# Polynomial solution of the equation $(1-x^2)y''-2xy'+6y=0$

Let $y$ be a polynomial solution of the differential equation $(1-x^2)y''-2xy'+6y=0$. If $y(1)=2$, then the value of the integral $\int_{-1}^{1}y^2dx$ is

1. $\frac{1}{5}$
2. $\frac{2}{5}$
3. $\frac{4}{5}$
4. $\frac{8}{5}$

First of all I can solve it by conventional method, which is long and provided this is a competitive exam question, the maximum time I can give is $2-3$ mins. Is there is shortcut method or trick to solve this kind of problems? I tried to integrate the equation though but ended up with a term of $y(-1)$ which has no value given. How can I do this in less time? Any help would be great. Thanks.

• Integration by parts? Wait, I'll try, but that the first I thought about when I saw the integral – Konstantin K Jan 11 '17 at 10:45
• Yes i tried that. – Kushal Bhuyan Jan 11 '17 at 10:46

Hint. Assume that a polynomial of the form $y(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$ satisfies the ODE, and pug it in the ODE. Then you shall find that $n=2$. Next, you shall find that $y=x^2-1/3$. Observe that if $y$ is a solution, then so is $cy$, for every $c\in \mathbb R$. Satisfaction of the condition $y(1)=2$, implies that the sought for solution is $$y(x)=3x^2-1.$$
A little mistake was there, as I did not see the initial condition. There is a property that Legender's Polynomial has that is it it's value is 1 at 1. Now since the initial condition is given as $y(1)=2$ implies $2P_2$ is a solution. Rest of the stuff will remain same.