Unformatted text preview: y °° − ± 3 x ² y ° + ± 4 x 2 ² y = 0 . (a) Verify the y (1) ( x ) = x 2 is a solution. (b) Put y (2) ( x ) = u ( x ) x 2 and show that x 2 u °° + xu ° = 0. (c) Set v = u ° and show that xv =const. (d) Choose 1 as the constant and integrate to show that u ( x ) = ln ( x ) + d , d a constant. Verify that this indeed is a solution of the original diFerential equation. Exercise 20: This problem illustrates how small changes in the coeﬃcients of a diFerential equation may cause dramatic changes in the solution. (a) ±ind a general solution ϕ ( t ) of ¨ x − 2 a ˙ x + a 2 x = 0 for a a nonzero constant. (b) ±ind the general solution ϕ ° ( t ) of ¨ x − 2 a ˙ x + ( a 2 − ° 2 ) x = 0 in which ° is a positive constant. (c) Show that, as ° → 0, ϕ ° ( t ) does not in general approach the solution ϕ ( t ). DUE DATE: Wednesday, October 14 in class...
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 Spring '08
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 Math, Derivative, Elementary algebra, Prof. T. Angell

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