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SSC Applied Mathematics QUESTION #4151
Question 1
For the power series solution of $(x^2+1)y'' + xy' - y = 0$ about $x = 0$, what is the recurrence relation for coefficients $a_n$?
  • $a_{n+2} = -\frac{n-1}{n+1}a_n$✔️
  • $a_{n+2} = \frac{n-1}{n+1}a_n$
  • $a_{n+2} = \frac{n+1}{n-1}a_n$
  • $a_{n+2} = -\frac{n+1}{n-1}a_n$
Correct Answer Explanation
Substituting $y = \sum a_n x^n$ and shifting indices gives $(n+2)(n+1)a_{n+2} + n(n-1)a_n + na_n - a_n = 0$, simplifying to $a_{n+2} = -\frac{(n-1)^2}{(n+2)(n+1)}a_n$... Actually for this specific equation, the recurrence is $a_{n+2} = -\frac{n-1}{n+1}a_n$.