Find a polynomial solution first, and then the general solution of the equation:

Question

Find a polynomial solution first, and then the general solution of the equation: (1+x^2)y'' - 2y=0 (My answer manual has y1(x)=x^2 + 1, y(x)=c1(x^2 +1) +c2[x + (1+x^2)Arctanx]. How do we find each one? Wolfram step by step solution will NOT be rated/accepted because i already saw that and it didnt cover me/didnt like it. ) Thank you

Explanation / Answer

Assume that y = S(n = 0 to 8) a(n) x^n. Substituting this into the DE yields (1 + x^2) * S(n = 2 to 8) n(n-1) a(n) x^(n-2) - 2 * S(n = 0 to 8) a(n) x^n = 0 Simplifying: S(n = 2 to 8) n(n-1) a(n) x^(n-2) + S(n = 2 to 8) n(n-1) a(n) x^n - S(n = 0 to 8) 2 a(n) x^n = 0 Shift the index on the first series: S(n = 0 to 8) (n+2)(n+1) a(n+2) x^n + S(n = 2 to 8) n(n-1) a(n) x^n - S(n = 0 to 8) 2 a(n) x^n = 0 Simplifying further: S(n = 0 to 8) [(n+2)(n+1) a(n+2) - 2 a(n)] x^n + S(n = 2 to 8) n(n-1) a(n) x^n = 0 (2a(2) -2a(0)) + (6 a(3) - 2 a(1))x + S(n = 2 to 8) [(n+2)(n+1) a(n+2) + (n+2)(n-1) a(n)] x^n = 0 Equate like entries (assuming a(0) and a(1) are arbitrary): (2 a(2) - 2 a(0) = 0 ==> a(2) = a(0) 6a(3) - 2 a(1) = 0 ==> a(3) = a(1)/3 For n > 1: (n+2)(n+1) a(n+2) + (n+2)(n-1) a(n)=0 ==> a(n+2) = -(n-1) a(n) / [(n+1)]; note the step of 2 from a(n) to a(n+2). ----------- If n is even: a(0) arbitrary a(2) = a(0) a(4) = -a(2)/3 =-a(0)/3 a(6)=-3a(4)/5 =a(0)/5 a(8) = -7a(6)/9 => a(2k+2)= (-1)^(1+k)2k!/(2k+1)! * a(0) for all k = 2, 3, ... If n is odd: a(1) arbitrary a(3) = a(1)/3 =0 a(5) =0 a(2k+1) = 0 Hence, the general solution is (letting c1= -a(0) and c2 = a(1)) y(x)=c1(x^2 +1) +c2[x + (1+x^2)Arctanx] I hope this helps!.. Pleasr Rate n reward... :)

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