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So I have learned about power series solution and Frobenius method in my Engineering Maths course in University, but i am quite confused about when to use which method to solve the 2nd order differential equation with variable coefficients. For example:

(1) $y''+2y'- \frac{4}{t^2}y=0$

(2) $y''+2ty'- 4y=0$

(3) $y''+\frac{2}{t^2}y'- 4y=0$

The question asks me to determine the singular point of each differential equation ( if exists) and then classify them as regular or irregular. I know that the (1) has x=0 as regular singular point and (3) has x=0 as irregular singular point. If I am correct, the (2) equation is ordinary and does not have a singular point.

Then, the question further asks if each of the equations can be solved by either normal Power Series solution or Frobenius method or not. Based on my understanding, normal Power series is used to solve ordinary 2nd order differential equation where the equation does not have any singularity. On the other hand, Frobenius method is used to solve 2nd order differential equation with regular singular point. So, if I am correct, then

(1) can be solved by Frobenius method but not normal power series solution

(2) can be solved by normal power series solution but not Frobenius method

(3) cannot be solved by both normal power series method and Frobenius method

But then I tried to solve the (2) equation using both power series solution and Frobenius method just to test my knowledge and understanding, and I get the same answer using both methods which is:

$$ y=a_0 (1+2x^2) +a_1 \biggl(x + \sum_{n=1}^∞ \frac{2^n(1)(-1)(-3)...(3-2n)} {(2n+1)!} x^{2n+1} \biggl)$$

So now I am wondering is this a sheer coincidence or is Frobenius method can be used to solve ordinary 2nd order differential equation too? Feel free to correct me if any of my said concept or used term is wrong because I do admit that my understanding about this whole power series solution is not very concrete. Thanks!

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    $\begingroup$ The Frobenius method contains, as you saw, power series as one of its possible results. There is no mutual exclusion. $\endgroup$ – Lutz Lehmann Dec 26 '19 at 21:24
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Yes, there's nothing to prevent you from using the Frobenius method when the equation has no singularity at $x=0$. The indicial roots will be $0$ and $1$, and there won't be any logarithmic term.

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  • $\begingroup$ which mean "(2) can be solved by normal power series solution but not Frobenius method" (which I written above) is wrong and actually should be "can be solved by both normal power series method and Frobenius method" right? $\endgroup$ – Tan xz Dec 26 '19 at 15:57

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