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I have one solution to $(1-x^2)y''-2xy'+\lambda^2 y=0$ and that is

$$

\lambda^2a_0 - (2 - \lambda^2)xa_1 + 2a_2 + 6xa_3 + \sum_{k = 2}^{\infty}\left[a_k\left(\lambda^2 - k - k^2\right) + a_{k + 2}\left(k^2 + 3k + 2\right)\right]x^k = 0.

$$

From this equation, we can conclude $\lambda^2a_0 + 2a_2 = 0$ and $-(2 - \lambda^2)xa_1 + 6xa_3 = 0$.

Therefore,

$$

a_2 = -\frac{\lambda^2}{2}a_0\quad\text{and}\quad a_3 = \frac{2 - \lambda^2}{6}a_1.

$$

Finally, we have $a_k\left(\lambda^2 - k - k^2\right) + a_{k + 2}\left(k^2 + 3k + 2\right) = 0$, i.e.

$$

a_{k + 2} = \frac{k(k + 1) - \lambda^2}{(k + 2)(k + 1)}a_k.

$$

Use the method of reduction of order to derive the following formula for the Legendre functions of the 2nd kind:

$$

Q_n(x) = P_n(x)\int\frac{1}{[P_n(x)]^2(1 - x^2)}dx, \quad n = 0,1,2,\ldots

$$

I basically need to solve

$$

y_2 = y_1\int\frac{\exp\left(\int\frac{2x}{1-x^2}dx\right)}{y_1^2}

$$

where $y_1$ is my solution above, correct?