legendre's polynomial problem 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 $\displaystyle \int_{-1}^{1}e^{y^2}dx$ is?
how to calculate this?

This is a Legendre differential equation. Its polynomial solution is $y=\alpha P_2(x) = \alpha \cdot \frac{1}{2}(3x^2-1)$.
If $y(1)=2$, then it follows that $y=3x^2-1$. And thus
$$\int_{-1}^1 e^{y^2}dx=\int_{-1}^1 e^{(3x^2-1)^2}dx$$
I leave it to others to evaluate this integral.
