# Frobenius Method general solution not providing an expected solution

I'm trying to solve the following ODE: $$3xy''+(3x+1)y'+y=0$$ So I started by assuming a solution(and its derivates) with the form:

$$\tag{1}y = \sum_{n=0}^{\infty } a_{n} x^{n + r}$$

$$y' = \sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1}$$

$$y'' = \sum_{n=0}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-2}$$ Replacing in original equation: $$3x\sum_{n=0}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-2} + 3x\sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1}+\sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1} + \sum_{n=0}^{\infty } a_{n} x^{n + r} = 0$$

$$3\sum_{n=0}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-1} + \sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1} + 3\sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r} + \sum_{n=0}^{\infty } a_{n} x^{n + r} = 0$$

To transform all $x^{n+r}$ into $x^{n+r-1}$ I've rewritten the equation as:

$$3\sum_{n=0}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-1} +\sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1} + 3\sum_{n=1}^{\infty } {(n+r-1)}a_{n-1} x^{n + r -1} + \sum_{n=1}^{\infty } a_{n-1} x^{n + r - 1} = 0$$

In order to make all sums start from the $n=1$:

$$\tag{2} 3r(r-1)a_{0}x^{r-1} + ra_{0}x^{r-1} + 3\sum_{n=1}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-1} + \sum_{n=1}^{\infty } {(n+r)}a_{n} x^{n + r-1} + 3\sum_{n=1}^{\infty } {(n+r-1)}a_{n-1} x^{n + r -1} + \sum_{n=1}^{\infty } a_{n-1} x^{n + r - 1} = 0$$

If I understood correctly, the indicial equation should then be: $$3r(r-1)a_{0}x^{r-1} + ra_{0}x^{r-1} = 0$$

and since $x^{r-1} \neq 0$ and $a_{0} \neq 0$:

$$3r(r-1) + r = 0$$ $$r_{1} = 0$$ $$r_{2} = \frac{2}{3}$$

From $(2)$:

$$[3(n+r)(n+r-1)+(n+r)]a_{n} + [3(n+r-1)+1]a_{n-1} = 0$$ which seems to give for both solutions of $r$: $$a_{n} = \frac{-a_{n-1}}{n}$$ It seems to me that this is in terms of $a_{0}$:

$$a_{n} = \frac{(-1)^{n}}{n!}a_{0}$$

So rewriting $(1)$ now would give me the solution and $y$ would be an infinite sum since all $a_{n} \neq 0$. And I know this can't be true because by deduction one can see that $y = e^{-x}$ is a solution to the ODE, which makes me think I made a mistake in some of the steps shown above, I'd be glad if anyone could show me where I made the mistake and give me some help in finding the correct general solution for this ODE.

• you seem to have calculated the expansion for $e^{-x}$, so one down and one to go. why are both solutions the same? Jun 25, 2017 at 3:32
• Exactly! I wasn't aware of that, thanks! And you are right, it seems the other solution gives $n(3n+2)a_{n}+3na_{n-1}=0$, is the general solution just the sum of these both particular solutions then? And I solved this for $a_{1}$, $a_{2}$ and $a_{3}$ and had trouble seeing a generalised form in terms of $a_{0}$. Jun 25, 2017 at 3:43
• The general solution is a sum of the two, but be sure that the two $a_0$ are not necessarily the same. As for the generalization of $y_2$ in terms of $a_0$ it is not always explicit, or one gets exhausted. When this is the case, it is customary to write a few terms, dot, dot, dot. Jun 25, 2017 at 6:23
• @reluctantmathematician thank you! Since your first comment is actually the answer to my original question, if you could post that as an answer I'd be glad to accept it. I think it's good not to leave unanswered questions in stackexchange. Jun 25, 2017 at 17:56
• confirmed ----- Jun 25, 2017 at 20:44

## 2 Answers

For solution $r_1$: $$a_n=\frac{(-1)^n}{n!}a_0$$$$y_1=a_0(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots)\ =\ a_0e^{-x}$$

For solution $r_2$: $$b_n=\frac{(-3)^n}{(5)(8)(11)\cdots(3n+2)}b_0$$ $$y_2=b_0(1-\frac{3}{5}x+\frac{9}{40}x^2-\frac{27}{440}x^3+\cdots)$$

The general solution is $\ y_1+y_2$.

Or try to restart the calculation with $y(x)=e^{-x}u(x)$, $$0=3x(u''-2u'+u)+(3x+1)(u'-u)+u=3xu''-(3x-1)u'$$ which gives $$\ln|u'|=\int \frac{3x-1}{3x}dx=x-\frac13\ln|x|+c \\~\\ u'=Cx^{-\frac13}e^x \\~\\ y=Ce^{-x}\int_0^x s^{-\frac13}e^s\,ds+De^{-x}$$ which you can then try to express in terms of the incomplete gamma function.