Given the ODE:

$2(x+1)y' = y$

How can I solve that using Power Series? I started to think about it:

$ \\2(x+1)\sum_{n=1}^{\infty}{nc_nx^{n-1}}-\sum_{n=0}^{\infty}{c_nx^n}=0 \\2\sum_{n=1}^{\infty}{nc_nx^{n}}+2\sum_{n=1}^{\infty}{nc_nx^{n-1}}-\sum_{n=0}^{\infty}{c_nx^n}=0 \\\sum_{n=0}^{\infty}{2nc_nx^{n}}+\sum_{n=0}^{\infty}{2(n+1)c_{n+1}x^{n}}-\sum_{n=0}^{\infty}{c_nx^n} = 0 \\\sum_{n=0}^{\infty}{[2nc_n + 2(n+1)c_{n+1} - c_n]x^n} = 0 $


$ \\2nc_{n}+2(n+1)c_{n+1}-c_n=0 \\c_{n+1}=\frac{c_n(1-2n)}{2(n+1)} $

Now, I should know what is the generic formula of $c_n$, but I can not see the pattern by assigning values to $n$. How can I proceed?

  • $\begingroup$ You should have $c_n$, not $c_{n+2}$ in the next to last equation, but it is fixed in the last. Also the last term there should be $c_n$, not $cn$. Your approach is good. $\endgroup$ – Ross Millikan Oct 1 '13 at 17:50
  • $\begingroup$ There is a pattern. I prefer $-\frac{1}{2}\frac{2n-1}{n+1}$. Let $c_1=a$. Then $c_2=a(-1/2)\frac{1}{2}$, So $c_3=a(-1/2)^2 \frac{1\cdot 3}{(2)(3)}$. Do not simplify! So $c_4=a(-1/2)^3 \frac{1\cdot 3\cdot 5}{(2)(3)(4)}$. Do not simplify. And so on. The numerator $1\cdot 3\cdot 5$ can be prettified by writing it as $\frac{6!}{2^3\cdot 3!}$. $\endgroup$ – André Nicolas Oct 1 '13 at 18:11
  • $\begingroup$ Could you explain why did you say: The numerator $1 \times 3 \times 5$ can be prettified by writing it as $ \frac{6!}{2^3\times 3!}$ ? $\endgroup$ – Richard Oct 1 '13 at 18:43

Let's say that you are given an initial value $f(0)=c_0$ (notice that this value corresponds exactly with the value of $c_0$ in the series). Then, as you have already worked out: $$c_n = c_{n-1}(-1)\frac{1}{2}\frac{2n-3}{n}$$ and so on going down to $n=0$, thus: $$c_n = \frac{(-1)^n}{2^n}\frac{\prod_{k=0}^n(2n-3)}{n!}c_0$$

  • $\begingroup$ Is there some way to avoid using that productory? $\endgroup$ – Richard Oct 1 '13 at 18:45
  • $\begingroup$ @Richard I don't think so... I also tried to find a closed form for the final function, but didn't get anywhere. Notice that having the first few terms is enough to get a very good approximation of the function around $0$, though (wichh is what you need in some contexts). $\endgroup$ – Daniel Robert-Nicoud Oct 1 '13 at 18:49
  • $\begingroup$ If $y=\sum_{n=0}^{\infty}{x^nc_n}$, could it be a binomial serie? Something like: $c_0(1+x)^{1/2}$. Could it be related? $\endgroup$ – Richard Oct 1 '13 at 18:55
  • $\begingroup$ @Richard Well, it should be, since the solution to the ODE is $y(x) = c_0(1+x)^{-\frac{1}{2}}$... You can try to show that the expansion of this function around $0$ gives you the series of coefficients of above. $\endgroup$ – Daniel Robert-Nicoud Oct 1 '13 at 19:41
  • $\begingroup$ Could you try to show it? $\endgroup$ – Richard Oct 2 '13 at 18:03

We can cheat and solve the ODE using separation of variables. The general solution for $x>-1$ is $$y(x)=c_0\sqrt{\mathstrut1+x}\ .$$ Using the formula for the binomial series therefore gives $$y(x)=c_0\sum_{k=0}^\infty{1/2\choose k}x^k\ .$$ Of course the numbers $a_k:={1/2\choose k}$ can be written in terms of factorials, if desired.


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