + q ( t ) y = g ( t ) . Example: Lets start simple: ay 00 + by + cy = 0 . ( CCH ) This equation has constant coefficients a,b,c, and is homogeneous [as g = 0 ]. What to do? The functions y ( t ) , y ( t ) , and y 00 ( t ) should be similar. So? 2 The functions y ( t ) , y ( t ) , and y 00 ( t ) should be similar. So? Well try y ( t ) = e rt . Then ay 00 + by + cy = ar 2 e rt + bre rt + ce rt = ( ar 2 + br + c ) e rt and y ( t ) = e rt solves ( CCH ) iff ar 2 + br + c = 0 , i.e. iff r = b b 2 4 ac 2 a , i.e. iff r = r 1 def = b + b 2 4 ac 2 a or r = r 2 def = b b 2 4 ac 2 a . 2 ar 2 + br + c is the characteristic polynomial of the equation ( CCH ) , and ar 2 + br + c = 0 is the characteristic equation . At first, this makes sense only if the Dis criminant b 2 4 ac is strictly positive, so we can take the square root of b 2 4 ac ; in this case we get two solutions y 1 ( t ) = e r 1 t...
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This note was uploaded on 09/07/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner

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