Recognizing recursion relation of series that is solutions of $y'' + y' + x^2 y = 0$ around $x_0 = 0$.

Question

I have been tried to find the general solution for the ODE above, but its recursion relation became so complex that I doesn't can construct its power series solution. Below is my attempt, assuming $y = \sum_{n = 0}^{\infty} a_n x^n$ is the solution.

\begin{equation*} y = \sum_{n = 0}^{\infty} a_n x^n; \quad y' = \sum_{n = 1}^{\infty} n a_n x^{n-1}; \quad y'' = \sum_{n = 2}^{\infty} n (n - 1) a_n x^{n - 2}. \end{equation*} \begin{equation*} \Rightarrow \sum_{n = 2}^{\infty} n (n - 1) a_n x^{n - 2} + \sum_{n = 1}^{\infty} n a_n x^{n-1} + x^2 \sum_{n = 0}^{\infty} a_n x^n = 0 \end{equation*} \begin{equation*} \sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n} + \sum_{n = 0}^{\infty} (n + 1) a_{n + 1} x^{n} + \sum_{n = 2}^{\infty} a_{n - 2} x^{n} = 0 \end{equation*} \begin{equation*} 2 \cdot 1 \cdot a_2 + 3 \cdot 2 a_3 \cdot x + \sum_{n = 2}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n} + 1 \cdot a_1 + 2 a_2 x + \sum_{n = 2}^{\infty} (n + 1) a_{n + 1} x^{n} + \sum_{n = 2}^{\infty} a_{n - 2} x^{n} = 0 \end{equation*} \begin{equation*} a_1 + 2 a_2 + x(2 a_2 + 3 \cdot 2 a_3) + \sum_{n = 2}^{\infty} [(n + 2) (n + 1) a_{n + 2} + (n + 1) a_{n + 1} + a_{n - 2}] x^n = 0 \end{equation*} And the recursion relation: \begin{equation*} \begin{split} & a_1 + 2 a_2 = 0 \Rightarrow a_1 = -2 a_2 \\ & 2 a_2 + 3 \cdot 2 a_3 = 0 \Rightarrow a_3 = - \frac{a_2}{3}\\ & a_{n + 2} = \frac{- (n + 1) a_{n + 1} - a_{n - 2}}{(n + 2) (n + 1)} = - \frac{(n + 1) a_{n + 1} + a_{n - 2}}{(n + 2) (n + 1)} \end{split} \end{equation*} Starting from here, I calculate the coefficients until $n = 7$( $a_9$ ) but its recognition is being difficult for me. The following coefficients are

$n = 2$: \begin{equation*} a_4 = \frac{- 3 a_3 - a_0}{4 \cdot 3} = \frac{a_2 - a_0}{4 \cdot 3} \end{equation*} $n = 3$: \begin{equation*} \begin{split} a_5 & = - \frac{4 a_4 + a_1}{5 \cdot 4} = - \frac{1}{5 \cdot 4} \bigg[4 \bigg( \frac{a_2 - a_0}{4 \cdot 3}\bigg) - 2 a_2\bigg] = - \frac{1}{5 \cdot 4 \cdot 3} \bigg[ a_2 - a_0 - 3 \cdot 2 a_2\bigg] \\ & = \frac{a_2 (3 \cdot 2 - 1) + a_0}{5 \cdot 4 \cdot 3} \end{split} \end{equation*} $n = 4$: \begin{equation*} \begin{split} a_6 & = - \frac{5 a_5 + a_2}{6 \cdot 5} = - \frac{1}{6 \cdot 5} \bigg[5 \bigg( \frac{a_2 (3 \cdot 2 - 1) + a_0}{5 \cdot 4 \cdot 3} \bigg) + a_2\bigg] = - \frac{a_2 (4 \cdot 3 + 3 \cdot 2 - 1) + a_0 }{6 \cdot 5 \cdot 4 \cdot 3} \end{split} \end{equation*} \begin{equation*} \vdots \end{equation*} $n = 7$: \begin{equation*} \begin{split} a_9 & = - \frac{8 a_8 + a_5}{9 \cdot 8} \\ & = - \frac{a_2 (7 \cdot 6 \cdot 3 \cdot 2 - 7 \cdot 6 + 6 \cdot 5 + 5 \cdot 4 + 4 \cdot 3 + 3 \cdot 2 - 1) + a_0 (1 - 6 \cdot 5 + 6 \cdot 7)}{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3} \end{split} \end{equation*} Note that in $n = 7$, the coefficient is similar, but it's getting different of the $n = 2, 3, 4$ because the factor of $a_0$ is growing since $n = 6$.

Thanks for your attention.

