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Unformatted text preview: y 3 x y + 4 x 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 coecients of a diFerential equation may cause dramatic changes in the solution. (a) ind a general solution ( t ) of x 2 a x + a 2 x = 0 for a a non-zero constant. (b) ind the general solution ( t ) of x 2 a x + ( a 2 2 ) x = 0 in which is a positive constant. (c) Show that, as 0, ( t ) does not in general approach the solution ( t ). DUE DATE: Wednesday, October 14 in class...
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- Spring '08