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.

  • $\begingroup$ Integration by parts? Wait, I'll try, but that the first I thought about when I saw the integral $\endgroup$ – Konstantin K Jan 11 '17 at 10:45
  • $\begingroup$ Yes i tried that. $\endgroup$ – 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 $$ y(x)=3x^2-1. $$

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See this is Legender's Equation, and the Legender's polynomial has some special properties. See Legendre's Differential Equation.

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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.

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  • $\begingroup$ your conclusion is in contradiction with the other answer. Can you please take a look $\endgroup$ – Kushal Bhuyan Jan 11 '17 at 11:12
  • $\begingroup$ Yes it is. There is a small mistake. I uploaded the answer again. $\endgroup$ – Sachchidanand Prasad Jan 11 '17 at 11:15

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