Solve ODE by series given a strange solution

Question

While studying for a test, I had this following question:

Solve with series substituion:

$$y''-x^2y'-3xy=0$$ So, doing $y=\sum_{n=0}^{\infty}a_nx^n$, I got: $2a_2+\sum_{n=0}^{\infty}[(n+3)(n+2)a_{n+3}-(n+3)a_n]x^{n+1}=0$ So, $a_2=0$ and $a_{n+3} = \frac{a_n}{n+2}$ Is this correct? I can't see any pattern in this terms. Thanks for attention.

Answer 1

Your solution is correct. The sequence starting with $a_1$ has a recognizable pattern: \begin{align} a_4&=\frac{a_1}{3}, \\ a_7&=\frac{a_4}{6}=\frac{a_1}{2\times 3^2}, \\ a_{10}&=\frac{a_7}{9}=\frac{a_1}{2\times 3\times 3^3}, \\ a_{13}&=\frac{a_{10}}{12}=\frac{a_1}{2\times 3\times 4\times 3^4}, \\ &\,\,\,\vdots \\ a_{3k+1}&=\frac{a_1}{k!\,3^k}. \tag{1} \end{align} The corresponding series is, therefore, $$ S(x)=a_1\sum_{k=0}^{\infty}\frac{x^{3k+1}}{k!\,3^k}=a_1x\exp\left(\frac{x^3}{3}\right). \tag{2} $$ You can easily check that $(2)$ indeed is a solution to the ODE $y''-x^2y'-3xy=0$.

Answer 2

Let's split up the indices into cases of $n\in\{3k,3k+1,3k+2\}$ for $k=0,1,2,\ldots$ :

$$\begin{align*} a_{3k} &= \frac{a_{3(k-1)}}{3k-1} \\ &= \frac{a_{3(k-2)}}{(3k-1)(3k-4)} \\ &= \frac{a_{3(k-3)}}{(3k-1)(3k-4)(3k-7)} \\ &~~\!\vdots \\ &= \frac{a_{3(k-\ell)}}{(3k-1)(3k-4)\cdots(3(k-\ell)-1)} \\[1ex] \implies a_{3k} &= \frac{a_0}{(3k-1)(3k-4)\cdots2} = a_0 \prod_{i=1}^{k}\frac1{3i-1} = \frac{a_0}{3^k} \cdot \frac{\Gamma\left(\frac23\right)}{\Gamma\left(k+\frac23\right)} \\[2ex] \hline a_{3k+1} &= \frac{a_{3k-2}}{3k} = \frac{a_{3(k-1)+1}}{3(k-0)}\\ &= \frac{a_{3k-5}}{3k(3k-3)} = \frac{a_{3(k-2)+1}}{3^2 (k-0)(k-1)} \\ &= \frac{a_{3k-8}}{3k(3k-3)(3k-6)} = \frac{a_{3(k-3)+1}}{3^3 (k-0)(k-1)(k-2)} \\ &~~\!\vdots \\ &= \frac{a_{3(k-\ell)+1}}{3^\ell k(k-1)\cdots(k-\ell+1)} \\[1ex] \implies a_{3k+1} &= \frac{a_1}{3^k k(k-1)\cdots1} = \frac{a_1}{3^k k!} \\[2ex] \hline a_{3k+2} &= \frac{a_{3k-1}}{3k+1} = \frac{a_{3(k-1)+2}}{3(k-0)+1} \\ &= \frac{a_{3k-4}}{(3k+1)(3k-2)} = \frac{a_{3(k-2)+2}}{(3(k-0)+1)(3(k-1)+1)} \\ &= \frac{a_{3k-7}}{(3k+1)(3k-2)(3k-5)} = \frac{a_{3(k-3)+2}}{(3(k-0)+1)(3(k-1)+1)(3(k-2)+1)} \\ &~~\!\vdots \\ &= \frac{a_{3(k-\ell)+2}}{(3k+1)(3k-2)\cdots(3(k-\ell+1)+1)} \\[1ex] \implies a_{3k+2} &= \frac{a_2}{(3k+1)(3k-2)\cdots4} = a_2 \prod_{i=1}^k \frac1{3i+1} = \frac{a_2}{3^k} \cdot \frac{\Gamma\left(\frac43\right)}{\Gamma\left(k+\frac43\right)} \end{align*}$$

Now, $a_2=0 \implies a_{3k+2}=0$ for all $k$, and the solution we're left with is

$$\begin{align*} y(x) &= \sum_{n=0}^\infty a_n x^n \\ &= \sum_{k=0}^\infty \left(a_{3k} x^{3k} + a_{3k+1} x^{3k+1} + a_{3k+2} x^{3k+2}\right) \\ &= a_0 \Gamma\left(\frac23\right) \sum_{k=0}^\infty \frac{1}{\Gamma\left(k+\frac23\right)} \left(\frac x{\sqrt[3]{3}}\right)^{3k} + a_1 \sum_{k=0}^\infty \frac{x^{3k+1}}{3^k k!} \end{align*}$$

The second sum is easily recognizable,

$$\sum_{k=0}^\infty \frac{x^{3k+1}}{3^k k!} = x \sum_{k=0}^\infty \frac1{k!} \left(\frac{x^3}3\right)^k = x e^{x^3/3}$$

while the first sum can be expressed in terms of the incomplete gamma function.