lecture_11 (dragged) 2 - ) . Then we have dy dt = lim n d...

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MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 6 3 Unfortunately, this is not entirely useful because both sides of the equation involve y = y ( t ). We use this idea to deFne a sequence of functions y n = y n ( t ) which will approximate the exact solution y = y ( t ). ±irst deFne the constant function y 0 ( t )= y 0 . Now deFne recursively the sequence of functions y n +1 ( t )= ° t t 0 G ± τ, y n ( τ ) ² + y 0 . Note that d dt ³ y n +1 ´ = G ± t, y n ² and y n +1 ( t 0 )= y 0 . Now consider the limit: y ( t )= lim n →∞ y n ( t
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Unformatted text preview: ) . Then we have dy dt = lim n d dt y n +1 = lim n G t, y n = G ( t, y ) , y ( t ) = lim n y n ( t ) = lim n y = y . Hence y = y ( t ) is the exact solution to the initial value problem. This technique of approximating the solution y = y ( t ) via a sequence y n = y n ( t ) is known as Picards Method ....
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue University-West Lafayette.

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