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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 xr1 y'' HxL = rHr  1L xr2 Substitute these into the differential equation to obtain x2 r Hr  1L xr2 + 2 x r xr1  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 = x2 . b.
Set yHxL = c1 x + c2 x2 yH1L = c1 + c2 = 1 fl c2 = 1  c1 y' HxL = c1  2 c2 x3 y' H1L = c1  2 c2 = 5 These two equations together yield c1 = 1, c2 = 2. So the solution is yHxL = 2 x2  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).
 Fall '07
 TERRELL,R
 Differential Equations, Equations

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