Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We're having trouble with this differential equation:

$xy'' + x^2y + y = 0$

We figured it is regular singular because there are no singular points. We assumed a frobenius solution:

$y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}$

and got:

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

from here we subbed k values to match the indices and got:

$\displaystyle \sum_{k=-1}^\infty c_{k+1}(k+r)(k+r+1)x^{k+r} + \sum_{k+1}^\infty c_{k-1}(k+r-1)x^{k+r} + \sum_{k=0}^\infty c_k x^{k+r} = 0$

From here we're a little fuzzy on how to find the r values.

share|cite|improve this question
up vote 1 down vote accepted

You have it right. Now just equate powers of $x$.

$x^{r-1}$: $r(r-1) c_0 = 0 \implies r=1$ or $r=0$. Assume for this solution that $r=1$.

$x^1$: $2 c_1+c_0 = 0$. This depends on $y(0)$.


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

$c_0$ and $c_1$ from above.

share|cite|improve this answer
Alright that makes sense. so once we have r = 1 or r = 0 we have two possibilities. Our book though says that if $r_1 - r_2 =$ a positive integer there exists two linearly independent solutions of the form $y_1(x) = \displaystyle \sum_{n=0}^\infty c_n x^{n+r_1}$ $y_2(x) = Cy_1(x)lnx + \sum_{n=0}^\infty b_n x^{n+r^2}, \quad b_0 \neq 0$ how would I go about finding these series? Would I just use the recursion formula for the two different r values? – Tyler McAtee Mar 20 '13 at 19:34
That is correct - use the recurrence. Yes, I do recall that $\log{x}$ 2nd solution, you get it deriving Frobenius series for orthogonal functions. – Ron Gordon Mar 20 '13 at 19:41
So I solved the recursive equation with the r and got: $c_{k+1} = -\frac{c_{k-1}(k+r-1) + c_k}{(k+r)(k+r+1)}$ Then I found the first three terms of the series (all our professor is asking of us) for r = 0 and r = 1: r = 0: $ c_{k+1} = - \frac{c_{k-1}(k+1) + c_k}{k(k+1)}$ $ c_0 = 1$ $ c_1 = 0$ $ c_2 = -1$ $ c_3 = -\frac{1}{6}$ and likewise for r = 1: $ c_{k+1} = -\frac{c_{k-1}k + c_k}{(k+1)(k+2)}$ $c_0 = 1$ $c_1 = 0$ $c_2 = -\frac{1}{6}$ $c_3 = \frac{1}{72}$ Now am I allowed to just assume $c_0 = 1, \quad c_1 = 0$ like that? Or is there another protocol for it? – Tyler McAtee Mar 20 '13 at 20:08
$c_0$ comes from an initial condition, or a normalization. $c_0=1$ is typical for standard special functions satisfying nonsingular differential equations. – Ron Gordon Mar 20 '13 at 20:11
yeah I believe we're assuming $c_0 = 1$ but where does $c_1$ come from? Apologies for asking so many questions! – Tyler McAtee Mar 20 '13 at 20:15

I do not think there is a closed form formula for the recurrence relation. However, one of solution can be expressed in terms of the HeunB function

$$x\,{{\rm e}^{-1/2\,{x}^{2}}}{\it HeunB} \left( 1,0,-1,2\,\sqrt {2},-\frac{\sqrt {2}x}{2} \right).$$

Note that, your ode is a special case of the Heun Biconfluent differential equation.

share|cite|improve this answer
HeunB functions is something I need to sit down tonight and go over, thank you! – Tyler McAtee Mar 20 '13 at 20:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.