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The ODE in question is: $$4xy''+2y'+y=0$$

Shifting the power series of each term so that they are all raised to the power $(n+r)$ will yield this recurrence relation: $$a_{n+1}={a_n\over (n+r+1)(-2-4(n+r))}$$ with $$r=1/2, 0$$

If you plug values of $n$ into this recurrence relation it is nearly impossible to find a pattern for $a_n$, unless I'm missing something.

Is there a way to continue to solve this ODE with the Frobenius method using this difficult recurrence relation, or any tricks to use earlier in the problem to avoid difficult recurrence relations?

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For every $c$, $$n+c=\frac{\Gamma(n+c+1)}{\Gamma(n+c)},$$ hence the recursion you arrived at can be rewritten as $$a_{n+1}=\frac{-a_n}{4(n+r+1)(n+r+\frac12)}=\frac{(-1)^n4^n\Gamma(n+r+1)\Gamma(n+r+\frac12)}{(-1)^{n+1}4^{n+1}\Gamma(n+r+2)\Gamma(n+r+\frac32)}a_n,$$ which immediately leads to $$ a_n=\frac{(-1)^n}{4^n}\frac{\Gamma(r+1)\Gamma(r+\frac12)}{\Gamma(n+r+1)\Gamma(n+r+\frac12)}a_0.$$ A basis of the space of solutions is given by these series when $r=0$ and when $r=\frac12$, that is, one can choose $\{y_0,y_{1/2}\}$, with $$y_r(x)\propto x^r\sum_{n\geqslant0}\frac{(-1)^nx^n}{4^n\Gamma(n+r+1)\Gamma(n+r+\frac12)}.$$ Legendre duplication formula reads$$4^{n+r}\,\Gamma(n+r+\tfrac12)\Gamma(n+r+1)=\sqrt{\pi}\,\Gamma(2n+2r+1),$$ which leads to a more familiar formulation of the basis of the space of solutions, namely, $$y_0(x)=\sum_{n\geqslant0}\frac{(-1)^nx^n}{(2n)!},\qquad y_{1/2}(x)=\sqrt{x}\sum_{n\geqslant0}\frac{(-1)^nx^n}{(2n+1)!},$$ also known as $$y_0(x)=\cos\sqrt{x},\qquad y_{1/2}(x)=\sin\sqrt{x}.$$

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Hint: take $z(x) = y(x^\alpha)$ with $\alpha>0$. Then try to write the differential equation on $z$ and try to guess a good $\alpha$.

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