Check That Y1(x) = X2 + 1 Is A Specialsolution Of(

Anonymous

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Check that y1(x) = x2 + 1 is a specialsolution of (x2 + 1)y'' - 2y = 0. find the general solution of the giver differentialequation

 

Answer No.1

Anonymous

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For the first part, y1 = 1+x^2 y1'=2x y1''=2 (x^2+1)y1''-2y1 = (x^2+1)(2) - 2(x^2+1) = 0 for the second part, rewrite equation as (x^2+1)y''+2xy'-2xy'-2y=0 d/dx((x^2+1)y') - d/dx(2xy) = 0 integrating (x^2+1)y' - 2xy = A (A=constant) you can rewrite this as (x^2+1)^2 d/dx(y/(x^2+1)) = A d/dx(y/(x^2+1)) = A/(x^2+1) integating y/(x^2+1) = int [Adx/(x^2+1)] = A[arctan x + B] (B=constant) y = A (1+x^2) arctan(x) + C (1+x^2) (I have rewritten AB as C) the special solution in first part is just the C(1+x^2) term.

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