How to solve linear, second order ODE with Frobenius method with a difficult recurrence relation?

Question

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?

Answer 1

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 $\alpha$.

Answer 2

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 $$y_0(x)=\cos\sqrt{x},\qquad y_{1/2}(x)=\sin\sqrt{x}.$$