# Running into trouble with this differential equation

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.

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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 $y(0)$.

$x^n$:

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

$c_0$ and $c_1$ from above.

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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.

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