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A. verify that y=2x^a solves the differential equation x^2y''=2y if the constant a satisfies the equation a^2-a-2=0. thus get the two solutions x^2 and x^-1. note that the first is valid on the whole interval -∞<X<∞ but the second on -∞<X<0 or 0<X<∞ only. this behavior is typical for a broad class of linear homogeneous equations know as equations of euler type. for this class the substitution y=x^a always lead to an algebraic equation for a.
B. the equation X^2y''''=2y'' admits a solution y=x^a, where a is a nonzero constant. what are the possible values of a?
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QA:
Let y=2x^a solve x^2y"=2y, then
x^2*2a(a-1)x^(a-2)=4x^a, 2a(a-1)x^a= 4x^a
2a(a-1)=2 or a^2-a-2=0, a=-1, 2
so that x^2, 1/x are sol. of x^2 y"=2y
QB:
Let y=x^a solve x^2 y^(4)=2y", then
a(a-1)(a-2)(a-3) x^(a-2)= 2a(a-1) x^(a-2)
Note:the above eq. holds for any constant number a,not only for nonzero.
a(a-1)(a-2)(a-3)=2a(a-1), then a(a-1)(a^2-5a+4)=0, thus a=0, 1, 1, 4
so that, y=1, x, x^4 satisfy x^2 y^(4)=2y"