Solution to Legendre eq in trig form

Question

Okay I'm having a little trouble in answering this question...

so the general solution is $y(x) = AP_n(x) + BQ_n(x)$

umm then what do I do?

Answer 1

When $x = cos\theta$, then $\frac{dx}{d\theta} = - sin\theta$, so the differential equation you give translates into $$[(1-x^2)y^{'}]^{'} + 6y =0$$ So, it's a Legendre equation with $n = 2$ (as $2(2+1) = 6$). Its two linearly independent solutions are \begin{cases} y = 3x^2 - 1 &( = 3 \cdot P_2) \\y = (3x^2 - 1)ln(\frac{1+x}{1-x}) - 6x &(= 4 \cdot Q_2) \end{cases} or, in terms of $\theta$, \begin{cases} y = 3cos^2\theta - 1 \\y = (3cos^2\theta - 1)ln(\frac{1+cos\theta}{1-cos\theta}) - 6cos\theta \end{cases} so the solution is $$y = A(3cos^2\theta - 1) + B\left((3cos^2\theta - 1)ln(\frac{1+cos\theta}{1-cos\theta}) - 6cos\theta\right)$$