I was trying to solve following question :
$$ (1-x^2)y''-2xy'+n(n+1)y=0 $$ has the $n^{th}$ degree polynomial solution $y_n(x)$ such that $y_n(1) =3$. We are given
$\int_{-1}^{1} [y_n^2(x) + y_{n-1}^2(x)] dx = 144/15$. Find the value of $n$.
My attemnpt:
We know that $\int_{-1}^{1} P_n^2(x) = \frac{2}{2n+1}$ where $P_n(x)$ is the $nth$ degree polynomial and solution of legenedre eq. Now using this fact in the given integral we get
$$ \frac{2}{2n+1} + \frac{2}{2(n-1)+1} =\frac{144}{15}$$
and this gives $$24(24n^2 -5n -6)= 0$$
From here I am unable to find the value of $n$.
My biggest doubt:
We know that $P_n(1) =1$ then how here in the questioned it is mentioned that $y_n(1) =3$.
Kindly help me. Am I going wrong somewhere?
The general solution of the differential equation $$(1-x^2)y''-2xy'+n(n+1)y=0$$ is given by $$y=c_1 P_n(x)+c_2 Q_n(x)$$ where appear Legendre polynomials of first and second kind (see here) but the second ones are not polynomials (so, I suppose that $c_2=0$ is a requirement).
On the other side, as you wrote, for any $n$, $P_n(1)=1$ which would make $c_1=3$.
So, we are left with $$\int_{-1}^{1} [y_n^2(x) + y_{n-1}^2(x)] dx = \frac{144} {15}$$ that is to say (almost as you wrote but you forgot the $3^2$ in front) $$9\left(\frac{2}{2n+1} + \frac{2}{2(n-1)+1}\right) =\frac{144}{15}$$
I am sure that you can easily take it from here.