Suppose $v \in \mathbb{R}$ and $$y\left(x\right)=x^v\left(1+\sum^{\infty}_{n=1}a_nx^n\right)$$ where the series converges for any $x \in \left(-r,r\right), r>0$

If $y\left(x\right)$ is a solution of the equation $$y''\left(x\right)+\left(2-\dfrac{1}{x}\right)y'\left(x\right)-\dfrac{5}{x^2}y\left(x\right)=0$$ what are the possible values for $v$? Using substitution of the solution into the equation, obtain a relationship between the coefficients $a_n$ and $a_{n-1}$. Hence for each value of $v$ found determine the corresponding coefficient $a_1$.

Struggling with this question as usually the solution is of the form $y\left(x\right)=\sum^{\infty}_{n=0}a_nx^{n+v}$. In this case I would use the method of Frobenius, i.e. differentiate $y\left(x\right)$ to find $y'\left(x\right)$ and $y''\left(x\right)$ and then substitute these into the equation, find the indicial equation to calculate the values of $v$ and then find recurrence relations.

Can someone please explain the method with this different form of solution.

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    $\begingroup$ If $a_0 =1$, then the given summation is the same as the usual one using the method of Frobenius $\endgroup$ – Alex Pavellas May 25 '16 at 14:12
  • $\begingroup$ So how do I find the values of v? Using usual method, I find the indicial equation to be $v^2-2v-5=0$ however the roots of this are not integer $\endgroup$ – Sophie Filer May 25 '16 at 14:34
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    $\begingroup$ That's fine, the roots don't have to be integers, and in fact rarely are $\endgroup$ – Alex Pavellas May 25 '16 at 14:36

Hint: your solution is of the form $y(x) = \sum_{n=0}^\infty a_n x^{n+v}$, with $a_0 = 1$: \begin{equation} x^v\left(1+\sum_{n=1}^\infty a_n x^n\right) = 1\cdot x^{0+v} + \sum_{n=1}^\infty a_n x^{n+v}. \end{equation}


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