Prelim1_5 - Prelim 1 Solution Problem 5 x2 y HxL 2 x y HxL 2 yHxL = 0 x> 0 a yHxL = xr y HxL = r xr-1 y HxL = rHr 1L xr-2 Substitute these into

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Unformatted text preview: Prelim 1 Solution Problem 5 x2 y'' HxL + 2 x y' HxL - 2 yHxL = 0 ; x > 0. a. yHxL = xr y' HxL = r xr-1 y'' HxL = rHr - 1L xr-2 Substitute these into the differential equation to obtain x2 r Hr - 1L xr-2 + 2 x r xr-1 - 2 xr = 0 xr H rHr - 1L + 2 r - 2L = 0. xr is not zero for x > 0 so we are left with rHr - 1L + 2 r - 2 = 0 r2 + r - 2 = 0 Hr - 1L Hr + 2L = 0 r = 1, -2. So solutions are yHxL = x yHxL = x-2 . b. Set yHxL = c1 x + c2 x-2 yH1L = c1 + c2 = 1 fl c2 = 1 - c1 y' HxL = c1 - 2 c2 x-3 y' H1L = c1 - 2 c2 = -5 These two equations together yield c1 = -1, c2 = 2. So the solution is yHxL = 2 x-2 - x ...
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This homework help was uploaded on 04/01/2008 for the course MATH 2930 taught by Professor Terrell,r during the Fall '07 term at Cornell University (Engineering School).

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