Answer 1

Since you have two arbitrary constants, let them to be $a_0$ and $a_1$. So, doing the same as you did $$a_2=-\frac {a_1}2\qquad a_3=\frac {a_1}6$$ and just continue with the recurrence relation you built $$a_{n+2}=- \frac{(n + 1)\, a_{n + 1} + a_{n - 2}}{(n + 2) (n + 1)}$$ which could "better" write $$a_n=-\frac{a_{n-1}} n-\frac{a_{n-4}}{n(n-1)}$$ Do not try to be more explicit. You would probably be wasting your time.

Answer 2

In here the general solution to this very ODE is found using different methods rather than by using a power series expansion. Firstly we eliminate the coefficient at the second derivative by substituting $y(x)=\exp(-x/2) v(x)$. Inserting this into the ODE gives: \begin{equation} v^{''}(x) + (-\frac{1}{4} + x^2) v(x)=0 \end{equation} which is a Schroedinger type of equation and as such we know how to solve it. We know that for big values of $x$ the function $v(x)$ has to have a Gaussian type of behavior (in order to cancel the term $x^2$ at the zeroth derivative) and therefore we write $v(x)=\exp(A x^2) u(x)$. We choose the constant $A$ appropriately so that the $x^2$ term cancels and we end up with an equation that is mapped to the confluent hypergeometric equation. In summary we have: \begin{equation} y(x) = \exp(-\frac{x}{2}-\imath \frac{x^2}{2})\cdot \left( C_1 H_{-\frac{1}{2}+\imath\frac{1}{8}}((-1)^{1/4} x) + C_2 F_{1,1}(1/4-\imath/16,1/2,\imath x^2)\right) \end{equation} where $H_n()$ are Hermite polynomials.

Now we have to express the constants $C_{1,2}$ through $y(0)=a_0$ and $y^{'}(0)=a_1$. This boils down to solving a pair of linear equations and yields: \begin{eqnarray} C_1&=&-\frac{(1-i) 2^{-2-\frac{i}{8}} \Gamma \left(\frac{1}{4}-\frac{i}{16}\right) (a_0+2 a_1)}{\sqrt{\pi }}\\ C_2&=&a_0+\frac{64 (-1)^{3/4} \Gamma \left(\frac{5}{4}-\frac{i}{16}\right) (a_0+2 a_1)}{17 \Gamma \left(-\frac{1}{4}-\frac{i}{16}\right)} \end{eqnarray}

and now we have a "closed form" for our recursion relation:

\begin{equation} a_n= \left.\frac{d^n}{d x^n} y(x)\right|_{x=0} \end{equation}

as the code below demonstrates:

In[35]:= y[x_] = 
  Exp[-x/2] Exp[-I/2 x^2] ( 
    C[1] HermiteH[-(1/2) + I/8, (-1)^(1/4) x] + 
     C[2] Hypergeometric1F1[1/4 - I/16, 1/2, I x^2]);
FullSimplify[(y''[x] + y'[x] + x^2 y[x])]

a0 =.; a1 =.;
subst = FullSimplify[
  ComplexExpand[Solve[{y[0], y'[0]} == {a0, a1}, {C[1], C[2]}]]]
l1 = First@
  CoefficientList[
   Collect[Normal[Series[y[x], {x, 0, 8}] /. subst], x, FullSimplify],
    x]


a = Table[0, {9}];
a[[1]] = a0; a[[2]] = a1;
a[[3]] = -a1/2;
a[[4]] = a1/6;
Do[
  a[[1 + n]] = -a[[n]]/n - a[[n - 3]]/(n (n - 1));
  , {n, 4, 8}];
l2 = Simplify[a]

Out[36]= 0

Out[38]= {{C[
    1] -> -(((1 - I) 2^(-2 - I/8) (a0 + 2 a1) Gamma[1/4 - I/16])/
    Sqrt[\[Pi]]), 
  C[2] -> a0 + (64 (-1)^(3/4) (a0 + 2 a1) Gamma[5/4 - I/16])/(
    17 Gamma[-(1/4) - I/16])}}

Out[39]= {a0, a1, -(a1/2), a1/6, 1/24 (-2 a0 - a1), 
 1/120 (2 a0 - 5 a1), 1/720 (-2 a0 + 17 a1), (2 a0 - 37 a1)/5040, (
 58 a0 + 67 a1)/40320}

Out[45]= {a0, a1, -(a1/2), a1/6, 1/24 (-2 a0 - a1), 
 1/120 (2 a0 - 5 a1), 1/720 (-2 a0 + 17 a1), (2 a0 - 37 a1)/5040, (
 58 a0 + 67 a1)/40320}

Note: Interestingly enough if we were to replace $x^2$ in the original ODE by $x^{2 n}$ then we would have gotten $y(x) =\exp(-x/2) \exp(\imath/(n+1) x^{n+1}) \cdot u(x)$ where the function $u(x)$ satisfies: \begin{equation} u^{''}(x)+2 \imath x^n u^{'}(x)+ (-\frac{1}{4} +\imath n x^{-1+n}) u(x) \end{equation} which in the case $n=2$ matches the tri-confluent Heun equation https://dlmf.nist.gov/31.12.