# solving the given differential equation $(1-x^2)y''+ 2xy' + ay = 0$ with $a$ a parameter

As trainingsexercise for my course I have got to solve the following problem:

the problem

Given the differential equation $$(1-x^2)y''+ 2xy' + ay = 0$$ with $$a$$ a parameter.

a) give the powerseries solution of this differential equation around $$x=0$$

b) find $$a$$ so that at least one of the linear independent solutions is a polysome

c) There are 2 values for $$a$$ so that both solutions become polysomes. find these two values and the polysomes

what I have tried so far

a) bringing the differential equation into its standard form, $$y'' + xP(x)y' + x^2Q(x)y =0$$ and with $$xP(x)$$ and $$x^2Q(x)$$ differentiable in $$x=0$$, means that I can find two linear independent solutions without using Frobenius-powerseries thus: $$y_1(x)= \sum_{n=0}^{\infty}{b_n x^n}$$ and $$y_2(x)= \sum_{n=0}^{\infty}{c_n x^n}$$

I used then $$y(x)= \sum_{n=0}^{\infty}{b_n x^n}$$ and filled it in the differential equation. Then I changed the indices so that we get all $$x$$ to the same powers and came thus up with this: $$\sum_{n=0}^{\infty}{(n+2)(n+1)c_{n+2} x^n}-\sum_{n=0}^{\infty}{n(n+1)c_n x^n}+2\sum_{n=0}^{\infty}{nc_n x^n}+a\sum_{n=0}^{\infty}{c_n x^n}=0$$

then I made a case distinction:

• $$n=0$$ : $$2c_2+ac_0=0 \implies c_2 = \frac{-a}{2}c_0$$
• $$n\geq1$$ : $$(n+2)(n+1)c_{n+2} - n(n-1)c_n+2nc_n +ac_n=0 \iff c_{n+2} = \frac{n(n-3)-a}{(n+2)(n+1)}c_n$$

Now we have always steps by two, and we can creat two linear independent solution by having once $$c_0=1,c_1=0$$ and once $$c_0=0,c_1=1$$ Doing this brings me to the formulas:

• $$n$$ is even: $$c_{2k} = \frac{2 \cdot \prod_{j=0}^{k-1}(2k-2j)(2k-3-2j)-a}{(2k+2)!}$$
• $$n$$ is odd: $$c_{2k+1} = \frac{6 \cdot \prod_{j=0}^{k-1}(2k+1-2j)(2k-2-2j)-a}{(2k+3)!}$$

and then I get: $$y_1(x)= 1+\sum_{k=0}^{\infty}{\frac{2 \cdot \prod_{j=0}^{k-1}(2k-2j)(2k-3-2j)-a}{(2k+2)!}\cdot x^{2k}}$$ and $$y_2(x)= x + \sum_{k=1}^{\infty}{\frac{6 \cdot \prod_{j=0}^{k-1}(2k+1-2j)(2k-2-2j)-a}{(2k+3)!}\cdot x^{2k+1}}$$

can anybody find my mistake?

for b) and c) I honestly don't know how I could solve this, any help would be very much appreciated. thank you in advance.

• For comparison, WolframAlpha indicates that the solutions can be expressed in terms of associated Legendre functions of order 2 (link. In particular, if one writes $a:=(\nu+2)(\nu-1)$ then the solutions are $(x^2-1)P_{\nu}^2(x),(x^2-1)Q_{\nu}^2(x)$. Commented Jan 16 at 15:39
• Commented Jan 16 at 18:44
– T_B
Commented Jan 16 at 19:46
• Also, one way to make the recurrence look slightly prettier is to take the power series as $\sum_n C_n x^n/n!$. This gives $C_n=c_n\cdot n!$ and so obtains $C_{n+2} = (n(n-3)-a)C_n$. In other words, it entirely eliminates the denominator. Commented Jan 16 at 20:39

There is no need to distinguish between $$n=0$$ and $$n \ge 1$$: the recurrence $$c_{n+2} = -\frac{a+3n-n^2}{(n+1)(n+2)} c_n$$ is true in both cases.
Hints for (b) and (c): Note that if $$a + 3 n - n^2 = 0$$, you'll get $$c_{n+2k} = 0$$ for all $$k \ge 1$$, and thus $$y_1$$ or $$y_2$$ (depending on whether $$n$$ is even or odd) is a polynomial. For both $$y_1$$ and $$y_2$$ to be a polynomial, you'll want $$a+3n-n^2$$ to be $$0$$ for both an even $$n$$ and an odd $$n$$.
• Thank you for your reply which helped me much further. So are my solutions for $y_1(x),y_2(x)$ not all that wrong?
• for (b) and (c) wouldn't I then have to set myself on a specific $n$, thus the order of the polynomial?
• If you look at the sequences of values of $3n - n^2$ for even $n$ and for odd $n$, you should find two that are in both sequences. Commented Jan 17 at 5:58